Sketch the titration curve for the titration of 0.15 m formic acid with 0.25 m naoh. you can start with any initial volume of the acid, as the volumes of base added will be proportional. the shape of the titration curve will not change significantly. all acid-base titration calculations start as limiting reactant problems, followed by an equilibrium or buffer calculation. you must calculate the ph at four regions of the titration curve to label your sketch: 1. the initial ph before any naoh has been added 2. the ph at some fraction of the equivalence point 3. the ph at the equivalence point 4. the ph at some volume past the equivalence point this will be covered in lab lecture and you will also find the examples in your textbook very helpful.
Answers
1. The initial pH of the solution is 1.89.
2. At half the equivalence point (20 mL of NaOH added), the pH of the solution is 3.26.
3. At the equivalence point (30 mL of NaOH added), the pH of the solution is 10.72.
4. At some volume past the equivalence point the pH of the solution is 12.30
In the sketch (in figure), the x-axis represents the volume of NaOH added and the y-axis represents the pH of the solution. The initial pH before any NaOH has been added is 1.89, which is the pH of the 0.15 M formic acid solution.
As NaOH is added, the pH increases slowly at first, but then increases more rapidly as the solution enters the buffer region. At the half-equivalence point (20 mL of NaOH added), the pH is 3.26. At the equivalence point (30 mL of NaOH added), the pH jumps up to 10.72 due to the complete reaction of the formic acid with NaOH.
After the equivalence point, the pH continues to increase as more NaOH is added. At 50 mL past the equivalence point, the pH is 12.30, which is close to the pH of a strong base.
The titration of 0.15 M formic acid (HCOOH) with 0.25 M NaOH can be represented by the following equation:
[tex]HCOOH + NaOH[/tex] → [tex]NaCOOH + H_2O[/tex]
Before any NaOH is added, the solution consists of 0.15 M formic acid, which is a weak acid. The initial pH of the solution can be calculated using the dissociation constant (Ka) of formic acid:
[tex]HCOOH + H_2O < = > H_3O^+ + HCOO^-[/tex]
[tex]Ka = [H_3O^{+}][HCOO^{-}]/[HCOOH][/tex]
Since formic acid is a weak acid, we can assume that [tex][H_3O^+][/tex] is equal to [tex][HCOO^-][/tex]. Let x be the concentration of [tex][H_3O^+][/tex] and [[tex][HCOO^-][/tex]] at equilibrium, then:
[tex]Ka = x^2 / (0.15 - x)[/tex]
At equilibrium, the concentration of HCOOH will be (0.15 - x) M.
Let's solve for x:
[tex]Ka = x^2 / (0.15 - x)[/tex]
[tex]1.77 * 10^{-4} = x^2 / (0.15 - x)[/tex]
x = 0.0129 M
1. Therefore, the initial pH of the solution is:
[tex]pH = -log[H_3O^+][/tex]
pH = -log(0.0129)
pH = 1.89
Now let's consider the pH at different points during the titration:
Before any NaOH has been added:
The initial pH of the solution is 1.89.
2. At some fraction of the equivalence point:
At the equivalence point, all of the formic acid will have reacted with an equal amount of NaOH. Since NaOH is a strong base, the solution will be basic after the equivalence point.
At some fraction of the equivalence point, we can assume that the solution is a buffer consisting of formic acid and its conjugate base, sodium formate (NaCOOH). We can use the Henderson-Hasselbalch equation to calculate the pH:
pH = pKa + log([NaCOOH] / [HCOOH])
At the fraction of the equivalence point, we can assume that the concentration of HCOOH and NaCOOH are equal, and the concentration of NaOH is equal to the fraction of the equivalence point times the initial concentration of formic acid. Thus:
[HCOOH] = 0.15 M - (fraction of equivalence point) × (volume of NaOH added)
[NaCOOH] = (fraction of equivalence point) × (volume of NaOH added)
[tex][H_3O^+] = Ka * [HCOOH] / [NaCOOH][/tex]
Let's assume that the fraction of the equivalence point is 0.5, which means that half of the initial concentration of formic acid has reacted with NaOH. Let's also assume that we have added 20 mL of NaOH so far:
[tex][HCOOH] = 0.15 M - 0.5 * 0.02 L * 0.25 M\\[HCOOH] = 0.14 M\\[NaCOOH] = 0.5 * 0.02 L * 0.25 M\\[NaCOOH] = 0.0025 M[/tex]
[tex][H_3O^+] = 1.77 * 10^{-4} * (0.14 / 0.0025)[/tex]
[tex][H_3O^+] = 9.88 * 10^{-3} M[/tex]
[tex]pH = pKa + log([NaCOOH] / [HCOOH])\\\\pH = 3.75 + log([0.0025 / 0.14])\\\\pH = 3.26[/tex]
Therefore, at half the equivalence point (20 mL of NaOH added), the pH of the solution is 3.26.
3. At the equivalence point:
The pH can be calculated using the hydrolysis constant (Kb) of sodium formate:
[tex]NaCOOH +[/tex] [tex]H_2O[/tex] ⇌ [tex]NaOH + HCOOH[/tex]
[tex]Kb = [NaOH][HCOOH]/[NaCOOH][/tex]
Let's assume that we have added 30 mL of NaOH, which is the equivalent amount to the initial concentration of formic acid:
[tex][NaOH] = [HCOOH] = 0.15 M\\[NaCOOH] = 0.5 * 0.03 L * 0.25 M\\[NaCOOH] = 0.00375 M\\\\Kb = [NaOH]^2 / [NaCOOH]\\\\Kb = (0.15)^2 / 0.00375\\\\Kb = 6\\\\pOH = -log[OH-]\\pOH = -log\sqrt{(Kb * [NaCOOH])} \\pOH = -log\sqrt{6 * 0.00375} \\pOH = 3.28\\pH = 14 - pOH\\pH = 10.72[/tex]
Therefore, at the equivalence point (30 mL of NaOH added), the pH of the solution is 10.72.
