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erastova [34]
4 years ago
14

What is the volume occupied by 0.103 mol of helium gas at a pressure of 0.95 atm and a temperature of 303 K ?

Chemistry
1 answer:
Finger [1]4 years ago
7 0

The pressure of He gas, P = 0.95 atm

Temperature, T = 303 K

Gas constant, R = 0.0821 L.atm.K⁻¹mol⁻¹

Number of moles of He gas, n = 0.103 mol

According to an ideal gas equation, PV = nRT

V = nRT/P

= (0.103 mol) (0.0821 L.atm.K⁻¹mol⁻¹) (303 K) / 0.95 atm

= 2.7 L

Thus, the volume of the He gas is 2.7 L.

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Answer:

C

In an exothermic reeaction, the energy of products is less than the energy of reactants.

Explanation:

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3 years ago
6.50 g of a certain Compound x, known to be made of carbon, hydrogen and perhaps oxygen, and to have a molecular molar mass of 1
gayaneshka [121]

Answer:

C_{10}H_8

Explanation:

Hello there!

In this case, according to the given information, it turns out possible for us to find the molecular formula of the given compound by firstly calculating both moles and grams of carbon in carbon dioxide and hydrogen in water, as the only sources of these elements derived from the compound x due to its combustion:

n_C=22.35gCO_2*\frac{1molCO_2}{44.01gCO_2}*\frac{1molC}{1molCO_2}=0.51molC\\\\m_C=0.51molC*\frac{12.01gC}{1molC}   =6.10gC

n_H=3.66gH_2O*\frac{1molH_2O}{18.02gH_2O}*\frac{2molH}{1molH_2O}=0.41molH\\\\m_H=0.41molH*\frac{1.01gH}{1molH}=0.41g

Now, since the addition of carbon and hydrogen is about 6.50 grams, we infer the compound has no oxygen, that is why we now set the mole ratios in the empirical formula for both C and H as shown below:

C:\frac{0.51mol}{0.41mol}= 1.24\\\\H:\frac{0.51mol}{0.51mol}= 1\\\\C_{1.24}H

Yet it cannot be decimal, that is why we multiply by 4 to get the correct whole-numbered empirical formula:

C_5H_4

Whose molar mass is 64.09 g/mol, which makes the ratio of molar masses:

\frac{128.g/mol}{64.09g/mol} =2

Therefore, the molecular formula is twice the empirical one:

C_{10}H_8

Regards!

8 0
3 years ago
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