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Leviafan [203]
3 years ago
9

A certain substance X has a normal boiling point of 124.2 °C and a molal boiling point elevation constant K,-06-0C-kg-mol ·A sol

ution is prepared by dissolving some urea ((N112)CO) in 650. g ofl. This solution boils at 124.7 oC, Calculate the mass of urea that was dissolved. Be sure your answer has the correct number of significant digits. 31.48 g
Chemistry
1 answer:
Mashutka [201]3 years ago
8 0

Answer:

32 g

Explanation:

The increase in the boiling point of X can be calculated using the following expression.

ΔT = Kb × b

where,

ΔT: increase in the boiling point

Kb: molal boiling point constant (0.62 °C.kg/mol) (I looked it on the web)

b: molality of the solute

ΔT = Kb × b

(124.7°C - 124.2°C) = (0.62 °C.kg/mol) × b

b = 0.81 mol/kg

The mass of X is 650 g (0.650 kg). Then, the moles of urea (solute) are:

\frac{0.81molUrea}{1kgX} .0.650kgX=0.53molUrea

The molar mass of urea is 60.06 g/mol. The mass of urea is:

0.53mol.\frac{60.06g}{mol} =32g

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Deffense [45]

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.

Finally, the q(rev) which is= nRT ln (v2/V1).

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).

7 0
3 years ago
Someone please walk me through this. I googled way too much this semester and I'm paying for it on this test. Just tell me how t
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Answer:

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A)The spring scale has a high level of precision and a low level of accuracy.

Explanation:

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4 years ago
How many moles of nitrogen, N, are in 61.0 g of nitrous oxide, N2O?
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Answer:

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Explanation:

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