4. At some volume past the equivalence point:
After the equivalence point, the solution will be basic due to the excess of NaOH. The pH can be calculated using the concentration of NaOH and the volume of NaOH added:
pOH = -log[OH-]
pOH = -log(0.25 × (volume of NaOH added - volume of NaOH at equivalence point))
pH = 14 - pOH
Let's assume that we have added 50 mL of NaOH past the equivalence point:
pOH = -log(0.25 × (0.05 L - 0.03 L))
pOH = 1.70
pH = 14 - pOH
pH = 12.30
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Related Questions
what is the expected major organic product from the treatment of 4-methyl-2-pentyne with excess hydrogen in the presence of a platinum catalyst? 4-methylpentane
Answers
4-methylpentane
The response of 4-methyl-2-pentyne with extra hydrogen in the presence of a platinum catalyst is a hydrogenation reaction, which entails the addition of hydrogen atoms across the triple bond of the alkyne. The expected foremost organic product is 4-methylpentane, which is formed through the complete discount of the triple bond to a single bond.
The hydrogenation of 4-methyl-2-pentyne proceeds through a stepwise addition of hydrogen atoms to the triple bond, forming an intermediate alkene and then a saturated alkane. However, the presence of extra hydrogen ensures that the alkene intermediate is quickly decreased to the alkane product, which is the extra thermodynamically secure form.
Therefore, the anticipated main organic product of the hydrogenation reaction of 4-methyl-2-pentyne with extra hydrogen in the presence of a platinum catalyst is 4-methylpentane
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things that happened to the organisms I tracked: (example: was eaten by other organisms)
Answers
Reproduced, transferred their genes to progeny, died from natural causes, fell ill or contracted a disease, became prey for a predator or were devoured by a scavenger, or moved to a new area.
What kind of non-living entity might be present in an ecosystem?Non-living things include items like rocks, water, the atmosphere, the climate, and natural occurrences like earthquakes and rockfalls. One of the qualities that characterises living beings is their capacity for reproduction.
What are five non-living examples?Its definition includes glass, the sun, water, sand, and rock as non-living objects. They show absolutely no signs of life. Some people define a non-living object as anything that once belonged to a live entity.
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a. what is the concentration of the unknown acid? b. what is the ka of the unknown acid? c. which of the indicators given below would be a good choice to use in the titration of this acid with the naoh?
Answers
a. To determine the concentration of the unknown acid, you would need to perform a titration with a known concentration of a strong base like NaOH. Follow these steps:
1. Record the volume of the unknown acid solution used.
2. Record the initial and final burette readings of NaOH.
3. Calculate the volume of NaOH used in the titration.
4. Write a balanced equation for the reaction between the unknown acid and NaOH.
5. Use stoichiometry to find the moles of acid reacted.
6. Divide the moles of acid by the volume of the unknown acid solution used (in liters) to find the concentration.
b. To determine the Ka of the unknown acid, you will need additional information, such as the pH at the half-equivalence point (where half of the acid has been neutralized) or the pH and concentration of the acid after a certain amount of NaOH has been added. Then, you can use the following steps:
1. Write the balanced equation for the dissociation of the acid in water.
2. Use the given pH information to find the concentration of H+ ions.
3. Use the stoichiometry of the reaction to find the concentration of the other species in the equation.
4. Write the expression for Ka, which is [products]/[reactants].
5. Plug the concentrations into the Ka expression and solve for Ka.
c. When choosing an indicator for a titration, you should select one whose pH range (the range over which the indicator changes color) coincides with the pH of the equivalence point (where the acid is completely neutralized). You will need to determine the pH of the equivalence point, then choose an indicator that changes color within that pH range.
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Both the e and z forms of the alkene will form in this reaction. if you only take into account product stability, which one would you expect to be the major product?
Answers
The trans isomer (E) would be the major product if you only take into account product stability.
As per the statement "Both the e and z forms of the alkene will form in this reaction. If you only take into account product stability, which one would you expect to be the major product?", the stability of the alkene is an important factor to determine the major product of a reaction. In the E/Z system, the two highest priority groups on each carbon atom in a double bond are placed in relation to each other.
The E/Z notation is based on the stereochemistry of alkenes or cycloalkenes. If the two highest priority groups are on the same side of the double bond, it is termed as Z (zusammen, German for "together"), while if they are on the opposite side, it is termed as E (entgegen, German for "opposite").For example:If you take into account product stability, then the major product would be the one with more stability.
In general, trans isomer (E) is more stable than the cis isomer (Z) because of the steric-hindrance caused by the substituent groups attached to the double bond. The greater the degree of steric hindrance, the lower the stability of the molecule.The trans isomer (E) has a linear arrangement of the carbon atoms around the double bond, whereas the cis isomer (Z) has a bent arrangement. The linear structure of the trans isomer is energetically more favorable than the bent structure of the cis isomer because it causes less steric hindrance. Hence, trans isomer is more stable than cis isomer.
