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Veronika [31]
2 years ago
10

A sealed 1.0L flask is filled with 0.500 mols of I_2 and 0.500 mols of Br_2. When the container achieves equilibrium the equilib

rium constant is 1.10x10^{-2}. What is the equilibrium concentration of the product, IBr?
Chemistry
1 answer:
DochEvi [55]2 years ago
7 0

Answer:

[IBr] = 0.049 M.

Explanation:

Hello there!

In this case, according to the balanced chemical reaction:

I_2+Br_2\rightarrow 2IBr

It is possible to set up the following equilibrium expression:

K=\frac{[IBr]^2}{[I_2][Br_2]} =0.0110

Whereas the the initial concentrations of both iodine and bromine are 0.50 M; and in terms of x (reaction extent) would be:

0.0110=\frac{(2x)^2}{(0.50-x)^2}

Which can be solved for x to obtain two possible results:

x_1=-0.0277M\\\\x_2=0.0245M

Whereas the correct result is 0.0245 M since negative results does not make any sense. Thus, the concentration of the product turns out:

[IBr]=2x=2*0.0249M=0.049M

Regards!

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7 0
3 years ago
4). One mole of monoclinic sulfur at 25C was placed in a constant-pressure calorimeter whose heat capacity (C) was 1620 J/K. T
andre [41]

<u>Answer:</u> The enthalpy change of the reaction is -243 J/mol

<u>Explanation:</u>

The heat released by the reaction is absorbed by the calorimeter and the solution.

The chemical equation used to calculate the heat released follows:

q=c\times \Delta T

where,

c = heat capacity of calorimeter = 1620 J/K

\Delta T = change in temperature = 0.150^oC=0.150K   (Change remains same)

Putting values in above equation, we get:

q=1620J/K\times 0.15K=243J

<u>Sign convention of heat:</u>

When heat is absorbed, the sign of heat is taken to be positive and when heat is released, the sign of heat is taken to be negative.

For the given chemical reaction:

S\text{ (monoclinic)}\rightarrow S\text{ (orthorhombic)}

We are given:

Moles of monoclinic sulfur = 1 mole

  • To calculate the enthalpy change of the reaction, we use the equation:

\Delta H_{rxn}=\frac{q}{n}

where,  

q = amount of heat released = -243 J

n = number of moles = 1 mole

\Delta H_{rxn} = enthalpy change of the reaction

Putting values in above equation, we get:

\Delta H_{rxn}=\frac{-243J}{1mol}=-243J/mol

Hence, the enthalpy change of the reaction is -243 J/mol

8 0
3 years ago
What is the wavelength of a wave having a frequency of 3.76x10^14
choli [55]

Answer:

3.76 x 1014 s-1? λ = c/ν = 3.00 x 108 m/s = 7.98 x 10-7 m 3.76 x 1014 s-1.

Explanation:

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Over the years, the thermite reaction has been used for welding railroad rails, in incendiary bombs, and to ignite solid-fuel ro
melamori03 [73]

<u>Answer:</u> The mass of nickel (II) oxide and aluminium that must be used is 18.8 g and 4.54 g respectively.

<u>Explanation:</u>

To calculate the number of moles, we use the equation:

\text{Number of moles}=\frac{\text{Given mass}}{\text{Molar mass}}      .....(1)

  • <u>For nickel:</u>

Given mass of nickel = 14.8 g

Molar mass of nickel = 58.7 g/mol

Putting values in equation 1, we get:

\text{Moles of nickel}=\frac{14.8g}{58.7g/mol}=0.252mol

For the given chemical reaction:

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  • <u>For nickel (II) oxide:</u>

By Stoichiometry of the reaction:

3 moles of nickel are produced from 3 moles of nickel (II) oxide

So, 0.252 moles of nickel will be produced from \frac{3}{3}\times 0.252=0.252mol of nickel (II) oxide

Now, calculating the mass of nickel (II) oxide by using equation 1:

Molar mass of nickel (II) oxide = 74.7 g/mol

Moles of nickel (II) oxide = 0.252 moles

Putting values in equation 1, we get:

0.252mol=\frac{\text{Mass of nickel (II) oxide}}{74.7g/mol}\\\\\text{Mass of nickel (II) oxide}=(0.252mol\times 74.7g/mol)=18.8g

  • <u>For aluminium:</u>

By Stoichiometry of the reaction:

3 moles of nickel are produced from 2 moles of aluminium

So, 0.252 moles of nickel will be produced from \frac{2}{3}\times 0.252=0.168mol of aluminium

Now, calculating the mass of aluminium by using equation 1:

Molar mass of aluminium = 27 g/mol

Moles of aluminium = 0.168 moles

Putting values in equation 1, we get:

0.168mol=\frac{\text{Mass of aluminium}}{27g/mol}\\\\\text{Mass of aluminium}=(0.168mol\times 27g/mol)=4.54g

Hence, the mass of nickel (II) oxide and aluminium that must be used is 18.8 g and 4.54 g respectively.

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What is the formula for manganese (ii) fluoride decahydrate? (you may use a * to represent the dot in the formula.)?
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4 0
2 years ago
Read 2 more answers
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