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elixir [45]
3 years ago
10

In a mixture of helium and chlorine, occupying a volume of 14.6 l at 871.7 mmhg and 28.6oc, it is found that the partial pressur

e of chlorine is 355 mmhg. what is the total mass of the sample?
Chemistry
1 answer:
Inga [223]3 years ago
3 0
First, let's find the total moles.

PV=nRT
(871.7 mmHg)(1 atm/760 mmHg)(14.6 L) = n(0.0821 L-atm/mol-K)(28.6 +273)
Solving for n,
n = 0.676 moles

Assuming ideal gas behavior, the pressure fraction is also equal to mole fraction.

Mole fraction of Chlorine = 355/871.7 = 0.407
Mole fraction of Helium = 1 - 0.407 = 0.593

Knowing Chlorine to be 35.45g/mol and Helium to be 4 g/mol.
Mass of Chlorine = (0.407)(0.676)(35.45 g/mol) = 9.75 g
Mass of Helium = (0.593)(0.676)(4 g/mol) = 1.6 g
Total Mass = 9.75+1.6 = <em>11.35 g</em>
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Which is a characteristic of nuclear fission
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3 years ago
Read 2 more answers
The change in entropy is related to the change in the number of moles of gas molecules. Determine the change in moles of gas for
tester [92]

The given question is incomplete. The complete question is:

The change in entropy is related to the change in the number of moles of gas molecules. Determine the change in moles of gas for each of the reactions and decide if the entropy increases decreases or has little to no change:

A. K(s)+O_2(g)\rightarrow KO_2(s)

B. CO(g)+3H_2(g)\rightarrow CH_4(g)+H_2O(g)

C. CH_4(g)+2O_2(g)\rightarrow CO_2(g)+2H_2O(g)

D. N_2O_2(g)\rightarrow 2NO(g)+O_2(g)

Answer: A. K(s)+O_2(g)\rightarrow KO_2(s) : decreases

B. CO(g)+3H_2(g)\rightarrow CH_4(g)+H_2(g) : decreases

C. CH_4(g)+2O_2(g)\rightarrow CO_2(g)+2H_2O(g): no change

D. N_2O_2(g)\rightarrow 2NO(g)+O_2(g) : increases

Explanation:

Entropy is defined as the randomness of the system.

Entropy is said to increase when the randomness of the system increase, is said to decrease when the randomness of the system decrease and is said to have no change when the randomness remains same.

In reaction K(s)+O_2(g)\rightarrow KO_2(s), as gaseous reactant is changed to solid product, entropy decreases.

In reaction CO(g)+3H_2(g)\rightarrow CH_4(g)+H_2O(g), as 4 moles of gaseous reactants is changed to 2 moles of gaseous product, entropy decreases.

In reaction CH_4(g)+2O_2(g)\rightarrow CO_2(g)+2H_2O(g), as 3 moles of gaseous reactants is changed to 3 moles of gaseous product, entropy has no change.

In reaction  N_2O_2(g)\rightarrow 2NO(g)+O_2(g) , as 1 mole of gaseous reactant is changed to 3 moles of gaseous product, entropy increases.

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3 years ago
Calculate how many grams of the product form when 16.7 g of calcium metal completely reacts. Assume that there is more than enou
swat32

39.96 g product form when 16.7 g of calcium metal completely reacts.

<h3>What is the stoichiometric process?</h3>

Stoichiometry is a section of chemistry that involves using relationships between reactants and/or products in a chemical reaction to determine desired quantitative data.

Equation:

Ca(s) + Cl_2(g) → CaCl_2(s)

In this case, for the undergoing reaction, we can compute the grams of the formed calcium chloride by noticing the 1:1 molar ratio between calcium and it (stoichiometric coefficients) and using their molar mass of 40 g/mol and 111 g/mol by using the following stoichiometric process:

m_{ca_C_l_2}= 16.7 g Ca x \frac{1 mol \;of \;Ca}{40g Ca} x \frac{1 mol \;of \;CaCl_2}{1 mol \;Ca} x \frac{111g of \;CaCl_2}{1 mol \;CaCl_2}

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Hence, 39.96 g product form when 16.7 g of calcium metal completely reacts.

Learn more about the stoichiometric process here:

brainly.com/question/15047541

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