What is the mass of 89.6 L of NH3(g) at STP?

Please help I don’t understand!

Answer: 4.4 x 10^-7

Explanation:

The dissociation equation for this reaction is:

MgCO3 (s) → Mg+2 (aq) + CO3-2 (aq)

[tex]\text { So, } k_{s p}=(x+0.08) x[/tex] (Here 0.08 >>> x )

[tex]\begin{aligned}\Rightarrow 3.5 \times 10^{-8} &=0.08 \times x \\\Rightarrow x &=\frac{3.5}{0.08} \times 10^{-8} \\&=4.37 \times 10^{-7} \\\Rightarrow x & \approx 4.4 \times 10^{-7} \mathrm{~M}\end{aligned}[/tex]

So the solubility MgCO₃ in a solution that containing 0.080 M Mg²⁺ is 4.4 x 10^-7

Answer:

b, and d

Explanation:

if the student did not aim correctly, and for d if the light waves travelled differently

Answer:

Cl2

Explanation:

Chlorine is a diatomic element  so it is   Cl 2

 the others are MOLECULES  (made up of atoms of ELEMENTS)

Dancing is a physical activity which strengthens the abdominal and calf muscles as well as improves flexibility.

A physical activity is any activity which involves movement of the body in order to strengthen the muscles and to be physically fit.

Dancing is a physical activity which involves intricate and repeated movements of the body.

Dancing is beneficial because;

Some muscles involved in dancing include;

Some risks in dancing include muscle injuries, dislocations and possibly fractures due to falls.

Therefore, a dancing routine strengthens the abdominal and calf muscles as well as improves flexibility.

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Answer:

This means that one mole of water molecules has a mass of 18.015g.

42.3g * 1 mole H2O/18.015g

= 2.3480 moles H2O

We are using the avogadros number to find the number of molecules of water

2.3480 H2O * 6.022*10^ 23moles/ 1mole of H2O

That's 2.3480 multiplied by 6.022*10^23 divided by 1 mole of H2O

Number of molecules = 1.4140 *10^24 molecules

The molarity of a new solution that has a volume of 25.0 mL of a 2.00 M HCl solution and diluted to 2.00 L is 0.025M.

The molarity of a solution can be calculated using the following formula:

C1V1 = C2V2

Where;

2 × 25 = 2000 × C2

50 = 2000C2

C2 = 50 ÷ 2000

C2 = 0.025M

Therefore, the molarity of a new solution that has a volume of 25.0 mL of a 2.00 M HCl solution and diluted to 2.00 L is 0.025M.

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From the balanced equation of the reaction, the coefficients of the reactants and products are 1:2:1.

A balanced equation of a chemical reaction is an equation which balances the atoms of each element on the reactants as well as the products side of a chemical reaction.

The balanced equation of the reaction that occurs when magnesium bromide is heated is shown below:

[tex]MgBr_2 \rightarrow Br_2+Mg [/tex]

Therefore, the coefficients of the reactants and products are 1:2:1.

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Answer:

C

Explanation:

Answer: D. Friction between two rock surfaces grinding away minerals.

Explanation: chemical weathering involves erosion from rocks/materials

The total number of Ca2+ ions and F- ions that are present in the cubic unit cell will be 4 and 8 respectively.

From the complete information, the number of Ca2+ at corners will be (8 × 1/8 = 1) and those a the center will be (6 × 1/2 = 3). Therefore, the total number of Ca2+ will be:

= 1 + 3 = 4

The total number of F- ions will be:

= 4 × 2 = 8

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Step 1: Write the balanced "molecular equation:

[tex]2HI(aq) + CrCO₃ (s) → CrI₂ (s) + H₂O(l) + CO₂(g)[/tex]

Step 2: Carbon dioxide and water are written in molecular form. Consult the solubility and net ionic equation rules on the information page to determine which of the other substances will dissociate:

[tex]2HI(aq) + CrCo₃ (s) → Crl₂ (s) +H₂O(l)+CO₂(g)[/tex]

yes no yes

Step 3: Dissociate all soluble salts, strong acids, and strong bases (except calcium hydroxide), Leave together all "not soluble" salts and weak acids or bases:

[tex] {2H₂O}^{ + } (aq) + {2I}^{ - } (aq) + CrCO₃ (s)→ {Cr}^{2 + } (aq) + {2I}^{ - }(aq) + 3H₂O(l) + CO₂(g)[/tex]

Step 4: Cross out "spectator lons that appear on both sides of the reaction (these lons do not participate in the chemistry) and rewrite the "net" reaction using the smallest possible coefficients:

[tex] {2H₂O}^{ + } (aq) + CrCO₃ (s) →3H₂O + CO₂ + {Cr}^{2 + } [/tex]

Answer:

.018 M

Explanation:

Mole weight of NaCl (from periodic chart) = 58.44 g /mole

4 g / 58.44 gm/mole = .06845 mole

.06845 mole / (3.8 liter ) = .018 M

Answer:

a) The molecular formula of oleic acid is: [tex]C_{18}H_{34}O_{2}[/tex]

b) There are 5.74 x 10^21 molecules of oleic acid in a 3.00 mL sample.

