Answer:
[tex]2.5\; \rm g[/tex] of [tex]\rm CaCO_{3}[/tex] would contain:
Approximately [tex]1.5 \times 10^{22}[/tex] calcium atoms (approximately [tex]0.025\; \rm mol[/tex],)Approximately [tex]1.5 \times 10^{22}[/tex] carbon atoms (approximately [tex]0.025\; \rm mol[/tex],) andApproximately [tex]4.5 \times 10^{22}[/tex] oxygen atoms (approximately [tex]0.075\; \rm mol[/tex].)Explanation:
Look up the Avogadro constant: [tex]N_{\rm A} \approx 6.022 \times 10^{23}\; \rm mol^{-1}[/tex].
For example, "[tex]1\; \rm mol[/tex] of carbon atoms" would contain [tex]N_{\rm A}[/tex] carbon atoms (approximately [tex]6.022\times 10^{23}[/tex]) by definition.
Look up the relative atomic mass of carbon, calcium, and oxygen on a modern periodic table:
Calcium: [tex]40.078[/tex].Carbon: [tex]12.011[/tex].Oxygen: [tex]15.999[/tex].In other words, the mass of [tex]1\; \rm mol[/tex] of calcium atoms would be [tex]40.078\; \rm g[/tex]. The mass of [tex]1\; \rm mol\![/tex] of carbon atoms would be [tex]12.011\; \rm g[/tex], and the mass of [tex]1\; \rm mol \!\![/tex] of oxygen atoms would be [tex]15.999\; \rm g[/tex].
As the formula [tex]\rm CaCO_{3}[/tex] suggests, every formula unit of this ionic compound includes one calcium atom, one carbon atom, and three oxygen atoms. The formula mass of [tex]\rm CaCO_{3}\![/tex] would give the mass of every mole of [tex]\rm CaCO_{3}\!\![/tex] formula units.
Calculate the formula mass of [tex]\rm CaCO_{3}[/tex] from the relative atomic mass data:
[tex]\begin{aligned} & M({\rm CaCO_{3}}) \\ =\; & 40.078\; \rm g \cdot mol^{-1} \\ & + 12.011\; \rm g \cdot mol^{-1} \\ & + 3 \times (15.999\; \rm g \cdot mol^{-1}) \\ =\; & 100.086\; \rm g \cdot mol^{-1}\end{aligned}[/tex].
Calculate the number of [tex]\rm CaCO_{3}[/tex] formula units in that [tex]2.5\; \rm g[/tex] of this compound:
[tex]\begin{aligned}& n({\rm CaCO_{3}}) \\ =\; & \frac{m({\rm CaCO_{3}})}{M({\rm CaCO_{3}})} \\ =\; & \frac{2.5\; \rm g}{100.086\; \rm g \cdot mol^{-1}} \\ \approx\; & 0.025\; \rm mol\end{aligned}[/tex].
In other words, [tex]2.5\; \rm g[/tex] of [tex]\rm CaCO_{3}[/tex] would contain approximately [tex]0.025\; \rm mol[/tex] [tex]\rm CaCO_{3}\![/tex] formula units.
Again, there are one calcium atom, one carbon atom, and one oxygen atom in every [tex]\rm CaCO_{3}[/tex] formula unit. That approximately [tex]0.025\; \rm mol[/tex] [tex]\rm CaCO_{3}\![/tex] formula units would thus contain:
Approximately [tex]1 \times 0.025\; \rm mol = 0.025\; \rm mol[/tex] calcium atoms, Approximately [tex]1 \times 0.025\; \rm mol = 0.025\; \rm mol[/tex] carbon atoms, andApproximately [tex]3 \times 0.025\; \rm mol = 0.075\; \rm mol[/tex] oxygen atoms.Make use of the Avogadro constant to convert the numbers.
For example, the number of calcium atoms in that approximately [tex]0.025\; \rm mol[/tex] of calcium atoms would be:
[tex]\begin{aligned} & N({\text{calcium}) \\ = \; & n({\text{calcium}) \cdot N_{\rm A} \\ \approx \; & 0.025\; \rm mol \times 6.022\times 10^{23} \cdot mol^{-1} \\ \approx \; & 1.5 \times 10^{22} \end{aligned}[/tex].
