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evablogger [386]
3 years ago
5

How much heat is required to raise the temperature of 65.8 grams of water from 31.5ºC to 46.9ºC?

Chemistry
1 answer:
vodomira [7]3 years ago
3 0

Answer: 1,013.32 cal × 4.18 J/cal = 4,235.68 J

Explanation:

1) Data:

Water ⇒ C = 1 cal/g°C

m = 65.8 g

Ti = 31.5°C

Tf = 36.9°C

Heat, Q = ?

2) Formula:

Q = mCΔT

3) Calculations:

Q = 65.8g × 1 cal/g°C × (46.9°C - 31.5°C) = 1,013.2 cal

4) You can convert from calories to Joules using the conversion factor:

1 cal = 4.18 J

⇒ 1,013.32 cal × 4.18 J/cal = 4,235.68 J

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Explanation:

Know the rules of multiplying wth perentheses.

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Energy is released when the nucleus of an atom splits and two smaller atoms are formed. What is the name of this process?
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A scientist digs up sample of arctic ice that is 458,000 years old. He takes it to his lab and finds that it contains 1.675 gram
Fiesta28 [93]

Answer:

6.70 grams of krypton-81 was present when the ice first formed

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3 0
3 years ago
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H is the answer :)
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4 0
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Lemur [1.5K]

<u>Answer:</u> The equation is given below.

<u>Explanation:</u>

Single replacement reactions are the chemical reactions in which more reactive metal displaces a less reactive metal from its chemical reaction. General equation for these reactions is given by the equation:

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Metal A is more reactive than metal B.

The reactivity of metals is judged with the help of reactivity series. In this series, the metals lying above are more reactive than the metals which lie below in the series.

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3 0
3 years ago
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