How can the rate of a reaction be increased?

A titration reaction requires 38.20 mL phosphoric acid solution to react with 71.00 mL of 0.348 mol/L calcium hydroxide to reach

NARA [144]

1a. The balanced equation for the reaction is:

1b. The number of mole of Ca(OH)₂ is 0.0247 mole

1c. The number of mole of H₃PO₄ is 0.0165 mole.

1d. The concentration of H₃PO₄ is 0.432 mol/L

2. The new concentration of the H₃PO₄ solution is 0.0432 mol/L

<h3>1a. The balanced equation for the reaction</h3>

<u>3</u>Ca(OH)₂ + <u>2</u>H₃PO₄ —> Ca₃(PO₄)₂ + <u>6</u>H₂O

<h3>1b. Determination of the mole of Ca(OH)₂</h3>

Volume of Ca(OH)₂ = 71 mL = 71 / 1000 = 0.071 L

Concentration of Ca(OH)₂ = 0.348 mol/L

Mole = Concentration × Volume

Mole = 0.348 × 0.071

<h3>1c. Determination of the mole of H₃PO₄. </h3>

3Ca(OH)₂ + 2H₃PO₄ —> Ca₃(PO₄)₂ + 6H₂O

From the balanced equation above,

3 moles of Ca(OH)₂ reacted with 2 moles of H₃PO₄.

Therefore,

0.0247 moles of Ca(OH)₂ will react with = $\frac{0.0247 * 2}{3}$ = 0.0165 mole of H₃PO₄.

Thus, the number of mole of H₃PO₄ is 0.0165 mole

<h3>1d. Determination of the concentration of H₃PO₄</h3>

Volume of H₃PO₄ = 38.20 mL = 38.20/ 1000 = 0.0382 L

Mole of H₃PO₄ = 0.0165 mole

<h3>Concentration of H₃PO₄ =?</h3>

$Concentration = \frac{mole}{volume} \\\\Concentration = \frac{0.0165}{0.0382}$

<h3>Concentration of H₃PO₄ = 0.432 mol/L</h3>

<h3>2. Determination of the new concentration of the H₃PO₄ solution.</h3>

Initial Volume (V₁) = 10 mL

Initial concentration (C₁) = 0.432 mol/L

New volume (V₂) = 100 mL

<h3>New concentration (C₂) =?</h3>

The new concentration of the H₃PO₄ solution can be obtained as follow:

0.432 × 10 = C₂ × 100

4.32 = C₂ × 100

Divide both side by 100

C₂ = $\frac{4.32}{100}\\$

Therefore, the new concentration of the H₃PO₄ solution is 0.0432 mol/L

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Để trung hoà 20ml dung dịch HCl 0.1M cần 10ml dung dịch NaOH nồng độ x mol/l . Giá trị của x là

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The question is: To neutralize 20ml of 0.1M HCl solution, 10ml of NaOH solution of concentration x mol/l is required. What is the value of x?

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Explanation:

Given: $V_{1}$ = 20 mL, $M_{1}$ = 0.1 M

$V_{2}$ = 10 mL, $M_{2}$ = x

Formula used is as follows.

$M_{1}V_{1} = M_{2}V_{2}\\0.1 M \times 20 mL = x \times 10 mL\\x = 0.2 M$

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Wind flow towards the low pressure and the air rises in the atmosphere. As the air increases, the water vapor within it solidifies, forming clouds and undergo precipitation. Low pressure formed in the center areas.

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Answer:

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