Likewise, the number of carbon atoms in that approximately [tex]0.025\; \rm mol[/tex] of carbon atoms would be:
[tex]\begin{aligned} & N({\text{carbon}) \\ = \; & n({\text{carbon}) \cdot N_{\rm A} \\ \approx \; & 0.025\; \rm mol \times 6.022\times 10^{23} \cdot mol^{-1} \\ \approx \; & 1.5 \times 10^{22} \end{aligned}[/tex].
The number of oxygen atoms in that approximately [tex]0.075\; \rm mol[/tex] of oxygen atoms would be:
[tex]\begin{aligned} & N({\text{oxygen}) \\ = \; & n({\text{oxygen}) \cdot N_{\rm A} \\ \approx \; & 0.075\; \rm mol \times 6.022\times 10^{23} \cdot mol^{-1} \\ \approx \; & 4.5 \times 10^{22} \end{aligned}[/tex].
The total charge on each atom is zero explain why?
Answer:
because it has the same number of protons as electrons.
PLeaseeeeeeee help me ┗( T﹏T )┛
How did you figure out the number of electrons in each atom?
Answer:
The atomic number of an atom [ number of protons ] is the number of electrons of an atom.
[tex].[/tex]
27. What is the mass of 2.34 x 1023 atoms of Sulfur?
a) 12.5 g
b) 1.25 x 1049
g
c) 1.25 x 1047 g
Answer:
[tex]\boxed {\boxed {\sf A. \ 12.5 \ grams}}[/tex]
Explanation:
We are asked to find the mass of 2.34 × 10²³ atoms of sulfur (S)
1. Convert Atoms to MolesFirst, we convert atoms to moles using Avogadro's Number of 6.022 × 10²³. This is the number of particles (atoms, molecules, formula units, etc.) in 1 mole of a substance. For this question, the particles are atoms of sulfur. There are 6.022 × 10²³ atoms of sulfur in 1 mole of sulfur.
We will convert using dimensional analysis, so we must set up a ratio.
[tex]\frac {6.022 \times 10^{23} \ atoms \ S}{1 \ mol \ S}[/tex]
We are converting 2.34 × 10²³ atoms of sulfur to moles, so we multiply the ratio by this value.
[tex]2.34 \times 10^{23} \ atoms \ S* \frac {6.022 \times 10^{23} \ atoms \ S}{1 \ mol \ S}[/tex]
Flip the ratio so the units of atoms of sulfur cancel.
[tex]2.34 \times 10^{23} \ atoms \ S* \frac{1 \ mol \ S} {6.022 \times 10^{23} \ atoms \ S}[/tex]
[tex]2.34 \times 10^{23}* \frac{1 \ mol \ S} {6.022 \times 10^{23}}[/tex]
[tex]\frac{2.34 \times 10^{23}} {6.022 \times 10^{23}} \ mol \ S[/tex]
[tex]0.3885752242 \ mol \ S[/tex]
2. Convert Moles to GramsNext, we convert moles to grams using the molar mass. This is the mass of 1 mole of a substance. The value is found on the Periodic Table as the atomic mass, but the units are grams per mole instead of atomic mass units. Look up sulfur's molar mass.
S: 32.07 g/molSet up another ratio.
[tex]\frac {32.07 \ g \ S}{ 1\ mol \ S}[/tex]
Multiply by the number of moles we calculated.
[tex]0.3885752242 \ mol \ S *\frac {32.07 \ g \ S}{ 1\ mol \ S}[/tex]
The units of moles of sulfur cancel.
[tex]0.3885752242 *\frac {32.07 \ g \ S}{ 1}[/tex]
[tex]0.3885752242 * {32.07 \ g \ S}[/tex]
[tex]12.46160744 \ g \ S[/tex]
3. RoundThe original value of atoms (2.34 × 10²³) has 3 significant figures (2,3, 4), so our answer must have the same. For the number we calculated, that is the tenths place. The 6 in the hundredth place tells us to round the 4 up to a 5.