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a concentration cell was set up at using two hydrogen electrodes. if the cell is generating a potential of , answer the following question: what is the concentration of in the cathode's half-cell solution, if the anode's half-cell is ?
Answers
The concentration of in the cathode's half-cell solution is 5.12 × 10^-6 M if the anode's half-cell is 1 M.
While answering questions on the Brainly platform, a question answering bot should always be factually accurate, professional, and friendly. Additionally, it should be concise and should not provide extraneous amounts of detail,
ignore any typos or irrelevant parts of the question, and use the terms mentioned in the question appropriately.In a concentration cell that was set up at using two hydrogen electrodes, the cell generates a potential of . The question is, what is the concentration of in the cathode's half-cell solution, if the anode's half-cell is ?
The given data is:Potential of the cell, °cell = 0.059 log10 [H+]cathode/[H+]anode= -0.0418 V (negative because H2 gas concentration is higher in the anode than in the cathode)Since we know the value of °cell,
we can calculate the cathode half-cell potential as:
E cathode = °cell - E anode = 0.0008 VAnd then using the Nernst equation, we can find the concentration of H+ in the cathode half-cell as follows:
E cathode = E° - (RT/nF) ln [H+]H+/H2The value of E° at room temperature,
T = 298K is zero, and the number of electrons involved,
n = 2. Hence,0.0008 V = -(0.0592/2) log10[H+]H+/H2[H+]H+/H2 = 5.12 × 10^-6 M.
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suppose 5.00 l of a gas is known to contain 0.965 mol. if the amount of gas is increased to 1.80 mol, what new volume will result
Answers
The volume of the result is 9.03 L when 5.00 l of a gas is known to contain 0.965 mol. if the amount of gas is increased to 1.80 mol.
The best gas regulation condition, PV = nRT, relates the strain (P), volume (V), measure of substance (n), and temperature (T) of a gas. Since the temperature is held consistent in this issue, we can utilize the accompanying type of the best gas regulation:
P1V1 = n1RT and P2V2 = n2RT
where P1, V1, n1, and P2, n2, and V2 are the underlying strain, volume, and measure of substance and last tension, volume, and measure of substance, individually.
We are given that the underlying volume V1 is 5.00 L and the underlying measure of substance n1 is 0.965 mol. We are likewise given that the last measure of substance n2 is 1.80 mol. To find the last volume V2, we can revamp the best gas regulation condition and address for V2:
V2 = (n2RT)/P2
We can utilize the underlying circumstances to find the underlying strain, which is:
P1 = (n1RT)/V1
We can then utilize the last measure of substance and the underlying strain to find the last tension, which is:
P2 = (n2RT)/V1
Subbing these qualities into the situation for V2 gives:
V2 = (n2RT * V1)/(n1RT + P2V1)
We can work on this articulation by offsetting the R and T terms, and connecting the given qualities:
V2 = (1.80 mol * 5.00 L)/(0.965 mol + (1.80 mol * 5.00 L * (0.965 mol/5.00 L)))
Subsequent to rearranging, we get:
V2 = 9.03 L
Thusly, the new volume will be 9.03 L when how much gas is expanded to 1.80 mol.
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a solution has 45.0 mg na2so4/ml. what is the na ion concentration (molarity) in this solution? a. 0.317 b. 0.634 c. 0.978 d. 0.714 e. 0.357
Answers
The Na+ ion concentration (molarity) in this solution is b) 0.634 M
To find the Na+ ion concentration (molarity) in the solution, first, determine the molar concentration of [tex]Na_{2} SO_{4}[/tex], then account for the fact that each [tex]Na_{2} SO_{4}[/tex] molecule contains two Na+ ions.
1. Convert the mass of Na2SO4 to moles:
45.0 mg [tex]Na_{2} SO_{4}[/tex] * (1 g / 1000 mg) * (1 mol / 142.04 g) = 0.000317 mol Na2SO4
2. Since there are 2 Na+ ions in each[tex]Na_{2} SO_{4}[/tex] molecule:
0.000317 mol [tex]Na_{2} SO_{4}[/tex] * 2 = 0.000634 mol Na+
3. Divide moles of Na+ by volume of the solution in liters:
0.000634 mol / 0.001 L = 0.634 M
So, the Na+ ion concentration (molarity) in this solution is 0.634 M. The correct answer is (b) 0.634.
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Which of the following reactions
is BALANCED and shows
INCOMPLETE combustion?
A. 2CH +40, - 5CO+6H,O
B. 2C₂H+40₂ → 7CO₂ + 6H₂O
C.
2C₂H+70₂4CO₂ + 6H₂O
D. 2C,H, +50, + 4CO + 6H₂0
Answers
The balanced reaction that shows incomplete combustion is 2C4H10 + 5O2 → 4CO + 6H2O.
option D.
What is balanced equation for incomplete combustion?The balanced equation for incomplete combustion involves a reactant, usually a hydrocarbon fuel, reacting with a limited supply of oxygen to produce carbon monoxide (CO) instead of carbon dioxide (CO2) and water (H2O).
Out of the options given, the balanced equation that shows incomplete combustion is:
D. 2C4H10 + 5O2 → 4CO + 6H2O
This equation shows incomplete combustion because it produces carbon monoxide (CO) instead of carbon dioxide (CO2). The equation is balanced because there are equal numbers of atoms of each element on both the reactant and product sides of the equation.
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Need help on my homework!! please
Answers
The dependent variable among the following components in the experiment is temperature as it is dependent on the others.
