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velikii [3]
3 years ago
5

It takes 167 s for an unknown gas to effuse through a porous wall and 99 s for the same volume of n2 gas to effuse at the same t

emperature and pressure. what is the molar mass of the unknown gas?
Chemistry
1 answer:
irakobra [83]3 years ago
3 0

Answer : The molar mass of the unknown gas will be 79.7 g/mol


Explanation : To solve this question we can use graham's law;


Now we can use nitrogen as the gas number 2, which travels faster than gas 1;

So, 167 / 99 = 1.687 So the nitrogen gas is 1.687 times faster that the unknown gas 1

We can compare the rates of both the gases;


So here, Rate of gas 2 / Rate of gas 1 = \sqrt{(molar mass 1 / molar mass 2)}

Now, 1.687 = square root [\sqrt{(molar mass 1) / (28.01 g/mol N_{2})} ]


When we square both the sides we get;


2.845 = (molar mass 1) / (28.01 g/mol N2)


On rearranging, we get,


2.845 X (28.01 g/mol N2) = Molar mass 1

So the molar mass of unknown gas will be = 79.7 g/mol

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Consider the cell pt | h2 (1 atm); h+ (? m) || hg2cl2(s); cl− (1 m) | hg 2 h+ + 2 e − → h2 e 0 = 0.00 v hg2cl2 + 2 e − → 2 hg +
Dmitry [639]

Answer:

1.39

Explanation:

[Hg2Cl2]= 1M

[H^+] = ????

E°cell= 0.35V

E= 0.268 V

Therefore E for the reaction must -0.082 V

n= 2 moles of electrons

From Nernst Equation:

E= E°cell- 0.0592/n log [Red]/[Ox]

0.0268= 0.35- 0.0592/2 log 1/[Ox]^2

-0.082= -0.0296 log 1/[Ox]^2

log 1/[Ox]^2= 0.082/0.0296

log 1/[Ox]^2= 2.77

1/[Ox]^2=Antilog (2.77)

[Ox]^2=1.698×10^-3

[Ox] = 0.0412 M

But pH= -log [H^+]= -log(0.0412)= 1.385

3 0
3 years ago
Read 2 more answers
How many times do u have to run back and forth across a football field to be meter
Taya2010 [7]
You would have to run 100 kilometers across it once. Then you will have to run 10 meters across it because a football field is 110 meters according to the NFL’s website
7 0
2 years ago
A tetraphenyl phosphonium chloride (TPPCl) powder (FW=342.39) is 94.0 percent pure. How many grams are needed to prepare 0.45 L
slega [8]

Answer:

5.41 g

Explanation:

Considering:

Molarity=\frac{Moles\ of\ solute}{Volume\ of\ the\ solution}

Or,

Moles =Molarity \times {Volume\ of\ the\ solution}

Given :

For tetraphenyl phosphonium chloride :

Molarity = 33.0 mM = 0.033 M (As, 1 mM = 0.001 M)

Volume = 0.45 L

Thus, moles of tetraphenyl phosphonium chloride :

Moles=0.033 \times {0.45}\ moles

Moles of TPPCl = 0.01485 moles

Molar mass of TPPCl = 342.39 g/mol

The formula for the calculation of moles is shown below:

moles = \frac{Mass\ taken}{Molar\ mass}

Thus,

0.01485\ g= \frac{Mass}{342.39\ g/mol}

Mass of TPPCl = 5.0845 g

Also,

TPPCl is 94.0 % pure.

It means that 94.0 g is present in 100 g of powder

5.0845 g is present in 5.41 g of the powder.

<u>Answer -  5.41 g</u>

5 0
3 years ago
How many grams of dry nh4cl need to be added to 2.50 l of a 0.800 m solution of ammonia, nh3, to prepare a buffer solution that
tia_tia [17]

Answer:

The correct answer is 574.59 grams.

Explanation:

Based on the given information, the number of moles of NH₃ will be,  

= 2.50 L × 0.800 mol/L

= 2 mol

The given pH of a buffer is 8.53

pH + pOH = 14.00

pOH = 14.00 - pH

pOH = 14.00 - 8.53

pOH = 5.47

The Kb of ammonia given is 1.8 * 10^-5. Now pKb = -logKb,  

= -log (1.8 ×10⁻⁵)

= 5.00 - log 1.8

= 5.00 - 0.26

= 4.74

Based on Henderson equation:  

pOH = pKb + log ([salt]/[base])

pOH = pKb + [NH₄⁺]/[NH₃]

5.47 = 4.74 + log ([NH₄⁺]/[NH₃])

log([NH₄⁺]/[NH₃]) = 5.47-4.74 = 0.73

[NH₄⁺]/[NH₃] = 10^0.73= 5.37

[NH₄⁺ = 5.37 × 2 mol = 10.74 mol

Now the mass of dry ammonium chloride required is,  

mass of NH₄Cl = 10.74 mol × 53.5 g/mol

= 574.59 grams.  

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3 years ago
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The alcohol on board is cognac 2
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