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likoan
2 months ago
15

NH4NO3, whose heat of solution is 25.7 kJ/mol, is one substance that can be used in cold pack. If the goal is to decrease the te

mperature from 25.0 °C to 5.0 °C, how many grams of NH4NO3 should we use for every 100.0 g of water in the cold pack? Assume no heat was lost outside of cold pack, and the specific heat of the resulted solution was the same as water, or 4.184 J/(g•°C).
Chemistry
1 answer:
lorasvet [2.7K]2 months ago
6 0

Answer:

To lower the temperature of the solution from 25.0°C to 5.0°C, it is necessary to use 35.2g of NH₄NO₃ for every 100.0g of water.

Explanation:

In order to cool down the solution, we need:

4.184 J/g°C × (5.0°C - 25.0°C) × (100.0g + X) = -Y

8368 J + 83.68 J/gX = Y (1)

Here, x represents the grams of NH₄NO₃ required, and Y represents the energy needed to remove heat.

Furthermore, the energy Y becomes:

Y = 25700 J/mol × \frac{1mol}{80,043g}X

Y = 321 J/g X (2)

Substituting (2) into (1)

8368 J + 83.68 J/g X = 321 J/g X

8363 J = 237.32 J/gX

X = 35.2g

This means 35.2g of NH₄NO₃ must be used for every 100.0g of water to achieve a temperature decrease from 25.0°C to 5.0°C.

I trust this information will be useful!

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The molecular formula is PbSO₄, indicating lead sulfate

Option c.

Explanation:

The percentage makeup shows that in 100 g of this compound, there are:

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To find the moles of each element, we divide by their molar masses:

68.3 g Pb / 207.2 g/mol = 0.329 moles Pb

10.6 g S / 32.06 g/mol = 0.331 moles S

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Next, we find the mole ratio by dividing each by the smallest number of moles:

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Thus, the molecular formula is PbSO₄, representing lead sulfate.

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