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Goshia [24]
3 years ago
13

Calculate the expansion work done on the system when exactly 1 mol of solid ammonium chloride, nh4cl, decomposes completely to y

ield gaseous ammonia, nh3 and hydrogen chloride, hcl at a temperature of 1250 k. treat the expansion as irreversible and the gases formed as perfect.
Chemistry
1 answer:
maria [59]3 years ago
8 0

The expansion work on an irreversible system is equal to:

W = - PV

 

Assuming that the gas is ideal, therefore:

PV = nRT

 

[P=pressure, V=volume, n=number of moles, R=gas constant, T=absolute temperature]

 

Therefore,

W = -nRT

W = -(1 mol) * (8.314 J / mol K) * (1250 K)

W = -10,392.50 J

 

Therefore about 10,392.50 J or 10.4 kJ of work is done on the system.

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Explanation:

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2 years ago
How many grams of phosphorus are in a sample of Ca3(PO4)2 that contains 50.0 g of calcium? Answer: ________ grams of phosphorus.
wel

<u>Answer:</u> The mass of phosphorus that is present for given amount of calcium is 28.53 g.

<u>Explanation:</u>

We are given:

Mass of calcium = 50 grams

The chemical formula of calcium phosphate is Ca_3(PO_4)_2

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So, 50 grams of calcium will combine with = \frac{62}{120}\times 50=28.53g of phosphorus.

Hence, the mass of phosphorus that is present for given amount of calcium is 28.53 g.

6 0
3 years ago
¿Cuál de las siguientes configuraciones globales corresponde a un elemento químico que se comporta como metal? 1 punto [He]2s2 2
yuradex [85]

Answer:

[Ne]3s2

Explanation:

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para la primera configuración, ns2 np6 corresponde a un gas noble.

para la segunda configuración ns2 np3 corresponde a un elemento no metálico del grupo 5.

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5 0
3 years ago
How much heat is absorbed when a 298.3 g piece of brass goes from 30.0 to 150
igor_vitrenko [27]

Answer:

Q = 1360.248 j

Explanation:

Given data:

Mass of brass = 298.3 g

Initial temperature = 30.0°C

Final temperature = 150°C

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Heat absorbed = ?

SOLUTION:

Formula:

Q = m.c. ΔT

Q = amount of heat absorbed or released

m = mass of given substance

c = specific heat capacity of substance

ΔT = change in temperature

ΔT = 150°C - 30.0°C  

ΔT = 120°C

Q = 298.3 g × 0.038 J/g.°C × 120°C

Q = 1360.248 j

3 0
3 years ago
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