Explanation:

a)

First, let's determine the empirical formula of oleic acid from the percentage composition provided by the question:

- Assuming a 100g sample of oleic acid, the correspondent mass of C, H and O would be:

mass of C = 76.54g

mass of H = 12.13g

mass of O = 11.33g

- We must determine the number of moles contained in each of the masses determined above. For that, let's use the atomic masses of C (12.01 u), H (1.007 u) and O (15.99 u) and the following equation:

[tex]number of moles = \frac{mass of sample}{molar mass}[/tex]

number of moles of C = 76.54g / 12.01 g/mol = 6.373 mol C

number of moles of H = 12.13g / 1.007 g/mol = 12.04 mol H

number of moles of O = 11.33g / 15.99 g/mol = 0.7086 mol O

- Next, we'll try to find whole-number ratios by dividing each number of moles obtained by the smallest value obtained (0.7086):

C = 6.373 / 0.7086 = 8.994 ≅ 9

H = 12.04 / 0.7086 = 16.99 ≅ 17

O = 0.7086 / 0.7086 = 1.000

- Now, we can write the empirical formula of oleic acid as: [tex]C_{9}H_{17}O[/tex]

Now that we have the empirical formula for oleic acid (C9H17O), let's check its molar mass and compare it to the experimental value provided by the question (282 g/mol):

molar mass (C9H17O) = 9 * 12.01 + 17 * 1.007 + 1 * 15.99 = 141.9 g/mol

Note that the molar mass obtained from the empirical formula (141.9 g/mol) and the experimental molar mass (282 g/mol) are different, thus we need to find an integer multiple to determine the molecular formula (keep in mind that molecular formulas always present kind and number of atoms of each element present in the molecular compound, while the empirical formula presents the smallest coefficients for each atom).

Molecular formula = n × Empirical formula

This integer multiple, n, can  be obtained by dividing the molar mass,  MM , of the compound by the empirical formula mass,  EFM  (the molar mass represented by the empirical formula):

n = MM / EFM

n = 282 / 141.9

n = 1.987 ≅ 2

Since the integer multiple is 2, we can multiply all coefficients in the empirical formula by 2 to find the molecular formula of oleic acid:

C9 H17 O x 2 = C18 H34 O2

Therefore, the molecular formula of oleic acid is: [tex]C_{18}H_{34}O_{2}[/tex]

b)

We need to calculate the amount of molecules of oleic acid in a 3.00mL sample, considering the density as 0.895 g/mL.

From the density provided, we can calculate the mass, in grams, of oleic acid contained in 3.00 mL of this compound:

1 mL ----------------------- 0.895 g oleic acid

3.00 mL ----------------- x

Solving for x, we have that there are 2.69 g of oleic acid in a 3.00 mL sample.

Next, considering the molar mass of oleic acid (282 g/mol), we can calculate the number of moles contained in 2.685g of the acid:

282 g -------------------- 1 mol oleic acid

2.69 g ----------------- y

Solving for y, we have that there are 0.00953 moles of oleic acid in a 3.00 mL sample.

Now, using the Avogadro's number (6.022 x 10^23 particles/mol), we can calculate the amount of molecules of oleic acid in the sample:

1 mol oleic acid ------------------------------- 6.022 x 10^23 molecules oleic acid

0.00953 mol oleic acid ------------------- z

Solving for z, we have that there are 5.74 x 10^21 molecules of oleic acid in a 3.00 mL sample.

Answer:

The answer is c.........

Answer:

About 4.63.

Explanation:

The reaction between ethanoic acid (CH₃COOH) and sodium hydroxide (NaOH) can be described by the following net ionic equation:

[tex]\displaystyle \text{CH$_3$COOH}_\text{(aq)} + \text{OH}^-_\text{(aq)}\longrightarrow \text{CH$_3$COO}^-_\text{(aq)} + \text{H$_2$O}_\text{($\ell$)}[/tex]

40.0 mL of 0.040 M NaOH contains:
[tex]\displaystyle 40.0\text{ mL} \cdot \frac{0.040\text{ mol NaOH}}{1000\text{ mL}} = 0.0016\text{ mol NaOH}[/tex]

Because NaOH is a strong base, it dissociates completely. Hence, the amount of OH⁻ present is 0.0016 mol. Na⁺ acts as a spectator ion.

