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3 years ago

Complete the following table.

Chemistry

2 answers:

Nutka1998 [239]3 years ago

6 0

Complete the following table.

Acid Molarity Moles of H⁺ released per liter

HCl 1

H2SO4 1

H3PO4 1

H2SO4 0.5

H3PO4 3

HNO3 2

[Answer]

olga_2 [115]3 years ago

5 0

Answer: see the last column in the final table of the explanation below.

Explanation:

1) Arrange the data to understand the question:

Acid Molarity Moles of H⁺ released per liter

HCl 1 ____

H2SO4 1 ____

H3PO4 1 ____

H2SO4 0.5 ____

H3PO4 3 ____

HNO3 2 ____

2) All the given acids are strong acids, so you can consider they dissociate completely and release all the hydrogens present in the chemical formula.

3) The molarity formula is M = n/V, where n is the number of moles of solute and V is the volume of the solution in liters

From that, n = M×V

Go one by one with that formula.


Acid Molarity Moles of solute number of H in Moles H⁺ per liter the chemical formula per liter

HCl 1 1M×1L = 1 1 1×1 = 1

H₂SO4₄ 1 1M×1L = 1 2 2×1 = 2
H₃PO₄ 1 1M×1L = 1 3 3×1 = 3

H₂SO₄ 0.5 0.5M×1L = 0.5 2 2×0.5 = 1

H₃PO₄ 3 3M×1L = 3 3 3×3 = 9

HNO₃ 2 2M×1L = 2 2 2×2 = 4

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1 year ago

150 mL of 0.25 mol/L magnesium chloride solution and 150 mL of 0.35 mol/L silver nitrate solution are mixed together. After reac

Murljashka [212]

Answer:

$0.175\; \rm mol \cdot L^{-1}$ .

Explanation:

Magnesium chloride and silver nitrate reacts at a $2:1$ ratio:

$\rm MgCl_2\, (aq) + 2\, AgNO_3\, (aq) \to Mg(NO_3)_2 \, (aq) + 2\, AgCl\, (s)$ .

In reality, the nitrate ion from silver nitrate did not take part in this reaction at all. Consider the ionic equation for this very reaction:

$\begin{aligned}& \rm Mg^{2+} + 2\, Cl^{-} + 2\, Ag^{+} + 2\, {NO_3}^{-} \\&\to \rm Mg^{2+} + 2\, {NO_3}^{-} + 2\, AgCl\, (s)\end{aligned}$ .

The precipitate silver chloride $\rm AgCl$ is insoluble in water and barely ionizes. Hence, $\rm AgCl\!$ isn't rewritten as ions.

Net ionic equation:

$\begin{aligned}& \rm Ag^{+} + Cl^{-} \to AgCl\, (s)\end{aligned}$ .

Calculate the initial quantity of nitrate ions in the mixture.

$\begin{aligned}n(\text{initial}) &= c(\text{initial}) \cdot V(\text{initial}) \\ &= 0.25\; \rm mol \cdot L^{-1} \times 0.150\; \rm L \\ &= 0.0375\; \rm mol \end{aligned}$ .

Since nitrate ions $\rm {NO_3}^{-}$ do not take part in any reaction in this mixture, the quantity of this ion would stay the same.

$n(\text{final}) = n(\text{initial}) = 0.0375\; \rm mol$ .

However, the volume of the new solution is twice that of the original nitrate solution. Hence, the concentration of nitrate ions in the new solution would be $(1/2)$ of the concentration in the original solution.

$\begin{aligned} c(\text{final}) &= \frac{n(\text{final})}{V(\text{final})} \\ &= \frac{0.0375\; \rm mol}{0.300\; \rm L} = 0.175\; \rm mol \cdot L^{-1}\end{aligned}$ .