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mario62 [17]
3 years ago
6

if a baloon filled with dry hydrogen weighs 35 gram,but weighs 440 grams when filled with the vapour of an organic compound. cal

culate the vapour density and RMM of the organic compound.​
Chemistry
1 answer:
Elan Coil [88]3 years ago
6 0

Answer:

1) The vapor density of the organic compound is approximately 12.57

2) The relative molar mass (RMM) of the organic compound is approximately 25.14 grams  

Explanation:

1) The mass of the balloon filled with dry hydrogen = 35 grams

The mass of the balloon filled with vapor of an organic compound = 440 grams

The vapor density = (Weight of a given volume of gas)/(Weight of equal volume of hydrogen)

The vapor density of the organic compound = (440)/(35) ≈ 12.57

The vapor density of the organic compound ≈ 12.57

2) The relative molar mass (RMM) = 2 × vapor density

The relative molar mass (RMM) of the organic compound = 2 × vapor density of the organic compound

The relative molar mass (RMM) of the organic compound ≈ 2 × 12.57 ≈ 25.14 grams  

The relative molar mass (RMM) of the organic compound ≈ 25.14 grams  

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Explanation:

1.  2.46\times 10^{21} molecules of CO₂

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One mole of an ideal gas with a volume of 1.0 L and a pressure of 5.0 atm is allowed to expand isothermally into an evacuated bu
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Answer:

w= - 1.7173 kJ, q= 1.7173 kJ, q(rev) = 1717.3 J = 1.7173 kJ.

Explanation:

Okay, from the question we are given the information below;

Number of moles, n= 1 mole; initial volume, v(1) = 1.0 litres (L); pressure (p) = 5atm, final volume(v2) = 2.0 Litres(L) ; the workdone, w= not given; the heat, q and q(rev)= not given and the gas was said to expand isothermally.

So, this question is a question from the part of chemistry known as thermodynamics. Therefore, grip yourself we are delving into thermodynamics 'waters' now.

For expansion isothermally; the workdone, w= -nRT ln v2/v1.

Where T= temperature= 25° C = 298 k and R= gas constant.

Therefore; workdone, w = - 1 × 8.314 × 298 × ln(2/1).

Workdone,w= - 1717.32204643. =

- 1717.3 Joules (J).

==> Workdone,w= - 1.7173 kJ.

Then, we are to find q. q can be solved by using the first law of thermodynamics, which by mathematical representation is:

∆U= q + w. Where ∆U= change in internal enegy. Since the question is dealing with isothermal expansion, there is this rule that says for an isothermal expansion ∆U = 0.

Hence, 0 =q + [- 1717.3 Joules (J)].

q=1717.3 J = 1.7173 kJ.

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q(rev) = 1 × 8.314 × 298 ln (2/1).

q(rev) = 1717.3 J = 1.7173 kJ.

PS: please note the negative signs in the workdone and the positive sign in the q(rev).

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