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2 years ago

How many moles are in a solution that has a molarity of 3.0M and a volume of 1500 ml.

Chemistry

1 answer:

Bad White [126]2 years ago

3 0

Answer:

1.25 Moles

Explanation:

1.25 moles of solute

Explanation:

Molarity is defined as the number of moles of solute per liter of solution.

Molarity = moles of solute / liter of solution

We are given the molarity and volume, both of which have the correct units. All we have to do is rearrange the equation to find the number of moles. You can do this by multiplying both sides of the equation by the volume to cancel it out on the right hand side. Afterwards, you should end up having the volume multiplied by the molarity equaling the number of moles of solute like so:

Moles of solute = Molarity * Volume

2.5M HCl * 0.5 L = 1.25 moles of HCl

I hope this made sense.

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Answer:

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For B: The $K_c$ for the given reaction is 1642.

Explanation:

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$2NO(g)+Cl_2(g)\rightleftharpoons 2NOCl(g)$

For A:

The expression of $K_p$ for the above reaction follows:

$K_p=\frac{(p_{NOCl})^2}{(p_{NO})^2\times p_{Cl_2}}$

We are given:

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Putting values in above equation, we get:

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For B:

Relation of $K_p$ with $K_c$ is given by the formula:

$K_p=K_c(RT)^{\Delta ng}$

where,

$K_p$ = equilibrium constant in terms of partial pressure = $4.0\times 10^1$

$K_c$ = equilibrium constant in terms of concentration = ?

R = Gas constant = $0.0821\text{ L atm }mol^{-1}K^{-1}$

T = temperature = 500 K

$\Delta ng$ = change in number of moles of gas particles = $n_{products}-n_{reactants}=2-3=-1$

Putting values in above equation, we get:

$4.0\times 10^1=K_c\times (0.0821\times 500)^{-1}\\\\K_c=\frac{4.0\times 10^1}{(0.0821\times 500)^{-1})}=1642$

Hence, the $K_c$ for the given reaction is 1642.

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