[tex]12.5 \ g \ S[/tex]
There are 12.5 grams of sulfur in 2.34 × 10²³ atoms of sulfur.
Unit of electro negativity
Answer:
There are no units with electro negativity. Linus Pauling designed a scale of electro negativity that ranks elements with respect to each other. So, for example, fluorine is a 4.0 in comparison to 0.7 for francium.
Explanation:
What is the specific heat of a metal with a mass of 14.0 g, heat of 3.45 kJ and a change in temperature of 6.2 ?
Answer:
Explanation:
0.0
NaClO4/ HCIO4 is a buffer system
Is NaClO4/HClO4 a buffer system
Answer:
In this type of solution, perchloric acid is a strong acid which means that its conjugate base is extremely weak. As such, the solution of HClO4 H C l O 4 and NaClO4 N a C l O 4 is not a buffer solution.
limewater turns cloudy or milk when a gas bubbled through it. what is the cause of its cloudiness?
hi <3
the cloudiness is caused by the production of carbon dioxide
hope this helps :)
Answer:
Lime water is calcium hydroxide with formula
[tex]{ \sf{Ca(OH)_2}}[/tex]
when calcium hydroxide reacts with the gas [ carbon dioxide ]
[tex]{ \sf{Ca(OH)_{2(aq)} + CO_{2(g)} → CaCO_{3(s)} + H_2O_{(l)}}}[/tex]
Calcium carbonate is white cloudy precipitate
Convert 50 kilograms into grams.show your work
Answer:
50000 g
Explanation:
1 kilogram (kg) is equal to 1000 grams (g).
1 kg = 1000 g
The mass m in grams (g) is equal to the mass m in kilograms (kg) times 1000:
m(g) = m(kg) × 1000
PLEASE ANSWER!!!!!!! 80 POINTS
Which statement describes the properties of lanthanoids and actinoids?(1 point)
A. They are malleable and lustrous, and they tend to conduct both electricity and thermal energy.
B. They are brittle and dull, and they tend to conduct both electricity and thermal energy.
C. They are brittle and dull, and they tend to be poor conductors of both electricity and thermal energy.
D. They are malleable and lustrous, and they tend to be poor conductors of both electricity and thermal energy.
Lanthanoids and actinoids are transition elements of the periodic table. These elements are lustrous and malleable and can conduct thermal energy and electricity. Thus, option A is correct.
What are lanthanoids and actinoids?Lanthanoids and actinoids are the terms used to refer to the elements of the f-block of the periodic table that are placed separately due to the difference in their electronic structures.
These elements show the property somewhat similar to that of the metals and hence are lustrous and malleable. Due to the transition electron phase, they are used in optical devices and cardiac pacemakers.
The lanthanoids and actinoids are good conductors of heat and electricity as their electrons can be transferred and carry charges like metals.
Therefore, option A. the lanthanoids and actinoids are conductors and are malleable.
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Determine the empirical formula of the compound formed when 1.2g of magnesium reacts with 3.55g of chlorine.Take the molar mass of magnesium and chlorine to be 24g and 35.5g respectively.
Explanation:
For Mg, (1.2 g Mg/24 g Mg) = 0.05 mol Mg.
For Cl, (3.55 g Cl/35.5 g Cl) = 0.1 mol Cl
So the ratio now is
Mg:Cl = 0.05 : 0.1 = 1:2
I got the 1:2 ratio by dividing both by the smallest number, which is 0.05 mol. Therefore, the empirical for formula of the substance is [tex]MgCl_2[/tex]
The empirical formula of the compound formed when magnesium reacts with chlorine is MgCl₂.
To determine the empirical formula of the compound formed when magnesium (Mg) reacts with chlorine (Cl), we need to find the ratio of the elements in terms of moles.
