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DiKsa [7]
3 years ago
5

How many moles of gas were added to a balloon that started with 2.3 moles of gas and a volume of 1.4 l given that the final volu

me was 7.2 l?
a.12
b.0.085
c.4.4
d.9.5
e.none of the above?
Chemistry
1 answer:
Ierofanga [76]3 years ago
6 0
<span>Ideal Gas Law is as follows: PV = nRT Now lets rearrange so we can get rid of some variables. V/n = RT/P Assuming temperature and pressure are the same (atmospheric pressure), we can get: PV/n = constant (k) Now since we are dealing with an initial state (2.3 moles, 1.4 liters) lets set that as V1 and n1 and the final state as V2 and n2 .V1/n1 = constant (k) V2/n2 = constant (k) Because the constant is the same (remember me making the assumptions above??) V1/n1 = V2/n2 now rearrangement: n1/V1/V2 = 1/n2 -1 = n2 Answer: 11.8 moles are in the final vessel. So to answer the question how much was ADDED, we need to subtract the initial mole number. 11.8 - 2.3 mols = 9.5 mols added</span>
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Answer:

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Explanation:

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\pi=iMRT

or,

\pi=i\times \frac{m_{solute}\times 1000}{M_{solute}\times V_{solution}\text{ (in mL)}}}\times RT

where,

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

m_{solute} = mass of protein = 400 mg = 0.4 g   (Conversion factor:  1 g = 1000 mg)

M_{solute} = molar mass of protein = ?

V_{solution} = Volume of solution = 5.00 mL

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Putting values in above equation, we get:

0.0861atm=1\times \frac{0.4g\times 1000}{M\times 100}\times 0.0821\text{ L. atm }mol^{-1}K^{-1}\times 298K\\\\M=1136.62g/mol=1.14\times 10^2g/mol

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4 0
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