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mash [69]
3 years ago
6

Calculate the volume occupied by 0.845 mole of nitrogen gas, the pressure, and the temperature. Find the volume.

Chemistry
2 answers:
Lyrx [107]3 years ago
4 0

<u>Answer:</u>

<em>The volume occupied by 0.845 mole of nitrogen gas is 15.9 L.</em>

<em></em>

<u>Explanation:</u>

Given

N = 0.845 mol

P = 1.37 atm

T = 315 k

We make use of the Ideal gas equation to solve this problem

Ideal gas law equation is

PV=nRT

Rearranging the formula                  

v= \frac {nRT}{V}

Plugging in the values given we have

v=\frac {(0.845mol \times 0.08206 L atmk^{-1} mol^{-1} \times 315K)}{1.37atm}

v = 15.9 L is the volume occupied by 0.845 mole of nitrogen gas.

NNADVOKAT [17]3 years ago
3 0

Answer:

V = 15.9512 dm³

Explanation:

Given data:

Pressure = P = 1.37 atm

Temperature = T= 315 K

Number of moles of nitrogen= n = 0.845 mol

Volume = V = ?

Formula:

PV = nRT

Now we will put the values in equation.

V = nRT/ P

V = ( 0.845 mol× 0.0821 dm³.atm.K⁻¹.mol⁻¹ × 315 K) / 1.37 atm

V = 21.853 dm³. atm/  1.37 atm

V = 15.9512 dm³

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Where E^{0}(reduction) and E^{0}(oxidation) represents standard reduction potential of reduction half cell and standard reduction potential of oxidation half cell.

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So, E^{0}=E_{Fe^{3+}\mid Fe}^{0}-E_{Mg^{2+}\mid Mg}^{0}=(-0.04+2.38)V=2.34V

Hence this pair will give spontaneous reaction.

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Hence this pair will give non-spontaneous reaction.

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Say I grab a bunch of aluminum foil (9.5g) and ball it up. Aluminum has a specific heat of .9J/g*C. How much would the aluminum’
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Solution:

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Q = m.c. ΔT

Q = amount of heat absorbed or released

m = mass of given substance

c = specific heat capacity of substance

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67 J = 9.5 g × 0.9 j/g.°C × ΔT

67 J = 85.5 j/°C × ΔT

ΔT = 67 J  /  85.5 j/°C

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Explanation:

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