Mg(OH)2 + 2HNO3 → Mg(NO3)2 + 2H20
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1a. The balanced equation for the reaction is:
<h3>3Ca(OH)₂ + 2H₃PO₄ —> Ca₃(PO₄)₂ + 6H₂O </h3>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
<h3>Mole of Ca(OH)₂ =? </h3>Mole = Concentration × Volume
Mole = 0.348 × 0.071
<h3>Mole of Ca(OH)₂ = 0.0247 mole </h3><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 = = 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>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:
<h3>C₁V₁ = C₂V₂</h3>0.432 × 10 = C₂ × 100
4.32 = C₂ × 100
Divide both side by 100
C₂ =
Therefore, the new concentration of the H₃PO₄ solution is 0.0432 mol/L
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Answer:
Explanation:
In multiplication and division problems, your answer can have no more significant figures than the number with the fewest significant figures.
(by my calculator)
There are three significant figures in 7.06 and two in 2.3.
You must round to significant figures and report the answer as 1.3 × 1.0⁷.