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3 years ago

The molarity of a solution of 5.0 g of kcl in 100. ml of solution is ________.

Chemistry

2 answers:

Sliva [168]3 years ago

8 0

Answer: The molarity of solution is 0.67 M

Explanation:

To calculate the molarity of solution, we use the equation:

$\text{Molarity of the solution}=\frac{\text{Mass of solute}\times 1000}{\text{Molar mass of solute}\times \text{Volume of solution (in mL)}}$

We are given:

Mass of solute (KCl) = 5.0 g

Molar mass of potassium chloride = 74.55 g/mol

Volume of solution = 100 mL

Putting values in above equation, we get:

$\text{Molarity of solution}=\frac{5g\times 1000}{74.55g/mol\times 100mL}\\\\\text{Molarity of solution}=0.67M$

Hence, the molarity of solution is 0.67 M

sdas [7]3 years ago

7 0

Molarity is defined as the number of moles of solute in 1 L of solution
mass fo KCl in the solution is - 5.0 g
number of moles of KCl - 5.0 g/ 74.5 g/mol = 0.067 mol
number of moles of KCl in 100 mL - 0.067 mol
therefore number of KCl moles in 1 L - 0.067 / 100 mL x 1000 mL = 0.67 M
molarity of KCl is 0.67 M

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Answer:

3.9.

Explanation:

(Molar mass of 342.15 g/mol).

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2050 gallons of water = 2050 × 3.785 = 7759.25 Litres.

After this we will now look for the Concentration aluminum sulfate using the formula below:

Concentration = number of moles ÷ volume. ------------------------------------(1).

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Therefore, slotting in the values of of number of moles and volume into the equation (1) above, we have;

Concentration= 8.8823 moles/ 7759.25 Litres. = 0.001145 M.

The next equation we are going to make use of today is that of the solubilty product that is;

Ksp = (Al^3+) (OH^-)^3.

So, we have the ksp = 1.3 × 10^-33 and the value of (Al^3+)= 0.001145 M.

Hence, making (OH^-) the subject of the formula;

(OH^-) = ( 1.3 × 10^-33 / 0.001145)^1/3.

(OH^-) = 8 × 10^-11 M.

Hence, pOH = - log (OH^-)

pOH = - log (8 × 10^-11).

pOH = 10.1

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pH + pOH = 14.

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Write the equilibrium expression for N2 (g) + 3 H2 (g) ↔ 2 NH3 (g) + heat I have no idea where to even start, due to the corona

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Answer:

The expression of an equilibrium constant will given as:

$K_c=\frac{[NH_3]^2}{[N_2][H_2]^3}$

Explanation:

Equilibrium constant is defined as the ratio of concentration of products to the concentration of reactants each raised to the power their stoichiometric ratios. It is expressed as $K_{eq}$

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The expression of an equilibrium constant will given as:

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Or n=m/MM

n = number of moles

m = mass

MM = molar mass

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