irsheffler irsheffler

07-04-2021
Chemistry

contestada

Question 6 of 10
How much energy is required to vaporize 2 kg of gold? Use the table below
and this equation: Q = mLvapor
Substance
Latent Heat
Fusion
(melting)
(kJ/kg)
Melting
Point
(°C)
Latent Heat
Vaporization
(boiling) (kJ/kg)
Boiling
Point
(°C)
Aluminum
400
660
1100
2450
Copper
207
1083
4730
2566
Gold
628
1063
1720
2808
Helium
52
-270
21
-269
Lead
24.5
327
871
1751
Mercury
11.4
-39
296
357
Water
335
0
2256
100

Respuesta :

urianaupaul urianaupaul

19-04-2021

Answer: Q = mlvap

Q = (2 kg)(1 kmol/197 kg)(1,000 mol/1 kmol)

Q = 10.15 kJ

Answer Link

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