Can you PLEASE help me with this Chemistry problem?

Question

HI, I am having some trouble with this problem. I need help, please. Can you please please please answer and explain it, that would be fantastic...thank you!

Heat is added to 2.27 kg of ice at -22 °C. How many kilocalories are required to change the ice to steam at 150 °C?

Answer 1

The specific heat of ice is 0.50 cal/g-oC, that means that 0.50 calories is needed to raise 1g of ice 1oC.



0.50 calories is needed to raise 1g of ice 1oC

We must raise the temp 22oC to reach the melting point.







2.27 *

1,000 =

2,270 TOT

2,270 g *

22 °C =

49,940 TOT

49,940 *

.5cal =

24,970 cal



The latent heat (the amount of energy to change the state of something from solid to liquid or liquig to gas. In this case solid water to liquid water.) is 80 cal/g



2,270g *

80cal =

181,600 cal



To raise the temp to the boiling point it takes 1cal/g so 2,270g x 1 = 2,270 calg x100°C = 227,000 cal



The latent heat for boiling/condensation is 540 cal/g so



2,270 *

540 =

1,225,800cal



Now add them up



1,225,800 +

227,000 =

1,452,800 TOT

1,452,800 +

181,600 =

1,634,400 TOT

1,634,400 +

24,970 =

1,659,370 TOT

1,659,370cal /

1,000 =

1,659.37 Kcal

Answer 2

This belongs to the physics department imo... just a tip ;)



Haven't done these kind of problems for 2 years, so can't help.

Answer 3

As many kilocalories as I ate for lunch - about two...

Answer 4

The GMW of hydogen is one million so as this skill that there is 29 g of Cl interior the given volume of HCl. you notice that 0.5 of the Cl is utilized in MnCl2 so, for this reason, 0.5 of the Cl is interior the gasoline Cl2. Your answer is d) 14.6g.