50.0 mL of 0.075 M CH₃COOH contains:
[tex]\displaystyle 50.0\text{ mL} \cdot \frac{0.075\text{ mol CH$_3$COOH}}{1000\text{ mL}} = 0.0038\text{ mol CH$_3$COOH}[/tex]

OH⁻ is the limiting reagent of the reaction. Therefore, as the reaction proceeds to completion:

Note that all species present are in a one-to-one ratio.

To find pH, we can use the Henderson-Hasselbalch equation:

[tex]\displaystyle \text{pH} = \text{p}K_a + \log\frac{[\text{Base}]}{[\text{Acid}]}[/tex]

The total volume of the solution is 40.0 mL + 50.0 mL = 90.0 mL.

Hence, find [CH₃COOH]:

[tex]\displaystyle \left[\text{CH$_3$COOH}\right] = \frac{0.0022\text{ mol}}{90.0\text{ mL}} \cdot \frac{1000\text{ mL}}{1\text{ L}} = 0.024\text{ L}[/tex]

Find [CH₃COO⁻]:
[tex]\displaystyle \left[\text{CH$_3$COO}^-\right] = \frac{0.0016\text{ mol}}{90.0\text{ mL}}\cdot \frac{1000\text{ mL}}{1\text{ L}} = 0.018\text{ M}[/tex]

Therefore, the pH of the resulting solution is:
[tex]\displaystyle \begin{aligned}\text{pH} & = \text{p}K_a + \log\frac{[\text{Base}]}{[\text{Acid}]} \\ \\ & = (4.75) + \log \frac{(0.018)}{(0.024)} \\ \\ & = 4.75 + (-0.12) \\ \\ & = 4.63\end{aligned}[/tex]

In conclusion, the pH of the resulting solution is about 4.63.

Here are the balanced equations

Done!

Apply Charles law

Answer:

C

Explanation:

It has the lowest Ksp, so it will have the lowest Molar Solubility

1. Acids have highest concentration of hydronium ions.

2. Bases have highest concentration of Hydroxide ions

3. a. less than 7 b. more than 7

4. Acids (Presence of hydronium ions)

You need Conclusions too?

Answer:

3.8 M

Explanation:

Mole weight of LiCl = 6.94 + 35.45 = 42.39    ( from periodic table)

175 g of this is then   175 / 42.39  = 4.13 moles

4.13 moles / (1085/1000) = 3.8 M

Answer:

All of the above :D

Explanation:

When rain falls, if it doesn't get absorbed by the dirt, it will go into streams, rivers, and ponds. That is actually how flooding is caused!!! :D

Have an amazing day!!

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Answer:

The answer is aluminum foil.

Answer:

Answer:

B + C ARE INCORRECT FROM THE ABOVE GIVEN CONVERSIONS.

Answer: can hydrogen produce as much energy as burning fossil fuels?

Explanation:

Answer:

Option B: Oxygen.

Explanation:

Hope this helped.

For yogurt production, probiotic bacteria and the Streptococcus thermophilus strain work together to produce a stable balance of bacteria in the culture.

The major product of the reaction between  acetic anhydride and CH3CH2CH2OH is CH3COOCH2CH2CH3. While, the major product of the reaction between acetic anhydride and NH3 is CH3CONH2.

Acetic anhydride is the compound (CH3COO)2. It is a precursor for the synthesis of many organic compounds.

Now let us consider the use of acetic anhydride in synthesis. The major product of the reaction between  acetic anhydride and CH3CH2CH2OH is CH3COOCH2CH2CH3. While, the major product of the reaction between acetic anhydride and NH3 is CH3CONH2.

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If a wildfire occurred in the Everglades, the insect which most likely to survive predator is gray-colored speckled moth.

Predators are those animals which for their survival will hunt or kill another animals for the fulfilment of their food need.

If wildfire occured in the everglades region then the insect would be the most likely to survive predators in an ash and soot covered habitat is gray-colored speckled moth which is also known as darwin moth as it takes part in evolution of natural process.

Hnece the required animal is gray-colored speckled moth.

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Answer:The larva might form a cocoon or a chrysalis.

Explanation:

Without conducting an experiment, one can determine if a species in a solution is acidic or basic by using a litmus paper.

Acidic species would turn a blue litmus paper to red but have no effects on red litmus paper.

Basic species would turn a red litmus paper to blues but have no effects on blue litmus paper.

Thus, in order to determine if a species in a solution is acidic or basic without conducting any experiment, dipping a litmus paper into the solution would give an indication.

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