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mixer [17]
3 years ago
14

How much heat must be removed from 25.0g of steam at 118.0C in order to form ice at 15C

Chemistry
1 answer:
NemiM [27]3 years ago
3 0

Answer:

-10778.95 J heat must be removed in order to form the ice at 15 °C.

Explanation:

Given data:

mass of steam = 25 g

Initial temperature = 118 °C

Final temperature = 15 °C

Heat released = ?

Solution:

Formula:

q = m . c . ΔT

we know that specific heat of water is 4.186 J/g.°C

ΔT = final temperature - initial temperature

ΔT = 15 °C - 118 °C

ΔT = -103 °C

now we will put the values in formula

q = m . c . ΔT

q = 25 g × 4.186 J/g.°C × -103 °C

q = -10778.95 J

so, -10778.95 J heat must be removed in order to form the ice at 15 °C.

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In the given examples, the difference in electronegativity is greatest between K  and Br i.e. 0.8 and 2.8 respectively with a difference of 2.0. This also makes sense since K and Br are on the extreme ends of the periodic table. Hence, potassium with a valence electron configuration of 4s1 will lose its s electron to Br (4s24p6) and form an ionic molecule K⁺Br⁻

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3 years ago
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valentina_108 [34]

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Explanation:

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To learn more about osmotic pressure

brainly.com/question/10046758

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amm1812

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