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Trava [24]
4 years ago
14

3. Commercial phosphoric acid, H3PO4, is often purchased as a 85.5 weight percent solution. Find the mg/L of H2SO4 and the molar

ity and normality of the solution. The concentrated sulfuric acid solution has a density of 1.71 g/ml at the temperature of the system.
Chemistry
1 answer:
Anni [7]4 years ago
8 0

Answer:

The answers to the question are

(a) The the mg/L of H₃PO₄is 1462.05 g

(b) The molarity  of H₃PO₄ in the solution is 14.92 mol/L

(c) The normality of H₃PO₄ in the solution 44.85 eq/L

Explanation:

(a) The commercially available H₃PO₄ has a concentration of 85.5 percent by weight, therefore

The mass of H₃PO₄ is found by

Density of water = 1.71 g/ml

mass of one liter of acid solution = 1.71 ×1000 = 1710 g

Therefore 85.5 % by weight of H₃PO₄ =1710×85.5/100 = 1462.05 g

Therefore we have 1462.05 g/L of H₃PO₄

Molar mass of H₃PO₄ = 97.994 g/mol

(b)Therefore the number of moles in 1462.05 g = 14.92 moles and the molarity = 14.92 mol/L

(c) The Normality = \frac{Grams of solute}{(Volume of solvents)(Equivalent weight)}  = 1465.05÷(1×97.994/3) = 44.85 eq/L

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We can use the heat equation,
Q = mcΔT 

where Q is the amount of energy transferred (J), m is the mass of the substance (kg), c is the specific heat (J g⁻¹ °C⁻¹) and ΔT is the temperature difference (°C).
Q = 11.2 kJ = 11200 J
m = <span>145 g
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ΔT = (67 - 22) °C = 45 °C
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11200 J = 145 g x c x 45 °C
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Hence, specific heat of benzene is 1.72 J g⁻¹ °C⁻¹.
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3 years ago
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The combustion of ethene in the presence of excess oxygen yields carbon dioxide and water: c2h4 (g) + 3o2 (g) → 2co2 (g) + 2h2o
meriva

Answer:

\boxed{-267.5}

Explanation:

You can calculate the entropy change of a reaction by using the standard molar entropies of reactants and products.

The formula is

\Delta_{r} S^{\circ} = \sum_n {nS_{\text{products}}^{\circ} - \sum_{m} {mS_{\text{reactants}}^{\circ}}}

The equation for the reaction is

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\Delta_{r} S^{\circ} = (2\times213.6 + 2\times69.9) - (1\times219.5 + 3\times205.0)\\\\= 567.0 - 834.5 = \boxed{-267.5 \text{ J}\cdot\text{K}^{-1} \text{mol}^{-1}}

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Answer: I need help

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