The temperature will be dependent on the other variables, such as the oxygen, vinegar, and steel wood. This experiment is designed to measure how the temperature changes when the other variables are altered. Temperature is the outcome that will be affected by changes in the independent variables, and changes in the dependent variable will provide insight into the effect of the independent variables. For example, when the oxygen is increased, the temperature may increase as well. Similarly when the vinegar is decreased, the temperature may decrease and the steel wood may also affect the temperature, though this will depend on the specific experiment.
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complete question: Julianne and Thabo are investigating temperature changes in chemical reactions. They have heard that vinegar removes the protective coating from steel wool, allowing it to rust. When iron combines with oxygen, heat is released. They use a Styrofoam cup with a lid to investigate this report.
Julianne records the initial temperature by wrapping the steel wool around the thermometer and placing it inside the empty Styrofoam cup. She records the temperature after 2 min. Thabo soaks the piece of steel wool in vinegar for 1 min and then squeezes out the excess vinegar. He wraps the wool around the thermometer, placing it back in the Styrofoam cup and seals the lid. He records the temperature after 15 min. The temperature has increased from 22 °C to 26 °C.
what is the dependent variable in this experiment?
A.) oxygen
B.) vinegar
C.) steel wood
D.) temperature
which one of the following substances, when dissolved in water at equal molar concentrations, will give the solution with the lowest electrical conductivity? a. cacl2 b. hno3 c. nh3 d. c6h12o6 (glucose) e. co2
Answers
The substance that will give the solution with the lowest electrical conductivity when dissolved in water at equal molar concentrations is C6H12O6 (glucose). The correct option is D.
a. CaCl2: Calcium chloride is an ionic compound that dissociates into ions (Ca2+ and 2Cl-) when dissolved in water, which increases electrical conductivity.
b. HNO3: Nitric acid is a strong acid that dissociates completely into ions (H+ and NO3-) when dissolved in water, which also increases electrical conductivity.
c. NH3: Ammonia is a weak base that partially forms ions (NH4+ and OH-) when dissolved in water, contributing to some electrical conductivity.
d. C6H12O6: Glucose is a covalent compound that does not dissociate into ions when dissolved in water, so it will not increase electrical conductivity.
e. CO2: Carbon dioxide is a covalent compound that dissolves in water to form a weak acid (H2CO3), which partially dissociates into ions (H+ and HCO3-), contributing to some electrical conductivity.
Since glucose (C6H12O6) does not dissociate into ions, it results in the lowest electrical conductivity among the listed substances.
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Complete the electron configuration for N. electron configuration: [He]
Answers
Answer: [He] 2s2 2p3
Explanation: im himithy
5.2643mol of ethane is held at 0.9035 atm and 506.25 °C. what is the volume in liters?
Answers
To find the volume of ethane , we can use the Ideal gas law. Which states -
[tex] \:\:\:\:\:\:\:\:\:\star\longrightarrow \sf \underline{PV=nRT} \\[/tex]
Where:-
P is the pressure measured in atmospheres V is the volume measured in litersn is the number of moles.R is the ideal gas constant (0.0821 L atm mol⁻¹ K⁻¹).T is the temperature measured in kelvin.As per question, we are given that-
P=0.9035 atm n=5.2643 moles T = 506.25°C = 506.25+273 = 779.25 KR = 0.0821 L atm mol⁻¹ K⁻¹Now that we have all the required values, so we can put them all in the Ideal gas law formula and solve for Volume -
[tex] \:\:\:\:\:\:\:\:\:\star\longrightarrow \sf \underline{PV=nRT} \\[/tex]
[tex] \:\:\:\:\:\:\:\:\:\longrightarrow \sf 0.9035 \times V= 5.2643\times 0.0821 \times 779.25\\[/tex]
[tex] \:\:\:\:\:\:\:\:\:\longrightarrow \sf 0.9035 \times V= 336.791\\[/tex]
[tex] \:\:\:\:\:\:\:\:\:\longrightarrow \sf V= \dfrac{336.791}{0.9035}\\[/tex]
[tex] \:\:\:\:\:\:\:\:\:\longrightarrow \sf V= 372.762......\\[/tex]
[tex] \:\:\:\:\:\:\:\:\:\longrightarrow \sf \underline{V= 372.762 \: L}\\[/tex]
Therefore, the volume of ethane is 372.762 L.what is the final volume of 5.31L of an ideal gas when heated from 200 K to 300 K at constant pressure?
Answers
Charles's Law-
[tex]\:\:\:\:\:\: \:\:\:\:\:\:\star\longrightarrow\sf \underline{\dfrac{V_1}{T_1}=\dfrac{V_2}{T_2}}\\[/tex]
Where:-
V₁ = Initial volumeT₁ = Initial temperatureV₂ = Final volumeT₂ = Final temperatureAs per question, we are given that -
V₁ =5.31LT₁ = 200KT₂ = 300KNow that we are given all the required values, so we can put them into the formula and solve for V₂ :-
[tex]\:\:\:\:\:\: \:\:\:\:\:\:\star\longrightarrow\sf \underline{\dfrac{V_1}{T_1}=\dfrac{V_2}{T_2}}\\[/tex]
[tex] \:\:\:\:\:\:\:\:\:\:\:\:\longrightarrow \sf V_2= \dfrac{V_1}{T_1}\times T_2\\[/tex]
[tex] \:\:\:\:\:\:\:\:\:\:\:\:\longrightarrow \sf V_2= \dfrac{5.31}{200}\times 300\\[/tex]
[tex] \:\:\:\:\:\:\:\:\:\:\:\:\longrightarrow \sf V_2 = 0.02655\times 300\\[/tex]
[tex]\:\:\:\:\:\: \:\:\:\:\:\:\longrightarrow \sf V_2=7.965 \\[/tex]
[tex] \:\:\:\:\:\:\:\:\:\:\:\:\longrightarrow \sf\underline{ V_2 = 7.965\: L}\\[/tex]
Therefore, the final volume of 5.31L of an ideal gas when heated from 200 K to 300 K at constant pressure will be 7.965 L.
