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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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From a quick observation

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What type of reaction is shown in the following chemical equation in NH3+HC1 NH4C1​
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Read 2 more answers
How many grams of water are theoretically produced for the following reaction given we have 2.6 moles of HCl and 1.4 moles of Ca
leva [86]

Answer:

The answer to your question is: 6.8 g of water

Explanation:

Data

2.6 moles of HCl

1.4 moles of Ca(OH)2

                           2HCl     +     Ca(OH)2    →        2H2O    +      CaCl2

MW                   2(36.5)               74                       36 g               111 g

                          73g                

                            1 mol of HCl ----------------  36.5 g

                           2.6 mol           --------------    x

                              x = (2.6 x 36.5) / 1   = 94.9 g

                           1 mol of Ca(OH)2 --------------   74 g

                         1.4 mol                  ---------------   x

                            x = (1.4 x 74) / 1  = 103.6 g

Grams of water

                        73 g of HCl ------------------   36g of H2O

                        94.9 g        -------------------    x

                     x = (94.9 x 36) / 73 = 46.8 g of water

6 0
2 years ago
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