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2 years ago

Using the balanced equation below.

Chemistry

1 answer:

exis [7]2 years ago

7 0

Answer:

28.0 grams CsF

Explanation:

To find the mass of CsF, you need to (1) convert grams CsXeF₇ to moles CsXeF₇ (via molar mass), then (2) convert moles CsXeF₇ to moles CsF (via mole-to-mole ratio from equation coefficients), and then (3) convert moles CsF to grams CsF (via molar mass). It is important to arrange the ratios/conversions in a way that allows for the cancellation of units. The final answer should have 3 sig figs to match the sig figs of the given value.

Molar Mass (CsF): 132.91 g/mol + 18.998 g/mol

Molar Mass (CsF): 151.908 g/mol

Molar Mass (CsXeF₇): 132.91 g/mol + 131.29 g/mol + 7(18.998 g/mol)

Molar Mass (CsXeF₇): 397.186 g/mol

1 CsF + XeF₆ -----> 1 CsXeF₇

73.1 g CsXeF₇ 1 mole 1 mole CsF 151.908 g
---------------------- x -------------------- x ------------------------ x ------------------ =
397.186 g 1 mole CsXeF₇ 1 mole

= 27.958 g CsF

= 28.0 g CsF

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How many molecules are in 41.8 g of sulfuric acid

Anton [14]

Answer

× 10²³ molecules are in 41.8 g of sulfuric acid

Explanation

The first step is to convert 41.8 g of sulfuric acid to moles by dividing the mass of sulfuric acid by its molar mass.

Molar mass of sulfuric acid, H₂SO₄ = 98.079 g/mol

$Mole=\frac{Mass}{Molar\text{ }mass}=\frac{41.8\text{ }g}{98.079\text{ }g\text{/}mol}=0.426187053\text{ }mol$

Finally, convert the moles of sulfuric acid to molecules using Avogadro's number.

Conversion factor: 1 mole of any substance = 6.022 × 10²³ molecules.

Therefore, 0.426187053 moles of sulfuric acid is equal

$\frac{0.426187053\text{ }mol}{1\text{ }mol}\times6.022×10²³\text{ }molecules=2.57\times10^{23}\text{ }molecules$

Thus, 2.57 × 10²³ molecules are in 41.8 g of sulfuric acid.

3 0

1 year ago

The vapor pressure of diethyl ether (ether) is 463.57 mm Hg at 25 °C. A nonvolatile, nonelectrolyte that dissolves in diethyl et

Alexxx [7]

Answer: The vapor pressure of solution is 459.17 mmHg

Explanation:

To calculate the number of moles, we use the equation:

$\text{Number of moles}=\frac{\text{Given mass}}{\text{Molar mass}}$ .....(1)

For testosterone:

Given mass of testosterone = 7.752 g

Molar mass of testosterone = 288.4 g/mol

Putting values in equation 1, we get:

$\text{Moles of testosterone}=\frac{7.752g}{288.4g/mol}=0.027mol$

For diethyl ether:

Given mass of diethyl ether = 208.0 g

Molar mass of diethyl ether = 74.12 g/mol

Putting values in equation 1, we get:

$\text{Moles of diethyl ether}=\frac{208.0g}{74.12g/mol}=2.81mol$

Mole fraction of a substance is calculated by using the equation:

$\chi_A=\frac{n_A}{n_A+n_B}$

$\chi_{\text{testosterone}}=\frac{n_{\text{testosterone}}}{n_{\text{testosterone}}+n_{\text{diethyl ether}}}$

$\chi_{\text{testosterone}}=\frac{0.027}{0.027+2.81}\\\\\chi_{\text{testosterone}}=0.0095$

The formula for relative lowering of vapor pressure will be:

$\frac{p^o-p_s}{p^o}=i\times \chi_{\text{solute}}$

where,

$p^o$ = vapor pressure of solvent (diethyl ether) = 463.57 mmHg

$p^s$ = vapor pressure of the solution = ?

i = Van't Hoff factor = 1 (for non electrolytes)

$\chi_{\text{solute}}$ = mole fraction of solute (testosterone) = 0.0095

Putting values in above equation, we get:

$\frac{463.57-p^s}{463.57}=1\times 0.0095\\\\p^s=459.17mmHg$

Hence, the vapor pressure of solution is 459.17 mmHg

7 0

4 years ago

How to determine the relative rates of reactant disappearance and product appearance using reaction stoichiometry

Lady_Fox [76]

Answer:

For example, the relative rate of a reaction at 20 seconds will be 1/20 or 0.05 s -1, while the average rate of reaction over the first 20 seconds will be the change in mass over that period, divided by the change in time.

Explanation:

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6 0

3 years ago

What does changing densities cause to occur

Dafna1 [17]

Answer:

Explanation:

Please Elaborate

8 0

3 years ago

How many moles in .5g of sodium bromide?

Genrish500 [490]

I'm not sure how many sign fig's you are required to have.
However I think the final answer would be 0.05 Moles, because of the .5g, that is considered 1 sign fig.

6 0

3 years ago