First, we calculate the number of moles of each element:
Moles of Mg = Mass of Mg / Molar mass of Mg
Moles of Mg = 1.2 g / 24 g/mol
Moles of Mg = 0.05 mol
Moles of Cl = Mass of Cl / Molar mass of Cl
Moles of Cl = 3.55 g / 35.5 g/mol
Moles of Cl = 0.1 mol
Next, we determine the simplest whole-number ratio between the moles of Mg and Cl:
Mg:Cl = 0.05 mol : 0.1 mol
To simplify the ratio, we divide both values by the smaller value (0.05 mol in this case):
Mg:Cl = 0.05 mol / 0.05 mol : 0.1 mol / 0.05 mol
Mg:Cl = 1 : 2
Therefore, the empirical formula of the compound formed when magnesium reacts with chlorine is MgCl₂.
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Write short Note On Valence shell
Answer:
The number of electrons in the outermost shell of a particular atom determines its reactivity, or tendency to form chemical bonds with other atoms. This outermost shell is known as the valence shell, and the electrons found in it are called valence electrons.
Explanation:
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please help me with this question i put 100 points on it and i will give brainlyest no links or files or i will report
The Earth’s tilt changes its position relative to the stars and constellations as the Earth rotates around its axis.
How do we measure the weight or mass of something
Explanation:
We can measure the weight or mass of something by using beam balance.
How many grams of H2O are produced when 35.0 g of NaOH reacts with 17.5 g of CO,?
2NaOH + CO2 -> Na2CO3 + H2O
Answer:
2NaOH + CO2 -> Na2CO3 + H2O
1) Find the moles of each substance
[tex]\eq n(NaOH)=\frac{35.0}{22.99+16.00+1.008\\}\ =\frac{35.0}{39.998} \ = 0.8750437522 moles\\n(CO_{2} ) = \frac{17.5}{12.01+32.00} = \frac{17.5}{44.01} = 0.3976369007 moles\\[/tex]
2) Determine the limitting reagent
[tex]\\NaOH = \frac{0.8750437522}{2} = 0.4375218761\\\\[/tex]
∴ Carbon dioxide is limitting as it has a smaller value.
3) multiply the limiting reagent by the mole ratio of unknown over known
n(H2O ) = 0.3976369007 × 1/2
= 0.1988184504 moles
4) Multiply the number of moles by the molar mass of the substance.
m = 0.1988184504 × (1.008 × 2 + 16.00)
= 0.1988184504 × 18.016
= 3.581913202 g
Explanation:
The correct answer is = 3.581913202 g.
n(H2O ) = 0.3976369007 × 1/2
= 0.1988184504 moles
m = 0.1988184504 × (1.008 × 2 + 16.00)
= 0.1988184504 × 18.016
= 3.581913202 g.
What does H2O stand for?H2O is the molecular formula of water, also called Dihydrogen monoxide. H2O symbolises two hydrogen atoms and one oxygen atom.
What is 1 mol H2O?
Water (H2O) is made from 2 atoms of hydrogen and 1 atom of oxygen. A mole of water molecules would be 2 moles of hydrogen atoms plus 1 mole of oxygen atoms. From the periodic table, we see the atomic weight of hydrogen is 1.0079 and the atomic weight of oxygen is 15.9994.
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The study of Environmental Science includes which of the following?
A.)human biology
B.)the relationship between the Earth and the moon
C.)interactions between humans and living and non-living things in the environment
D.)the study of organisms living in the ocean
Answer:
Hi, There! the Correct Answer is below!
C.)interactions between humans and living and non-living things in the environment
Explanation:
Environmental scientists study subjects like the understanding of earth processes, evaluating alternative energy systems, pollution control and mitigation, natural resource management, and the effects of global climate change. ... Environmental scientists bring a systems approach to the analysis of environmental problems.
____________________________________________________________
Have a wonderful Day :)
~Darkspirit~
Answer:
interactions between human and living and non living things in the environment
What are the products of combining iron (II) and Copper (II) oxide in a Chemical reaction?
oxide in a chemical reaction
CuFe2O4 are the products of combining iron (II) and Copper (II) oxide in a Chemical reaction.
What is chemical reaction?The conversion of one or more chemicals into new one or more chemicals is called chemical reaction.
Example: iron reacts with oxygen to form rust.
What is product?The final substance of chemical reaction is called product.
Example: CuFe2O4 is the final substance called product when iron (II) and Copper (II) oxide react with each other.
