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Vikki [24]
3 years ago
8

How many grams of water can be cooled from 33 ∘c to 15 ∘c by the evaporation of 50 g of water? (the heat of vaporization of wate

r in this temperature range is 2.4 kj/g. the specific heat of water is 4.18 j/g⋅k.)?
Chemistry
1 answer:
Eva8 [605]3 years ago
3 0

Answer: -

1595 g

Explanation: -

Heat of vaporization = 2.4 Kj/ g

Mass of water to be vaporized = 50 g

Heat released = Mass of water to be vaporized x Heat of vaporization

= 50 g x 2.4 KJ /g

= 120 KJ

= 120000 J

Initial temperature= 33+273= 306 K

Final temperature =15+273=288 K

Change in temperature required = T = 306 - 288 = 18 K

specific heat of water is 4.18 J / g K

Mass of water that can be cooled = Total heat / (specific heat of water x Change in temperature)

= 120000 J / ( 4.18 J / g K x 18 K)

= 1595 g

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2 years ago
f 4.2 moles of copper metal react with 6.3 moles of silver nitrate, how many moles of silver metal can be formed, and how many m
alex41 [277]
<span>1. Balance the reaction
Cu + 2AgNO3 → Cu(NO3)2 + 2Ag 
2. Find the limiting [divide moles of reactant by their balancing numbers]
Cu: 4.2/1 =4.2 
AgNO3: 6.3/2= 3.15 
AGNO3 is the limiting
How many moles of silver metal will be formed?
The ratio between the limiting reactant and silver is 2:2 or 1:1
therefore 6.3 moles of silver will be formed.

How many moles of excess reactant will be left?
Since each mole of copper is equal to 2 moles of AgNO3. divide 6.3 by 2 then minus it from coppers moles
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4.2-3.15 = 1.05 moles of copper will be left
</span>
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3 years ago
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