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GarryVolchara [31]
3 years ago
5

How much heat would be evolved (released) if 50.0g of steam is at 114°C was cooled and condensed into water at 87°C?

Chemistry
1 answer:
LuckyWell [14K]3 years ago
6 0

Answer:

Q ≈ 1 x 10⁵ joules (1 sig. fig. based on 50g sample given)

Explanation:

Three heat transitions need be considered. Cooling the steam to 100°C, condensing the steam to water and cooling the water from 100°C to 87°C.

Total Heat exchange (Q) = ∑ heat transitions

Q = (m·c·ΔT)steam + (m·ΔH)condensation + (m·c·ΔT)water cooling

= (50g x 0.48j/g·°C x 14°C) + (50g x 2259j/g) + (50g x 4.184j/g·°C)

= 24j + 112,950j + 209.2j

= 113,183.2j ≈ 1 x 10⁵ joules (1 sig. fig. based on 50g sample given)

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Answer:

142.82 g

Explanation:

The following data were obtained from the question:

Volume of water = 12 mL

Volume of water + gold = 19.4 mL

Density of gol= 19.3 g/cm³

Mass of gold =.?

Next, we shall determine the volume of the gold. This can be obtained as follow:

Volume of water = 12 mL

Volume of water + gold = 19.4 mL

Volume of gold =.?

Volume of gold = (Volume of water + gold) – (Volume of water)

Volume of gold = 19.4 – 12

Volume of gold = 7.4 mL

Finally, we shall determine the mass of the gold as follow:

Note: 1 mL is equivalent to 1 cm³

Volume of gold = 7.4 mL

Density of gol= 19.3 g/cm³ = 19.3 g/mL

Mass of gold =?

Density = mass /volume

19.3 = mass of gold /7.4

Cross multiply

Mass of gold = 19.3 × 7.4

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6 0
3 years ago
Under what conditions do real gases behave most ideally.
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Under high temperatures and low pressure, gases behave the most ideal.

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