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Dieldrin is a pesticide that is effective in killing?
Answer:
fddoiiijjjjjjjhhggggrrddffgghuuiii
Convert a length of 20.0m to inches
Answer:
787.401575 inches
How many grams of Co, are produced when 3.60 g of Fe 0 reacts with excess CO?
Fe2O3 + CO -> Fe + CO2
Answer:
2.98 g CO2
Explanation:
1. Balance the chemical equation
1 Fe2O3 + 3CO -> 2Fe + 3CO2
2. Use periodic table to find mass from each CO2 and Fe2O3
3. Calculate
3.60 g Fe2O3 x [tex]\frac{1 mol Fe2O3}{159.69 g Fe2O3}x\frac{3 mol CO2}{1 mol Fe2O3}[/tex] x[tex]\frac{44.01 g CO2}{1 mol CO2}[/tex] = 2.98 g CO2
The grams of CO, produced when 3.60 g of Fe 0 reacts with excess CO is 2.98 grams Co2.
What is mass?Mass is the quantity of matter of a physical body.
The balanced equation is
[tex]\rm Fe_2O_3 + CO - > Fe + CO_2[/tex]
The calculation is
[tex]3.60 g \;Fe_2O_3 \times \dfrac{ 1 mol\;Fe_2O_3}{159.69} \times \dfrac{3 mol CO_2}{1 mol \;Fe_2O_3 } \times \dfrac{44.01\;CO_2}{1 mol\; CO_2} = 2.98 g\;CO_2[/tex]
Thus, the grams of CO, produced when 3.60 g of Fe 0 reacts with excess CO is 2.98 grams Co2.
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The starting material is a benzene ring with a hydroxy group on carbon 1 and an N H 2 on carbon 4. This reacts with one equivalent of acetic anhydride, which is an oxygen flanked by two carbonyls, each bonded to a methyl group.
The final product of the reaction is 4-amino-3-hydroxy-1-acetylbenzene, also known as N-acetylanthranilic acid.
What is catalyst?A catalyst is a substance that accelerates or reduces the temperature or pressure required to initiate a chemical reaction without being consumed in the process.
The Friedel-Crafts acylation reaction occurs when benzene reacts with acetic anhydride in the presence of an acid catalyst.
The acid catalyst activates the acetic anhydride to form an acylium ion intermediate in this reaction. This intermediate reacts with the benzene ring, giving it an acetyl group.
On carbon 1, there is a hydroxy group, and on carbon 4, there is an NH2 group.
Both of these functional groups are activating groups, which means they donate electrons to the benzene ring and increase its reactivity to electrophilic aromatic substitution reactions like Friedel-Crafts acylation.
Acetic anhydride adds an acetyl group to the benzene ring in a regioselective manner, which means it only adds to one position on the ring.
Thus, in this case, it adds to carbon 1, which is more reactive than carbon 4 due to the presence of the hydroxy group.
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Where do you find elements that have energy levels that are all full of electrons? (1 point)
across the bottom of the periodic table
on the right side of the periodic table
on the left side of the periodic table
across the top of the periodic table
Answer:
पेरिोदिछ ताबले माछकिनौय
Answer:
on the right side of the periodic table
Please show the working for question 1 to 8
Answer:
5) C
6) D
7) D
8) C
Explanation:
Which of the following is an example of a change from gravitational potential
energy to kinetic energy?
Answer:
The speed of a roller coaster increases as gravity pulls it downhill and is greatest at its lowest point. Viewed in terms of energy, the roller-coaster-Earth system's gravitational potential energy is converted to kinetic energy
Explanation:
a ball held in the air is the example of a change from gravitational potential energy to kinetic energy.
2. Based on the following balanced equation:
2 NH3 + 202 N2O + 3 H2O
a. How many moles of O2 are required to react completely with 3.00 moles of NH3?
1.00 1.50 2.00 2.50 3.00 3.50 4.00 4.50 5.00
b. How many moles of H20 are produced after complete reaction of 3.00 moles of NH3?
1.00
1.50 2.00 2.50 3.00
3.50 4.00 4.50 5.00
c. How many moles of N O are produced after complete reaction of 3.00 moles of NH3?
