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Svetllana [295]
3 years ago
12

Ethane, C2H6, can be formed by reacting acetylene, C2H2, with hydrogen gas as follows: C2H2(g) H2(g) ⇌ C2H6(g) Exothermic What c

hange will be observed if the temperature of the reaction mixture at equilibrium were increased
Chemistry
1 answer:
Pavel [41]3 years ago
8 0

Answer:

Equilibrium shifts to the right

Explanation:

An exothermic reaction is one in which temperature is released to the environment. Hence, if the reaction vessel housing an exothermic reaction is touched after reaction completion, we will notice that the reaction vessel e.g beaker is hot.

To consider the equilibrium response to temperature changes, we need to consider if the reaction is exothermic or endothermic. In the case of this particular question, it has been established that the reaction is exothermic.

Heat is released to the surroundings as the reactants are at a higher energy level compared to the products. Hence, increasing the temperature will favor the formation of more reactants and as such, the equilibrium position will shift to the left to pave way for the formation of more reactants. Thus , more acetylene and hydrogen would be yielded

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A 0.216 g sample of an aluminium compound X reacts with an excess of water to produce a single hydrocarbon gas. This gas burns c
makkiz [27]
0.216g of aluminium compound X  react with an excess of water water to produce gas. this gas burn completely  in O2  to form H2O and 108cm^3of CO2 only . the volume of CO2 was measured at room temperature and pressure

0.108 / n  =  24 / 1 
n = 0.0045 mole ( CO2 >>0.0045 mole 
0.216 - 0.0045 = 0.2115
so Al =   0.2115 / 27  =>  0.0078 mole 
C = 0.0045 * 1000 => 4.5    and Al  = 0.0078 * 1000 = 7.8 

7 0
3 years ago
How many grams of oxygen gas are there in a 2.3L tank at 7.5 atm and 24° C?
jolli1 [7]

Answer:

22.656 grams of oxygen gas are there in a 2.3L tank at 7.5 atm and 24° C

Explanation:

An ideal gas is characterized by three state variables: absolute pressure (P), volume (V), and absolute temperature (T). The relationship between them constitutes the ideal gas law:

P * V = n * R * T

where R is the molar constant of the gases and n the number of moles.

In this case you know:

  • P= 7.5 atm
  • V= 2.3 L
  • n= ?
  • R= 0.082 \frac{atm*L}{mol*K}
  • T= 24 °C= 297 °K (being 0°C=273°K)

Replacing:

7.5 atm* 2.3 L=n*0.082 \frac{atm*L}{mol*K} *297K

Solving:

n=\frac{7.5 atm* 2.3 L}{0.082 \frac{atm*L}{mol*K} *297K}

n=0.708 moles

Knowing that oxygen gas is a diatomic gas of molecular form O₂ and its mass is 32 g / mole, you can apply the following rule of three: if 1 mole contains 32 grams, 0.708 moles, how much mass will it have?

mass=\frac{0.708 moles*32 grams}{1mole}

mass= 22.656 grams

<u><em>22.656 grams of oxygen gas are there in a 2.3L tank at 7.5 atm and 24° C</em></u>

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