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Tatiana [17]
3 years ago
8

What volume in liters of carbon monoxide will be required to produce 18.9 L of nitrogen in the reaction below

Chemistry
1 answer:
mina [271]3 years ago
8 0

Answer:

37.8 L OF CARBON MONOXIDE IS REQUIRED TO PRODUCE 18.9 L OF NITROGEN.

Explanation:

Equation for the reaction:

2 CO + 2 NO ------> N2 + 2 CO2

2 moles of carbon monoxide reacts with 2 moles of NO to form 1 mole of nitrogen

At standard temperature and pressure, 1 mole of a gas contains 22.4 dm3 volume.

So therefore, we can say:

2 * 22.4 L of CO produces  22.4 L of N2

44.8 L of CO produces 22.4 L of N2

Since, 18.9 L of Nitrogen is produced, the volume of CO needed is:

44.8 L of CO = 22.4 L of N

x L = 18.9 L

x L = 18.9 * 44.8 / 22.4

x L = 18.9 * 2

x = 37.8 L

The volume of Carbon monoxide required to produce 18.9 L of N2 is 37.8 L

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lapo4ka [179]

Answer:

0.1988 J/g°C

Explanation:

-Qmetal = Qwater

Q = mc∆T

Where;

Q = amount of heat

m = mass of substance

c = specific heat of substance

∆T = change in temperature

Hence;

-{mc∆T} of metal = {mc∆T} of water

From the information provided in this question, For water; m= 22.0g, ∆T = (24°C-19°C), c = 4.18J/g°C.

For metal; m= 34.0g, ∆T = (24°C-92°C), c = ?

Note that, the final temperature of water and the metal = 24°C

-{34 × c × (24°C-92°C)} = 22 × 4.18 × (24°C-19°C)

-{34 × c × (-68°C)} = 459.8

-{34 × c × -68} = 459.8

-{-2312c} = 459.8

+2312c = 459.8

c = 459.8/2312

c = 0.1988

The specific heat capacity of the metal is 0.1988 J/g°C

6 0
2 years ago
Would you expect hydrogen chloride to be a gas, a liquid or a solid, at room temperature and pressure? Explain your answer
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nasty-shy [4]

<u>Answer:</u> The total pressure inside the container is 77.9 kPa

<u>Explanation:</u>

Dalton's law of partial pressure states that the total pressure of the system is equal to the sum of partial pressure of each component present in it.

To calculate the total pressure inside the container, we use the law given by Dalton, which is:

P_T=p_{N_2}+p_{O_2}+p_{Ar}

We are given:

Vapor pressure of oxygen gas, p_{O_2} = 40.9 kPa

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Putting values in above equation, we get:

p_T=23.3+40.9+13.7\\\\p_{T}=77.9kPa

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