A balloon is filled with air. It has a volume of 720 mL at a temperature of 22° C. You put the
balloon inside your hot oven where the temperature is now 109° C. What is the new volume
of the balloon?
mL
Answers
At a temperature of 109° C, the balloon's new volume is roughly 932.6 mL.
What happens to the balloon's volume as the temperature rises?The gas particles take in more heat as the temperature rises. They accelerate and advance apart from one another. Hence, an increase in volume is brought on by a rise in temperature.
We can use Charles's Law to solve this question,
V1/T1 = V2/T2
where, V1 = initial volume
T1 = initial temperature
V2 = final volume
T2 = final temperature,
Now, we have to convert the temperatures to the absolute scale, which is Kelvin (K).
T1 = 22 + 273.15 = 295.15 K
T2 = 109 + 273.15 = 382.15 K
Now, we can substitute values;
V1/T1 = V2/T2
720/295.15 = V2/382.15
Solving this equation,
V2 = (720/295.15) x 382.15
V2 = 932.6 mL
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A piece of silver releases 113.8 Joules of heat while cooling 45.0 ° C. What is the mass of the sample? Silver has a specific heat of 0.240 J/g°C.
Answers
We can use the formula for heat released:
Q = mcΔT
where Q is the heat released, m is the mass of the sample, c is the specific heat of the material, and ΔT is the change in temperature.
Plugging in the given values, we have:
113.8 J = m(0.240 J/g°C)(-45.0°C)
Simplifying:
m = 113.8 J / (0.240 J/g°C * -45.0°C)
m = 106.5 g
Therefore, the mass of the silver sample is 106.5 g.
ellular respiration is a chemical process in cells that releases energy the cells need to function. What statement below is true
bout this reaction. (1 point)
O
The process of cellular respiration releases energy because the energy that is released when the bonds are
formed in CO₂ and water is equal to the energy required to break the bonds of sugar and oxygen.
O
The process of cellular respiration releases energy because the energy that is released when the bonds
that are formed in CO2 and water is lost when bonds of glucose and oxygen are broken.
O
The process of cellular respiration releases energy because the energy that is released when the bonds are
formed in CO₂ and water is greater than the energy required to break the bonds of sugar and oxygen.
O
The process of cellular respiration releases energy because the energy that is released when the bonds are
formed in CO₂ and water is less than the energy required to break the bonds of sugar and oxygen.
Answers
Because more energy is released when water and carbon dioxide molecules are formed than when sugar and oxygen bonds are broken, the process of cellular respiration releases energy.
Where does the chemical energy that is released during cellular respiration come from?Cellular respiration is the process by which chemical energy from food is transformed into adenosine triphosphate (ATP), and afterwards waste products are expelled.
What kind of chemical process is cellular respiration?Through a sequence of chemical processes called cellular respiration, glucose is broken down to create ATP, which may then be used as an energy source for a variety of bodily functions.
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a buffer that contains 0.49 m of a base, b and 0.45 m of its conjugate acid bh , has a ph of 8.87. what is the ph after 0.024 mol of ba(oh)2 are added to 0.62 l of the solution?
Answers
The pH of the buffer solution after 0.024 mol of Ba(OH)₂ are added is approximately 4.90.
To solve this problem, we need to use the Henderson-Hasselbalch equation, which relates the pH of a buffer solution to the concentrations of the weak acid and its conjugate base;
pH = pKa + log([base]/[acid])
where pKa will be the dissociation constant of the weak acid, [base] is the concentration of the conjugate base, as well as [acid] is the concentration of the weak acid.
From the given information, we can determine the pKa of the weak acid;
pH = 8.87 = pKa + log([base]/[acid])
pKa = 8.87 - log([base]/[acid])
pKa = 4.83
We can also determine the initial concentrations of the weak acid and its conjugate base.
[base] = 0.49 M
[acid] = 0.45 M
Now, we can use the balanced chemical equation for the reaction of Ba(OH)₂ with BH;
Ba(OH)₂ + 2BH → BaB₂ + 2H₂O
The moles of BH that react with the added Ba(OH)₂ is equal to half the moles of Ba(OH)₂ added, because the stoichiometry of the reaction is 2:1. Therefore, moles of BH consumed = 0.024 mol / 2 = 0.012 mol
The volume of the solution is given as 0.62 L, so the new concentration of BH is;
[acid] = (0.45 M x 0.62 L - 0.012 mol) / 0.62 L
[acid] = 0.441 M
The new concentration of B can be determined from the conservation of mass equation:
[base] + [BaB2] = constant
[base] + 0.012 mol / 0.62 L = constant
[base] = 0.49 M - 0.012 mol / 0.62 L
[base] = 0.470 M
Now, we can use the Henderson-Hasselbalch equation to determine the new pH of the buffer solution
pH = pKa + log([base]/[acid])
pH = 4.83 + log(0.470/0.441)
pH = 4.90
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GEN CHEM 2 PLEASE HELP
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To find the pH of 0.33 M HCOOK, we need to first determine the concentration of HCOO- ions in the solution. When HCOOK dissolves in water, it dissociates to form HCOO- and K+ ions.