1.00 1.50 2.00 2.50 3.00
3.50
4.00 4.50 500
To do this, just take the given miles and multiply it with the ratio of molecules in the chemical equation to get your answer.
How many moles of aluminum oxide (Al2O3) are in a sample with a mass of 204.0 g?
Answer:
[tex]\boxed {\boxed {\sf 2.001 \ mol \ Al_2O_3}}[/tex]
Explanation:
We are asked to convert grams of aluminum oxide to moles. We will do this using the molar mass or mass of 1 mole of a substance. The values are found on the Periodic Table. They are equal to the atomic masses, but the units are grams per mole instead of atomic mass units.
We have the compound aluminum oxide or Al₂O₃. We must look up the molar masses of the individual elements: aluminum and oxygen.
Al: 26.981538 g/mol O: 15.999 g/molThe formula of the compound has subscripts (a subscript of 2 after Al and subscript of 3 after O). There are 2 moles of aluminum and 3 moles of oxygen in 1 mole of aluminum oxide. We must multiply the molar mass by its subscript.
Al₂ : 26.981538 * 2=53.963076 g/mol O₃: 15.999 * 3=47.997 g/mol Al₂O₃ =53.963076 + 47.997 = 101.960076 g/molWe convert using dimensional analysis, so we set up a ratio using the molar mass.
[tex]\frac {101.960076 \ g \ Al_2O_3}{ 1 \ mol \ Al_2O_3}[/tex]
We are converting 204.0 grams to moles, so we multiply by this value.
[tex]204.0 \ g \ Al_2O_3 *\frac {101.960076 \ g \ Al_2O_3}{ 1 \ mol \ Al_2O_3}[/tex]
Flip the ratio so the units of grams of aluminum oxide cancel.
[tex]204.0 \ g \ Al_2O_3 *\frac { 1 \ mol \ Al_2O_3}{101.960076 \ g \ Al_2O_3}[/tex]
[tex]204.0 *\frac { 1 \ mol \ Al_2O_3}{101.960076}[/tex]
[tex]\frac {204.0}{101.960076} \ mol \ Al_2O_3[/tex]
[tex]2.00078313 \ mol \ Al_2O_3[/tex]
The original measurement of grams has 4 significant figures, so our answer must have the same. For the number we found, that is the thousandth place. The 7 in ten-thousandth place tells us to round the 0 up to a 1.
[tex]2.001 \ mol \ Al_2O_3[/tex]
204.04 grams of aluminum oxide is equal to approximately 2.01 moles of aluminum oxide.
solve it and get 50 points
Answer:
Not sure but I think it is C!
Explanation:
In nature, Chlorine-35 isotopes (35 amu) make up 75
% of all chlorine. Chlorine-37 (37 amu) isotopes make
up 25 % of all chlorine,
Show your calculation for the AVG. Atomic Mass of
chlorine:
* The 'emf generated by
a copper-constantan thermometer is 15 mv. If the cold junction is at a temperature of 20 degrees celcius.Calculate the temperature of the hot junction when sensitivity
of the thermocouple is 0.03mv/degree celcius
Thermometers are used to measure temperatures. The temperature of the hot junction of the thermocouple is [tex]520^oC[/tex]
Given that:
[tex]EMF = 15mv[/tex]
[tex]T_1 =20^oC[/tex] --- Temperature at cold junction
[tex]s = 0.03mv/^oC[/tex] ---- Sensitivity
The temperature [tex]T_2[/tex] at the hot junction is calculated as follows:
[tex]T_2 = \frac{EMF}{s} + T_1[/tex]
So, we have:
[tex]T_2 = \frac{15mv}{0.03mv/^oC} + 20^oC[/tex]
[tex]T_2 = \frac{15}{0.03/^oC} + 20^oC[/tex]
[tex]T_2 = \frac{15}{0.03}^oC + 20^oC[/tex]
[tex]T_2 = 500^oC + 20^oC[/tex]
[tex]T_2 = 520^oC[/tex]
Hence, the temperature of the hot junction of the thermocouple is [tex]520^oC[/tex]
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