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IRISSAK [1]
3 years ago
14

For the reaction 2SO2+O2 ---> 2SO3 write the forward and reverse reactions

Chemistry
1 answer:
cupoosta [38]3 years ago
3 0
2SO2(g) + O2(g) ---> 2SO3(g) • 2SO2(g) + O2(g) <--- 2SO3(g) • 2SO2(g) + O2(g) <---> 2SO3(g) • 2SO2(g) + O2(g) <--->> 2SO3(g) • Note: The rate of change reaches an equilibrium. The concentration of reactants and products is not equa
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What is the concentration (in M) of a 225ml potassium sulfate solution that contains 4.15g of potassium?
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The concentration of solution in M or mol/L can be calculated using the following formula:

C=\frac{n}{V} .... (1)

Here, n is number of moles and V is volume of solution in L.

The molecular formula of potassium sulfate is K_{2}SO_{4} thus, there are 2 moles of potassium in 1 mol of potassium sulfate.

1 mol of potassium will be there in 0.5 mol of potassium sulfate.

Mass of potassium is 4.15 g, molar mass is 39.1 g/mol.

Number of moles can be calculated as follows:

n=\frac{m}{M}

Here, m is mass and M is molar mass

Putting the values,

n=\frac{(4.15 g}{(39.1 g/mol}=0.1061 mol

Thus, number of moles of  K_{2}SO_{4} will be 0.1061\times 0.5=0.053 mol.

The volume of solution is 225 mL, converting this into L,

1 mL=10^{-3}L

Thus,

225 mL=0.225 L

Putting the values in equation (1),

C=\frac{(0.053 mol}{0.225 L}=0.236 M

Therefore, concentration of potassium sulfate solution is 0.236 M.


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3 years ago
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