Since HCOOK is a salt of a weak acid, it will undergo hydrolysis in water, and the HCOO- ions will react with water to form HCOOH and OH- ions.
The balanced equation for this reaction is: HCOO- (aq) + H2O (l) ⇌ HCOOH (aq) + OH- (aq)
Using the given K₂ value for HCOOH, we can calculate the equilibrium concentrations of HCOO- and HCOOH:
K₂ = [HCOOH][OH-] / [HCOO-]
1.78 x 10⁻⁴ = [HCOOH][OH-] / [0.33]
[HCOOH][OH-] = 5.874 x 10⁻⁵
Assuming x is the concentration of HCOOH and OH- formed, we can set up an ICE table:
HCOO⁻ (aq) + H2O (aq) = HCOOH (aq) + OH^- (aq)
I 0.33 M 0 0
C - x x x
E 0.33 - x x x
Substituting the equilibrium concentrations into the K₂ expression, we get:
1.78 x 10⁻⁴ = x⁻² / (0.33 - x)
Since x is small compared to 0.33, we can approximate (0.33 - x) as 0.33:
1.78 x 10⁻⁴ = x⁻² / 0.33
Solving for x, we get x = 2.49 x 10⁻³ M
So, [HCOOH] = [OH-] = 2.49 x 10⁻³ M
To calculate the pH, we can use the equation: pH = 14 - pOH
pOH = -log[OH-] = -log(2.49 x 10⁻³) = 2.60
Therefore, pH = 14 - 2.60 = 11.40 (rounded to two significant figures)
Hence, the pH of 0.33 M HCOOK is 11.40.
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as a chemical reaction is taking place, the student notices that the temperature has dropped. this indicates that the system has increased in energy. what type of process is this?
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A decrease in temperature during a chemical reaction indicates that the system has lost energy to the surroundings and is because the energy released during the reaction is transferred from the system to the surroundings in the form of heat.
The type of process that causes a decrease in temperature during a chemical reaction is known as an exothermic process. In exothermic reactions, energy is released from the system in the form of heat and transferred to the surroundings, resulting in a decrease in the energy of the system. Exothermic reactions are commonly observed in combustion reactions, where a fuel reacts with oxygen to produce heat and light energy.
It is important to note that the amount of energy released during an exothermic reaction can vary depending on the specific reactants and conditions involved. In some cases, exothermic reactions can release a significant amount of energy, such as in the case of explosive reactions. In other cases, the energy released may be relatively small and not readily apparent, such as in the case of rusting of iron.
In summary, a decrease in temperature during a chemical reaction is an indication of an exothermic process, where energy is released from the system to the surroundings in the form of heat.
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A cylinder is filled with 10.0 L of gas and a piston is put into it. The initial pressure of the gas is measured to be 250 kPa. The piston is now pulled up, expanding the gas, until the gas has a final volume of 21.0 L . Calculate the final pressure of the gas. Be sure your answer has the correct number of significant digits.
Answers
The gas will have a final pressure of 26.8 kPa.
The pressure of a gas is inversely related to its volume at constant temperature, as well as its number of moles, according to Boyle's law.Using the ideal gas law, the final pressure can be calculated as follows:
[tex]P_1V_1=P_2V_2[/tex]
[tex]P_1[/tex]= the gas's initial pressure is 209 kPa.
[tex]P_2[/tex] =gas's final pressure =?
[tex]V_1[/tex] = 10.0 L, which is the gas's initial volume.
[tex]V_2[/tex] =78.0 L, which is the gas's final volume.
Now that all the needed variables have been entered, we can apply this formula to determine the final gas pressure.
209 kPa*10.0L= [tex]P_2[/tex] *78.0L
[tex]P_2 =26.8 kPa[/tex]
Therefore, the final pressure of the gas is 26.8kPa.
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the ph at the endpoint of a titration of hn3 with naoh is 8.74. which indicators would be a good choice for this titration?
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Thymol blue would be a good choice for this titration. As thymol blue has a pH range of 8 to 9.6, its color would change from yellow to green towards the end, making it the most obvious selection out of all of them.
If an indicator changes color at the end point, the end point has been reached and the indicator is being used to measure it. The conclusion of the titration is assumed to be at 8.74. Hence, the pH range of the indicator should be such that at 8.74, the color changes and is visible to the eye. As previously mentioned, a suitable indicator should have a pKin value that is quite near to the anticipated pH at the equivalence point. The choice of the indicator is not very important for a strong acid–strong base titration because of the significant pH shift that takes place at the equivalence point.
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a buffer was prepared by mixing 0.50 mole of hx acid and 0.50 mole of nax to form an aqueous solution with a total volume of 1.00 liter. the ph of this buffer was 4.925. then, to 400 ml of this buffer solution was added 25.0 ml of 2.0 m hcl. what is the ph of this new solution?
Answers
The pH of the new solution after adding the HCl is 2.82.
How to calculate pH after adding the HCl ?
To calculate the new pH of the buffer solution after adding the HCl, we need to use the Henderson-Hasselbalch equation, which relates the pH of a buffer solution to the pKa and the ratio of the concentrations of the conjugate acid and base.
