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melisa1 [442]
2 years ago
8

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

Chemistry
1 answer:
LekaFEV [45]2 years ago
7 0
<h3>The system will absorb 24.9 KJ of heat energy.  </h3>

<h3>Step 1:</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>

<h3>Step 2</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.

<h3>Step 3</h3>

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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Air containing 0.04% carbon dioxide is pumped into a room whose volume is 6000 ft3. The air is pumped in at a rate of 2000 ft3/m
koban [17]

Here is the full question:

Air containing 0.04% carbon dioxide is pumped into a room whose volume is 6000 ft3. The air is pumped in at a rate of 2000 ft3/min, and the circulated air is then pumped out at the same rate. If there is an initial concentration of 0.2% carbon dioxide, determine the subsequent amount in the room at any time.

What is the concentration at 10 minutes? (Round your answer to three decimal places.

Answer:

0.046 %

Explanation:

The rate-in;

R_{in} = \frac{0.04}{100}*2000

R_{in} = 0.8

The rate-out

R_{out} = \frac{A}{6000}*2000

R_{out} = \frac{A}{3}

We can say that:

\frac{dA}{dt}= 0.8-\frac{A}{3}

where;

A(0)= 0.2% × 6000

A(0)= 0.002 × 6000

A(0)= 12

\frac{dA}{dt} +\frac{A}{3} =0.8

Integration of the above linear equation =

e^{\int\limits \frac {1}{3}dt } = e^{\frac{1}{3}t

so we have:

e^{\frac{1}{3}t}\frac{dA}{dt}} +\frac{1}{3}e^{\frac{1}{3}t}A = 0.8e^{\frac{1}{3}t

\frac{d}{dt}[e^{\frac{1}{3}t}A] = 0.8e^{\frac{1}{3}t

Ae^{\frac{1}{3}t} =2.4e\frac{1}{3}t +C

∴ A(t) = 2.4 +Ce^{-\frac{1}{3}t

Since A(0) = 12

Then;

12 =2.4 + Ce^{-\frac{1}{3}}(0)

C= 12-2.4

C =9.6

Hence;

A(t) = 2.4 +9.6e^{-\frac{t}{3}}

A(0) = 2.4 +9.6e^{-\frac{10}{3}}

A(t) = 2.74

∴ the concentration at 10 minutes is ;

=  \frac{2.74}{6000}*100%

= 0.0456667 %

= 0.046% to three decimal places

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1.1 * 10^7 km

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The solubility product for an insoluble salt with the formula M2X3 is written as ________, where s is the molar solubility
masha68 [24]

Answer:

36s^5

Explanation:

We have;

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If [M^3+(aq)] = [X^2-(aq)] = s

We then have;

Ksp = (2s)^2 * (3s)^3

Ksp = 4s^2 * 9s^3

Ksp = 36s^5

Note that Ksp is known as the solubility product. It is an equilibrum equation that shows the solubility of a solute in water.

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