<span>Gallium-72 that is what i got when i did the math for it</span>
The 7160 cal energy is required to melt 10. 0 g of ice at 0. 0°C, warm it to 100. 0°C and completely vaporize the sample.
Calculation,
Given data,
Mass of the ice = 10 g
Temperature of ice = 0. 0°C
- The ice at 0. 0°C is to be converted into water at 0. 0°C
Heat required at this stage = mas of the ice ×latent heat of fusion of ice
Heat required at this stage = 10 g×80 = 800 cal
- The temperature of the water is to be increased from 0. 0°C to 100. 0°C
Heat required for this = mass of the ice×rise in temperature×specific heat of water
Heat required for this = 10 g×100× 1 = 1000 cal
- This water at 100. 0°C is to be converted into vapor.
Heat required for this = Mass of water× latent heat
Heat required for this = 10g ×536 =5360 cal
Total energy or heat required = sum of all heat = 800 +1000+ 5360 = 7160 cal
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The awnser would be c i got you fam
We are given the equation to use which is:
ΔG = ΔH - TΔS
We are also given that:
ΔG = 173.3 kJ
T = 303 degrees kelvin
ΔH = 180.7 kJ
Substitute with these givens in the above equation to get ΔS as follows:
ΔG = ΔH - TΔS
173.3 = 180.7 - 303ΔS
303ΔS = 180.7 - 173.3
303ΔS = 7.4
ΔS = 7.4 / 303 = 0.02442 kJ/K which is equivalent to 24.42 J/k
Based on the above calculations, the correct choice is:
D. 24.42 J/K
