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vaieri [72.5K]
3 years ago
5

The sample of 15.0 g of KCl is dissolved into a solution with a total volume of 250.0 mL. What is the molarity of KCl in the sol

ution?
Chemistry
1 answer:
lina2011 [118]3 years ago
3 0

Answer:

0.805 M.

Explanation:

Hello!

In this case, since the molarity of a solution is computing by dividing the moles of solute over the volume of solution in liters (M=n/V), for 15.0 g of potassium chloride (74.55 g/mol) we compute the corresponding moles:

n=15.0gKCl*\frac{1molKCl}{74.55gKCl}=0.201molKCl

Next, since the volume is 0.2500 in liters, the molarity turns out:

M=\frac{0.201mol}{0.2500L} \\\\M=0.805M

Best regards!

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A solution prepared by mixing 10 ml of 1 m hcl and 10 ml of 1.2 m naoh has a ph of
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Explanation:

pH is the measure of acidity or alkalinity of a solution.

Moles of H^+ ion = Molarity\times {\text {Volume in L}}=1M\times 0.01L=0.01mol

Moles of OH^- ion = Molarity\times {\text {Volume in L}}=1.2M\times 0.01L=0.012mol

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For neutralization:

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Thus (0.012-0.01)= 0.002 moles of OH^- are left in 20 ml or 0.02 L of solution.

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A 35.6 g sample of ethanol (C2H5OH) is burned in a bomb calorimeter, according to the following reaction. If the temperature ros
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<h3>Answer:</h3>

A) -1.24 × 10^3 kJ/Mol

<h3>Explanation:</h3>

we are given;

Mass of ethanol, m = 35.6 g

Temperature change, Δt(35.0 to 76.0°C) = 41 °C

Specific heat capacity of the calorimeter, c = 23.3 kJ/°C

Molar mass of ethanol = 46.07 g/mol

We are required to the heat change of the reaction.

  • We need to note that the reaction is an exothermic reaction since there is an increase in temperature which means heat was lost to the surroundings.

Therefore; we are going to use the following steps;

<h3>Step 1 : Moles of ethanol </h3>

We know, Moles = Mass ÷ molar mass

Thus, moles of ethanol = 35.6 g ÷ 46.07 g/mol

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<h3>Step 2: Enthalpy change or heat change for the reaction.</h3>

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but we are given s[pecific heat capacity in Kj/°C and we require heat change in kJ/mol

Therefore;

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Therefore, values of ΔH of the reaction is -1.24 × 10^3 kJ/Mol

7 0
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