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OLEGan [10]
4 years ago
8

Duncan takes a break from studying and goes to the gym to swim laps if swimming Burns 8.15 x 10 to the 5th power calories per ho

ur how many kilojoules does swimming burn in that same amount of time
Chemistry
2 answers:
Marat540 [252]4 years ago
8 0
3411.427 kilojoules

8.15 x10^5 cal x (4.18 joules/ 1cal) x (1kJ/1000J) = 3,406.7 kJ
Naddika [18.5K]4 years ago
8 0

<u>Answer:</u> Duncan burns 3409.96 kJ in the same amount of time.

<u>Explanation:</u>

We are given that Duncan burns 8.15\times 10^5 Calories per hour and we need to find the amount she burns in kilojoules.

For that we use the conversion factors:

1Cal=4.184J\\\\1kJ=1000J

Converting 8.15\times 10^5cal/hr into kilojoules per hour, we get:

\Rightarrow (\frac{8.15\times 10^5Cal}{1hr})(\frac{4.184J}{1Cal})(\frac{1kJ}{1000J})\\\\\Rightarrow \frac{3409.96kJ}{hr}

Hence, duncan burns 3409.96 kJ in the same amount of time.

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Answer:

As, the temperature decreased from 40.0 °C to 0.0 °C an amount will be recrystallized and precipitated as solid crystals in the water (51.0 g - 14.0 g = 37.0 g) and 14.0 g will be dissolved in water.

Explanation:

  • Firstly, we must mention that:

The solubility of KNO₃ per 100.0 g of water at 40.0 °C = 63.0 g.

The solubility of KNO₃ per 100.0 g of water at 0.0 °C = 14.0 g.

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  • <em>As, the temperature decreased from 40.0 °C to 0.0 °C an amount will be recrystallized and precipitated as solid crystals in the water (51.0 g - 14.0 g = 37.0 g) and 14.0 g will be dissolved in water.</em>
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The gas in a balloon occupies 2.25 L at 298 K at what temperature will the balloon expand to 3.50 L
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<em><u> calculation</u></em>

by use of Charles law formula

that is  V1/T1 = V2/T2   where;

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T1= 298 k

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T2=?

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T2= (3.50 l x 298 K)  /2.25 L  = 463.56 K

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