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aleksley [76]
3 years ago
13

In the reaction below, 3.46 atm each of H2 and Cl2 were placed into a 1.00 L flask and allowed to react: H2(g) + Cl2(g) <=&gt

; 2 HCl(g) Given that Kc = 50.1, calculate the equilibrium pressure of HCl. Enter your answer to 2 decimal places.
Chemistry
1 answer:
Novosadov [1.4K]3 years ago
6 0

Answer:

The equilibrium pressure of HCl is 5.40 atm

Explanation:

Kc = [HCl]^2/[H2][Cl2]

Initial pressure of H2 = 3.46 atm and Cl2 = 3.46 atm

Let the equilibrium pressure of HCl be y

From the equation of reaction, mole ratio of H2 to HCl is 1:2, equilibrium pressure of H2 is (3.46 - 0.5y)

Also, from the y of reaction, mole ratio of Cl2 to HCl is 1:2, equilibrium pressure of Cl2 is (3.46 - 0.5y)

Kc = 50.1

50.1 = y^2/(3.46 -y)(3.46 - y)

50.1(11.9716 - 3.46y + 0.25y^2) = y^2

12.525y^2 - 173.346y + 599.77716 = y^2

11.525y^2 - 173.346y + 599.77716 = 0

y^2 - 15.04y + 52.04 = 0

The value of y is obtained using the quadratic formula

y = [15.04 - (sqrt (-15.04^2 - 4×1×52.04))] ÷ 2(1) = (15.04 - 4.25) ÷ 2 = 10.79 ÷ 2 = 5.40 atm (to 2 decimal places)

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8 0
3 years ago
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Compare the boiling points of alcohol, ethers and alkanes and explain one reason for the difference
scoundrel [369]

Alcohols have higher boiling points than do ethers and alkanes of similar molar masses.

<h3>What is meant by Boiling Point ?</h3>

The temperature at which the vapor pressure of a liquid equals the pressure surrounding the liquid is called the boiling point of the substance.

Approximately boiling point of Ethanol: 78.4°C, ether: 34.6°C, ethanes: 68°C

The reason for the difference is that it takes more energy to separate alcohol molecules then it does to separate alkane molecules and ether molecules.

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Therefore Boiling point are in the order ;

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8 0
2 years ago
The decomposition of copper(II) nitrate on heating is endothermic reaction. 2Cu(NO3)2(s) → 2C10(s) + 4NO2(g) + O2(g) Calculate t
Basile [38]

Answer:

The enthalpy change for the given reaction is 424 kJ.

Explanation:

2Cu(NO_3)_2(s)\rightarrow 2CuO(s) + 4NO_2(g) + O_2(g),\Delta H_{rxn}=?

We have :

Enthalpy changes of formation of following s:

\Delta H_{f,Cu(NO_3)_2}=-302.9 kJ/mol

\Delta H_{f,CuO}=-157.3 kJ/mol

\Delta H_{f,NO_2}= 33.2 kJ/mol

\Delta H_{f,O_2}= 0 kJ/mol (standard state)

\Delta H_{rxn}=\sum [\Delta H_f(product)]-\sum [\Delta H_f(reactant)]

The equation for the enthalpy change of the given reaction is:

\Delta H_{rxn} =

=(2 mol\times \Delta H_{f,CuO}+4\times \Delta H_{f,NO_2}+1 mol\times \Delta H_{f,O_2})-(2mol\times \Delta H_{f,Cu(NO_3)_2})

\Delta H_{rxn}=

(2mol\times (-157.3 kJ/mol)+4\times 33.2 kJ/mol=1 mol\times 0 kJ/mol)-(2 mol\times (-302.9 kJ/mol)

\Delta H_{rxn}=424 kJ

The enthalpy change for the given reaction is 424 kJ.

6 0
3 years ago
How many calories of heat are required to raise the temperature of 1.00kg of water from 10.2 degrees Celsius to 26.8 degrees Cel
Vika [28.1K]

Answer:

Explanation:

we know that specific heat is the amount of heat required to raise the  temperature of substance by one degree mathmeticaly

Q=mcΔT

ΔT=T2-T1

ΔT=26.8-10.2=16.6

C for water is 4.184

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Q=1.00*4.184*16.6

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now we have to covert joule into calorie

1 calorie =4.2 j

x calorie=69.4 j/2

so 69.4 j =34.7 calorie thats why 34.7 calorie heat is required to raise the temperature of water from 10.2 to 26.8 degree celsius

6 0
3 years ago
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LenKa [72]

Answer:

Explanation:

Let the number of moles of oxygen = x

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Since the balance number for oxygen is 1 and the balance number for water is 2, you must set up a proportion. (Those balance numbers represent the number of moles).

1/x = 2 / 13.3                      Cross Multiply

2*x = 13.3                          Divide both sides by 2

2x/2 = 13.3/2

x = 6.65

You need 6.65 moles of oxygen.

8 0
2 years ago
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