First, we need to find the concentrations of the acid and its conjugate base in the buffer solution. We know that the total volume of the buffer solution is 1.00 L, and the number of moles of HX and NaX are equal, so each one has a concentration of:
C = n/V = 0.50 mol / 1.00 L = 0.50 M
Since HX is an acid, it will donate a proton (H+) to form its conjugate base (X-). Therefore, the acid and base concentrations in the buffer solution are:
[HX] = 0.50 M
[X-] = 0.50 M
The pKa of HX can be calculated from the pH and the acid dissociation constant (Ka):
pH = pKa + log([X-]/[HX])
4.925 = pKa + log(0.50/0.50)
4.925 = pKa
Now we can use the Henderson-Hasselbalch equation to find the pH of the buffer solution after adding the HCl:
pH = pKa + log([X-]/[HX])
pH = 4.925 + log(0.50/0.50)
pH = 4.925
This means that the pH of the original buffer solution is unchanged by the addition of the HCl.
However, if we want to calculate the pH of the new solution after adding the HCl, we need to use the stoichiometry of the reaction:
HX + HCl -> H2O + XCl
We know that 25.0 mL of 2.0 M HCl were added to the buffer solution, so the number of moles of HCl added is:
n(HCl) = C x V = 2.0 mol/L x 0.025 L = 0.050 mol
Since HX and HCl react in a 1:1 ratio, the number of moles of HX consumed is also 0.050 mol. This leaves us with:
n(HX) = 0.50 mol - 0.050 mol = 0.450 mol
The new volume of the solution is 400 mL + 25.0 mL = 0.425 L. Therefore, the new concentration of HX is:
[HX] = n(HX) / V = 0.450 mol / 0.425 L = 1.06 M
The new concentration of X- can be calculated from the buffer equation:
Ka = [H+][X-] / [HX]
Ka = 10⁻pKa = 10⁻⁴.925 = 7.09 x 10⁻⁵
[H+] = Ka [HX] / [X-] = 7.09 x 10⁻⁵ x 1.06 M / 0.50 M = 0.151 mM
[pH = -log[H+] = 2.82]
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2 X + 3Y ----> 4 Z
If 2.50 moles of X were reacted with Y, how many moles of Z would be produced?
Answers
5.00 moles of Z would be produced when 2.50 moles of X are reacted with Y.
From the balanced chemical equation:
2 X + 3Y → 4 Z
We can see that the molar ratio of X to Z is 2:4 or 1:2.
Therefore, if 2.50 moles of X are reacted, we can calculate the number of moles of Z produced using this ratio:
2.50 moles X × (2 moles Z / 1 mole X) = 5.00 moles Z.
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Calculate the weight of 6.023*10²⁴ molecules of SO2.
Answers
Answer:
64g
Explanation:
refer attachment
Answer:
Explanation: To calculate the weight of sulphur and oxygen,
6.023*10^23 (Avogadro's number) is the number of molecules in a mole. Therefore, if there are 6.023*10^24 molecules, there are 10 moles of SO2.
To calculate the weight, we need to take the molar mass of sulphur and oxygen.
There will be 320.2 gms of sulphur and 320 gms of oxygen.
suppose a vessel contains n2o5 at a concentration of 1.08m. calculate how long it takes for the concentration of n2o5 to decrease by 86.0%. you may assume no other reaction is important.
Answers
It would take approximately 17.2 hours for the concentration of N2O5 to decrease by 86.0% in the given vessel.
To calculate the time it takes for the concentration of N2O5 to decrease by 86.0%, we need to use the first-order rate equation:
ln([N2O5]t/[N2O5]0) = -kt
Where [N2O5]t is the concentration of N2O5 at time t, [N2O5]0 is the initial concentration of N2O5, k is the rate constant, and t is the time.
Rearranging the equation to solve for t, we get:
t = (ln([N2O5]t/[N2O5]0))/(-k)
Since the reaction is first-order, the rate constant can be determined from the half-life of the reaction using the equation:
t1/2 = ln2/k
where t1/2 is the half-life.
Given that the concentration of N2O5 decreases by 86.0%, the final concentration ([N2O5]t) is:
[N2O5]t = 0.14[N2O5]0
Substituting the given values into the equation for t, we get:
t = (ln(0.14))/(-k)
To determine k, we need to find the half-life of the reaction. Since the reaction is first-order, the half-life can be calculated using the equation:
t1/2 = ln2/k
Substituting the given concentration of N2O5 and the rate constant into the equation, we get:
t1/2 = (ln2)/k = (ln2)/(2.303*0.108) = 5.45 hours
Finally, substituting the values for k and t1/2 into the equation for t, we get:
t = (ln(0.14))/(-0.108)
t = 17.2 hours
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consider an unknown compound with the formula c h i . given that the compound is composed of 54.53 % c, 9.15% h and 36.32% o what is the empirical formula of the compound?
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Think about an unidentified substance with the formula c h i. it contains 54.53% C, 9.15% H, and 36.32% O. The empirical formula of the compound is C2H4O.
To find the empirical formula of a compound from the given percentage composition, we need to convert the percentages into the number of moles of each element. We can assume a convenient mass for the compound, such as 100 g, and use the molar masses of each element to convert the percentages into moles.
