Need help with Chemistry Homework. Please!!

[Gotvibe]

1. A 5 inch piece of cake provides 1790 kJ of energy. What is this in Calories?

 

5. Mercury has a freezing point of -38.8 degrees Celsius. How much heat energy (in joules) must be released by Hg if 1.40mL is cooled from 23.0 degrees Celsius to -38.8 degrees Celsius and then frozen to a solid? (The density of Hg is 13.6 g/cm3, its specific heat is 0.140 J/g – K and its heat of fusion is 11.4 J/g.)

 

8. Using the following reactions, calculate the enthalpy change for the formation of PbO(s) from Pb(s) and O(g).

 

Pb(s) + CO(g) -> PbO(s) + C(s) H= -106.8 kJ

 

2 C(s) + O2(g) -> 2 CO(g) H= -221.0 kJ

 

Pb(s) + O2(g) -> PbO(s) H= ? kJ

 

9. You wish to know the enthalpy change for the formation of PCl3(l) based on the equations below. If 1.10 mol of PCl3(l) from phosphorus and chlorine.

 

P4(s) + 6 Cl2(g) -> 4 PCl3(l) H= ? kJ

 

P4(s) + 10 Cl2(g) -> 4 PCl5(s) H= -1774.0 kJ

 

PCl3(l) + Cl2(g) -> PCl5(s) H= -123.8 kJ

 

11. CaO is mixed with H2O to give Ca(OH)2, which reacted with CO2 to give limestone.

 

Ca(OH)2(s) + CO2(g) -> CaCO3(s) + H2O(g)

 

(a) Calculate the enthalpy change

(:umno: What quantity of heat is evolved or absorbed if 2.30 kg of Ca(OH)2 is allowed to react with a stoichiometric amount of CO2?

 

14. A 9.85g piece of metal is heated to 98.6 degrees Celsius and then the temperature drops to 0.0 degrees Celsius, it is found that 0.43g of ice melted. What is the metal’s specific heat?

 

16. Suppose you add 100.0g of H2O at 65.0 degrees Celsius to 100.0g of ice at 0.0 degrees Celsius. The water cools to 0.0 degrees Celsius. When the ice and water temperature get to a uniform temperature of 0.0 degrees Celsius, how much ice has melted?

 

 

 

Im not asking for answers, Im just asking how to setup the problems. I've already failed the test, I just want to understand these concepts for the final. Help me out, anyone, please!! :hihi:

[hallenrm]

What? you want us to complete your homework. No way, young kids like you cannot be encouraged to become lazy. Do some work, look through your text books, even the if some point perplexes you, ask here.

 

Completing your homework, no way! :umno:

[InfiniteNow]

Im not asking for answers, Im just asking how to setup the problems.

What? you want us to complete your homework. No way, young kids like you cannot be encouraged to become lazy. Do some work, look through your text books...

:umno:

[Mercedes Benzene]

I've moved this thread from chemistry to the Science Projects and Homework forum.

I'll be glad to help you "set up" your problems, but we don't give answers here at Hypography.

[Gotvibe]

I've moved this thread from chemistry to the Science Projects and Homework forum.

I'll be glad to help you "set up" your problems, but we don't give answers here at Hypography.

 

Id greatly appreciate it!!

 

I mean, I ahve a few ideas, it's just that webassign isnt accepting my answers, so Im obviously doing SOMETHING wrong..

[Jay-qu]

well I will get you started, Calories and kJ are just two ways of measuring energy. One calorie is defined at the energy taken to raise 1 gram of water by 1 degree celcius. This amount of energy is 4.18J.

 

That should be enough info to work out the first one :phones:

[Gotvibe]

So,

 

1790 kJ x Cal/4.18J = 428. Cal

 

Got that one.

 

I need help with number 14.

 

I realize I need to use c=Q/mT

 

m= 9.85g

delta T= -98.6 degrees C

I need help finding Q. Can I just look up the Q in the book as H2O(s) ?

[Gotvibe]

Ok, I've figured them out except for #5, 11, 14, & 16.

 

#14 & 16 are the same principal,

 

Q1+Q2=0

 

, but my equations just arent working out.

 

Someone, anyone, help me set these up. Please!!

[eric l]

5. Mercury has a freezing point of -38.8 degrees Celsius. How much heat energy (in joules) must be released by Hg if 1.40mL is cooled from 23.0 degrees Celsius to -38.8 degrees Celsius and then frozen to a solid? (The density of Hg is 13.6 g/cm3, its specific heat is 0.140 J/g – K and its heat of fusion is 11.4 J/g.)

 

 

 

 

Im not asking for answers, Im just asking how to setup the problems. I've already failed the test, I just want to understand these concepts for the final. Help me out, anyone, please!! :lol:

 

Specific heath and heath of fusion are give in J/g, so we must first convert the volume of merucry into mass or weight. Density is 13.6 g/cm³, 1 mL = 1 cm³, so 1.40 ml = 1.40*13.6 g = 19.04 g.

 

Cooling from + 23.0°C to -38.8°C releases [23.0 -(-38.8)]*0.140 J/g. or 19.04*61.8*0.140 J for the given amount of mercury.

 

Freezing releases 19.04g*11.4J/g for the same amount.

 

Please make the multiplications and additions for yourself, I'll start on the other questions in the meanwhile.

[eric l]

8. Using the following reactions, calculate the enthalpy change for the formation of PbO(s) from Pb(s) and O(g).

 

Pb(s) + CO(g) -> PbO(s) + C(s) H= -106.8 kJ

 

2 C(s) + O2(g) -> 2 CO(g) H= -221.0 kJ

 

Pb(s) + O2(g) -> PbO(s) H= ? kJ

 

:lol:

 

In this one, we have to get rid of the CO. Equation 2 gives us 2 CO, so we must take (2*equation 1) - equation2 - wimple algebra actually !

 

2[Pb + CO -> PbO + C + 106.8 kJ] H = -213.6 kJ

[2C + O2 -> 2 CO + 221.0 kJ] H = - 221.0 kJ

substracting gives

[2Pb + O2 -> 2 PbO - 7.4 kJ] H = 7.4 kJ

 

Anyway, something seems wrong with the reactions : it is

• either Pb + O2 -> PbO2
  
• or 2Pb + O2 -> 2 PbO
  

altough you can write Pb + 1/2 O2 -> PbO

[eric l]

14. A 9.85g piece of metal is heated to 98.6 degrees Celsius and then the temperature drops to 0.0 degrees Celsius, it is found that 0.43g of ice melted. What is the metal’s specific heat?

 

16. Suppose you add 100.0g of H2O at 65.0 degrees Celsius to 100.0g of ice at 0.0 degrees Celsius. The water cools to 0.0 degrees Celsius. When the ice and water temperature get to a uniform temperature of 0.0 degrees Celsius, how much ice has melted?

 

:lol:

 

I suppose somewhere in the test you must be given the heat of fusion for water. I remember that it is roughly 0.25 J/g - which was precise enough for estimations, but I suppose you were given a more precise number.

 

Question 16

For the temperature drop from 65°C to 0°C you need the specific heat of water - this again is converting calories into Joules (or vice versa).

 

0.1 kg of water cooling from 65°C to 0°C will give you 4.187*0.1*65 J or 27,2155 J; this will melt (aproximately) 108 g of ice.

 

Question 14

Melting 0.43 g of ice would require about 0.11 J. This would come from cooling 9.85 g of metal over 98.6°C. The specific heat would be 0.11J/(9.85g*98.6°C) or about 0.01J/g*°C (which seems extremely low to me - are you sure of all the data ?)

 

I hope I have been of some help