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3 years ago

Consider the reaction: NaNO3(s) + H2SO4(l) NaHSO4(s) + HNO3(g) ΔH° = 21.2 kJ

Chemistry

1 answer:

LekaFEV [45]3 years ago

7 0

<h3>The system will absorb 24.9 KJ of heat energy. </h3>

Data obtained from the question

NaNO₃ + H₂SO₄ —> NaHSO₄(s) + HNO₃ ΔH° = 21.2 KJ

<h3>Heat absorbed by converting 100 g of NaNO₃ =? </h3>

Determination of the mass of NaNO₃ that was converted from the balanced equation.

Molar mass of NaNO₃ = 23 + 14 + (16×3)

= 23 + 14 + 48

= 85 g/mol

Mass of NaNO₃ from the balanced equation = 1 × 85 = 85 g

Thus, 85 g of NaNO₃ were converted from the balanced equation.

Determination of the heat absorbed by converting 100 g of NaNO₃.

From the balanced equation above,

When 85 g of NaNO₃ were converted, 21.2 KJ were absorbed by the system.

Thus, the conversion of 100 g of NaNO₃ will cause the system to absorb = $\frac{100 * 21.2}{85} =$ 24.9 KJ of heat energy.

Therefore, the system will absorb 24.9 KJ of heat energy.

Learn more: brainly.com/question/13728390

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Calculate each of the following quantities.<br> (a) mass in kilograms of 3.7 x 1020 molecules of NO2

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Answer:

The answer to your question is: 0.028 kg of NO2

Explanation:

Data

3.7 x 10²⁰ molecules of NO2 in kg

MW of NO2 = 14 + (16 x 2) = 14 + 32 = 46 kg

1 mol of NO2 --------------------- 6.023 x 10 ²³ molecules

x --------------------- 3.7 x 10²⁰ molecules

x = 3.7 x 10²⁰ x 1 / 6.023 x 10 ²³

x = 0.00061 mol

1 mol of NO2 --------------------- 46 kg of NO2

0.00061 mol ------------------ x

x = 0.00061 x 46/1

x = 0.028 kg of NO2

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Matter that is made up of only one kind of Atom

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Explanation:

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A 0.8870 g sample of a mixture of NaCl and KCl is dissolved in water, and the solution is then treated with an excess of AgNO3 t

bagirrra123 [75]

Answer : The percent by mass of NaCl and KCl are, 92.22 % and 7.78 % respectively.

Explanation :

As we know that when a mixture of NaCl and KCl react with excess $AgNO_3$ then the silver ion react with the chloride ion in both NaCl and KCl to form silver chloride.

Let the mass of NaCl be, 'x' grams and the mass of KCl will be, (0.8870 - x) grams.

The molar mass of NaCl and KCl are, 58.5 and 74.5 g/mole respectively.

First we have to calculate the moles of NaCl and KCl.

$\text{Moles of }NaCl=\frac{\text{Mass of }NaCl}{\text{Molar mass of }NaCl}=\frac{xg}{58.5g/mole}=\frac{x}{58.5}moles$

$\text{Moles of }KCl=\frac{\text{Mass of }KCl}{\text{Molar mass of }KCl}=\frac{(0.8870-x)g}{74.5g/mole}=\frac{(0.8870-x)}{74.5}moles$

As, each mole of NaCl and KCl gives one mole of chloride ions.

So, moles of chloride ions in NaCl = $\frac{x}{58.5}moles$

Moles of chloride ions in KCl = $\frac{(0.8870-x)}{74.5}moles$

The total moles of chloride ions = $\frac{x}{58.5}moles+\frac{(0.8870-x)}{74.5}moles$

Now we have to calculate the moles of AgCl.

As we know that, this amount of chloride ion is same as the amount chloride ion present in the AgCl precipitate. That means,

Moles of AgCl = Moles of chloride ion = $\frac{x}{58.5}moles+\frac{(0.8870-x)}{74.5}moles$

Now we have to calculate the moles of AgCl.

The molar mass of AgCl = 143.32 g/mole

$\text{Moles of }AgCl=\frac{\text{Mass of }AgCl}{\text{Molar mass of }AgCl}=\frac{2.142g}{143.32g/mole}=0.0149moles$

Now we have to determine the value of 'x'.

Moles of AgCl = $\frac{x}{58.5}moles+\frac{(0.8870-x)}{74.5}moles$

0.0149 mole = $\frac{x}{58.5}moles+\frac{(0.8870-x)}{74.5}moles$

By solving the term, we get the value of 'x'.

$x=0.818g$

The mass of NaCl = x = 0.818 g

The mass of KCl = (0.8870 - x) = 0.8870 - 0.818 = 0.069 g

Now we have to calculate the mass percent of NaCl and KCl.

$\text{Mass percent of }NaCl=\frac{\text{Mass of }NaCl}{\text{Total mass of mixture}}\times 100=\frac{0.818g}{0.8870g}\times 100=92.22\%$

$\text{Mass percent of }KCl=\frac{\text{Mass of }KCl}{\text{Total mass of mixture}}\times 100=\frac{0.069g}{0.8870g}\times 100=7.78\%$

Therefore, the percent by mass of NaCl and KCl are, 92.22 % and 7.78 % respectively.

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3 years ago