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sasho [114]
3 years ago
14

If 42 grams of carbon in 52 grams of oxygen are used how many grams of CO2 will be produced. I need to show work.

Chemistry
1 answer:
UNO [17]3 years ago
4 0

Answer:

How many moles of oxygen gas are required to make 8.33 moles of carbon dioxide? ... be used to produce 1.99 grams of water. 1.99 mg H2O X. 1mol H2O. 18.0g X ... c. If the reaction produces 5.3 mg of carbon dioxide how many grams of water ... X. 25mol O2. 2mol C8H18. X. 32.0g O2. 1mol O2. = 4.80 x 103g O2. Answer ...

Explanation:

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Define Mole Concept give 3 Significance of Mole concept​
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The mole is important because it allows chemist to work with a subatomic world with macro world units and amount. Atoms molecules and formula units are very small and very difficult to work with usually. However the mole allows a chemist to work with amount large enough to use.

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Identify the state of matter for A, B, and C.
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A) particles are close together in random positions with about equal kinetic energy and intermolecular forces.

These points are about liquid state.

B) particles are close together in fixed positions with low kinetic energy

These points satisfy the qualities of Solid state

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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]

<u>Answer:</u> The vapor pressure of solution is 459.17 mmHg

<u>Explanation:</u>

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

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

  • <u>For testosterone:</u>

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

  • <u>For diethyl ether:</u>

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
3 years ago
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