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3 years ago

Convert the following to Kelvin 12) 0° C 13) -50° C 14) 90° C 15) -20° C

Chemistry

2 answers:

Gekata [30.6K]3 years ago

8 0

Answer:

12)273.15 kelvin

13)223.15 kelvin

14)363.15 kelvin

15)253.15 kelvin

Explanation:

thank u!!!

densk [106]3 years ago

5 0

Answer:

12) 273.15

13) 223.15

14) 363.15

15) 253.15

Explanation:

Formula:

_C+273.15=_K

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3 years ago

Two gases, A and B, are at equilibrium in a sealed cylinder. Individually, gas A is colorless, while gas B is dark colored. The

Dmitriy789 [7]

Answer:

(a) color

(b) endothermic

(c)

b The ΔH value would have the same magnitude value but opposite sign.

c The K expression would be inverted.

Explanation:

Let's consider the following reaction at equilibrium.

A(g) ⇄ 2 B(g)

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(a) The cylinder should appear (color or colorless)

At equilibrium, there is a mixture of A and B, so the cylinder should appear colored.

(b) When the system is cooled, the cylinder's appearance becomes very light colored. Therefore, the reaction must be (endothermic or exothermic)

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Suppose the reaction equation were written as follows: 2 B(g) ⇄ A(g)

(c) Which of these statements would then be true?

a The value of K would not change. FALSE. The new K would be the inverse of the direct K.

b The ΔH value would have the same magnitude value but opposite sign. TRUE. This is stated by Lavoisier-Laplace Law.

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3 years ago

what is the minimum mass of ethylene glycol that must be dissolved in 14.5 kg of water to prevent the solution from freexing at

lbvjy [14]

Answer:

The minimum mass of ethylene glycol = 6.641 Kg

Explanation:

$\Delta T_f= T_f-T_f'\\T_f'=T_f-\Delta T_f$

Where T_f = freezing point of pure solvent water, 0°C

T_f'= Freezing point of solvent after mixture

K_f = Freezing point depression constant = 1.86 °C/m

Moecular weight of ethylene glycol = 60 g/mol

Weight of ethylene glycol = 14.5 Kg= 14.5×10^3 g

molality of ethylene glycol

$m = \frac{weight}{mol.wt} \times\frac{1000}{V}$

Substitute the values to calculate m

$m = \frac{w}{60} \times\frac{1000}{14.5\times1000}$

by formula

$0-(-14.2) =1.86\frac{w}{60} \times\frac{1000}{14.5\times1000}$

calculating we get w = 6641.93 g

Therefore, The minimum mass of ethylene glycol = 6.641 Kg

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3 years ago