For this unknown compound, if we assume 100 g of the compound, we have:
Mass of C = 54.53 g
Mass of H = 9.15 g
Mass of O = 36.32 g
Using the molar masses of each element (12.01 g/mol for C, 1.01 g/mol for H, and 16.00 g/mol for O), we can convert these masses into moles:
Moles of C = 54.53 g / 12.01 g/mol = 4.54 mol
Moles of H = 9.15 g / 1.01 g/mol = 9.06 mol
Moles of O = 36.32 g / 16.00 g/mol = 2.27 mol
Next, we divide each of the mole values by the smallest mole value to get the smallest whole number ratio of the atoms:
C: 4.54 mol / 2.27 mol = 2
H: 9.06 mol / 2.27 mol = 4
O: 2.27 mol / 2.27 mol = 1
Therefore, the empirical formula of the compound is C2H4O.
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mammoth cave np contains deposits of potassium nitrate (kno3) that were used in the past to produce saltpeter, an important ingredient in gunpowder. these nitrate deposits were exploited extensively (and virtually exhausted) to support military action during:
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Mammoth Cave NP contains deposits of potassium nitrate (KNO3) that were used in the past to produce saltpeter, an important ingredient in gunpowder. These nitrate deposits were exploited extensively (and virtually exhausted) to support military action during the War of 1812.
Mammoth Cave NP contains deposits of potassium nitrate (KNO3) that were used in the past to produce saltpeter, an important ingredient in gunpowder. These nitrate deposits were exploited extensively (and virtually exhausted) to support military action during the War of 1812.The park is 211.26 square kilometers in size and has approximately 400 miles of surveyed caves. The longest cave system in the world is located here.
Mammoth Cave was established in 1941 as a national park. Since then, scientists have discovered that Mammoth Cave National Park's cave system is among the most biologically diverse in the world. Mammoth Cave National Park has been designated a UNESCO World Heritage Site and a Biosphere Reserve as well. So therefore these nitrate deposits were exploited extensively (and virtually exhausted) to support military action is during the War of 1812.
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what are humic substances? group of answer choices carbonates and phosphates glucose molecules clay particles silt and sand recalcitrant organic molecules
Answers
Humic substances are recalcitrant organic molecules. Option E is correct.
Humic substances are complex organic compounds that are formed by the decomposition of dead plant and animal matter. They are found in soil, water, and sediments, and are an important component of organic matter in the environment. Humic substances are recalcitrant organic molecules that are resistant to further decomposition, and play a critical role in nutrient cycling, water retention, and soil structure.
They are composed of three main fractions: humic acid, fulvic acid, and humin. Humic acid is the largest and most complex fraction, and is insoluble in water at acidic pH. Fulvic acid is the smallest and most soluble fraction, and is soluble in water at all pH values.
Humin is the fraction that remains after the extraction of humic and fulvic acids, and is relatively insoluble in both water and organic solvents. Humic substances are known to have a wide range of beneficial properties, including soil improvement, plant growth enhancement, and water quality improvement.
Hence, E. is the correct option.
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--The given question is incomplete, the complete question is
"What are humic substances? group of answer choices A) carbonates and phosphates B) glucose molecules C) clay particles D) silt and sand E) recalcitrant organic molecules."--
A student is to prepare 250.0 mL of 0.100 M CuSO4 from a 0.500 M stock solution. What volume of stock solution is needed?
Answers
The volume of stock solution needed is 50mL when a student is set to prepare 250.0 mL of 0.100 M [tex]CuSO_4[/tex] from a 0.500 M stock solution.
Given the initial concentration of [tex]CuSO_4[/tex] solution (M1) = 250mL
The initial volume of [tex]CuSO_4[/tex] solution (M1) = 0.100M
The final concentration of [tex]CuSO_4[/tex] = V2
The final concentration of [tex]CuSO_4[/tex] (M2) = 0.500M
We know that molarity also known as molar concentration, is a measure of the concentration of a solute in terms of moles per liter of solution. It is also useful for comparing different solutions and for calculating the amount of a reactant needed in a reaction.
M1*V1 = M2*V2 = constant such that:
0.25 * 0.100 = 0.500 * V2
V2 = 0.05L = 50mL
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What volume of O₂ at 10.0° C and .890 atm would be required to generate 55.2 grams of NO₂?
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The chemical reaction between oxygen and nitrogen dioxide has the following balanced chemical equation: 2Nitrogen dioxide + Oxygen -> 2Nitrogen dioxide
What is volume of oxygen required at 0 degree celsius and 1 atm to burn completely?Hence volume of oxygen gas measured at 0oC and 1atm, needed to burn completely 1L of propane gas under the same conditions is 5L.
According to the equation, 2 moles of Nitrogen dioxide and 1 mole of Oxygen combine to form 2 moles of Nitrogen dioxide.
We must first determine how many moles of Nitrogen dioxide 55.2 grams will produce:
mass / molar mass equals moles of Nitrogen dioxide.
Nitrogen dioxide moles are equal to 55.2 g/46.0055 g/mol.
1.200 mol Equals 1 mole of Nitrogen dioxide.
We just require half as many moles of Oxygen since 1 mole of Oxygen produces 2 moles of Nitrogen dioxide:
1.200 moles of Oxygen are equal to 0.600 moles when divided by two.
Now, we can calculate the volume of Oxygen needed using the ideal gas law:
PV = nRT
P equals pressure where= 0.890 atm
V = volume (in liters)
n = number of moles = 0.600 mol
R = gas constant = 0.08206 L·atm/mol·K
T = temperature = 10.0 + 273.15 = 283.15 K
Solving for V:
V = nRT / P
V = (0.600 mol)(0.08206 L·atm/mol·K)(283.15 K) / 0.890 atm
V = 14.2 L
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