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prohojiy [21]
4 years ago
5

In an experiment, an unknown gas effuses at one-half the speed of oxygen gas, which has a molar mass of 32 g/mol. which might be

the unknown gas?
Chemistry
1 answer:
goblinko [34]4 years ago
5 0
If we assume the rate of diffusion of oxygen is 1 , then that of the unknown gas is 1/2.
From the Grahams law of diffusion;
R1/R2= √mm2/√mm1
 1/0.5 = √mm2/√32
   4 = mm2/32
mm2 = 128
Therefore the molecular mass of the unknown gas i 128 g/mol
I therefore think the gas is Hydrogen iodide
since, H=1, I= 127 , thus HI = 128 g/mol
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4 0
2 years ago
A or B i need help fast
Bezzdna [24]

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3 years ago
400. mg of an unknown protein are dissolved in enough solvent to make 5.00 mL of solution. The osmotic pressure of this solution
ch4aika [34]

<u>Answer:</u> The molecular weight of protein is 1.14\times 10^2g/mol

<u>Explanation:</u>

To calculate the concentration of solute, we use the equation for osmotic pressure, which is:

\pi=iMRT

or,

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

where,

\pi = Osmotic pressure of the solution = 0.0861 atm

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

R = Gas constant = 0.0821\text{ L atm }mol^{-1}K^{-1}

T = temperature of the solution = 25^oC=[25+273]K=298K

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

Hence, the molecular weight of protein is 1.14\times 10^2g/mol

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3 years ago
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3 years ago
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