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3 years ago

A sample of nitrogen gas expands in volume from 16 L to 5.4 L at constant

Chemistry

1 answer:

xxTIMURxx [149]3 years ago

6 0

a. W = 0 J

b. W = - 308.028 J

<h3>Further explanation</h3>

Given

Nitrogen gas expands in volume from 1.6 L to 5.4 L

Required

The work done

Solution

Isothermal :

W = -P . ΔV

Input the value :

a. At a vacuum, P = 0

So W = 0

b. At pressure = 0.8 atm

W = - 0.8 x ( 5.4 - 1.6)

W = -3.04 L.atm ( 1 L.atm = 101.325 J)

W = - 3.04 x 101.325

W = - 308.028 J

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Answer:

$4.86\times10^{-7}\ \text{m}$

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$\lambda^{-1}=R\left(\dfrac{1}{n_1^2}-\dfrac{1}{n_2^2}\right)\\\Rightarrow \lambda^{-1}=1.09677583\times 10^7\left(\dfrac{1}{2^2}-\dfrac{1}{4^2}\right)\\\Rightarrow \lambda=\left(1.09677583\times 10^7\left(\dfrac{1}{2^2}-\dfrac{1}{4^2}\right)\right)^{-1}\\\Rightarrow \lambda=4.86\times10^{-7}\ \text{m}$

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3 years ago

How many mL of a .132 M aqueous solution of sodium chloride, NaCl, must be taken to obtain 3.59 grams of the salt?

Sever21 [200]

Answer:

465mL

Explanation:

Volume of a solution, V =Mass of substance, m/(Molarity of the solution of the substance, M × molar mass of the substance, M.m)

Given in the question,

M=.132M

M.m=23+35.5 = 58.5g/mol

m=3.59g

V= 3.59/(.132×58.5)

V = 0.465L

Volume in mL = volume in L × 1000

= 0.465 × 1000 = 465mL

Therefore, 465mL of a .132M aqueous solution of sodium chloride, NaCl, must be taken to obtain 3.59 grams of the salt

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3 years ago