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IRISSAK [1]
3 years ago
10

The heat capacity of liquid water is 4.18 J/g·°C and the heat of vaporization is 40.7 kJ/mol. How many kilojoules of heat must b

e provided to convert 1.00 g of liquid water at 67°C into 1.00 g of steam at 100°C?
Chemistry
1 answer:
pychu [463]3 years ago
3 0

Answer:

The correct answer would be - 2.4KJ or, 2400J

Explanation:

Given:

heat capacity of liquid water - 4.18 J/g·°C

heat of vaporization - 40.7 kJ/mol

Mass of water = 1g

Moles of water = mass/molar mass

= 1g/18.016g

= 0.055 moles

Then,

Total heat required = q1(to raise the temperature to 100) + q2(change from the liquid phase to gas/steam)

= m *s*dt + moles * heat of vaporization

= (1g * 4.18 j/gc * (100-67)) + 0.055* 40.7 KJ

= 137.94J + 2.26KJ

=0.138KJ + 2.26KJ

=2.4KJ or, 2400J

Thus, the correct answer would be - 2.4KJ or, 2400J

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Mr. Aguilar has ice on his wrist, literally. If the ice on Mr. Aguilar’s wrist (H2O) weighs 546 grams, how many ice particles ar
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Answer:

The answer to your question is  1.83 x 10²⁵ particles

Explanation:

Data

particles of H₂O = ?

mass of H₂O = 546 g

Process

1.- Calculate the molar mass of Water

Molar mass = (2 x 1) + (1 x 16)

                   = 2 + 16

                   = 18 g

2.- Use proportions to find the number of particles. Use Avogadro's number.

                 18 g ---------------- 6.023 x 10²³ particles

                546 g ---------------   x

                  x = (546 x 6.023 x 10²³) / 18

3.- Simplification

                  x = 3.289 x 10²⁶ / 18

4.- Result

                  x = 1.83 x 10²⁵ particles

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