Question

Can someone help?
http://oi63.tinypic.com/2cp6wsi.jpg

Answer 1

@LeibyStrauss

Answer 2

@whpalmer4

Answer 3

@Data_LG2

Answer 4

For the b portion of it I got 4.17x10^2

Answer 5

@CShrix

Answer 6

What is the "b portion" ? O.o

anyway, I'm not sure how to explain it but I'll try.

First you have to calculate the amt of heat released form the combustion of benzoic acid using its enthalpy of combustion (3226.7 kJ/mol) and the amt of benzoic acid burned in the calorimeter (3.0365g). Convert the amt of burned benzoic acid into moles and multiply it by the enthalpy of combustion.

Answer 7

Oh whoops. That's for another question I'm on. I have two left I'm trying to get done.

Answer 8

But I need help with the one I posted too.

Answer 9

How do you do the atm?

Answer 10

http://oi64.tinypic.com/2eq7u2u.jpg

Answer 11

atm? amt means amount , i just shortened it :3

Answer 12

Oh!

Answer 13

let's stick with the first one that you posted first

Answer 14

Alright

Answer 15

So do I need to do q=C deltaT + ms deltaT

Answer 16

no, not yet... we are calculating for the heat released from the combustion of benzoic acid first. Follow what I wrote above. Then, using that formula, we will solve for the heat energy of water and we will subtract it from combustion of the acid. This will equal to the heat energy of the bomb calorimeter, which we will then use to calculate for the heat capacity :P


So what did you get for the heat released?

Answer 17

3.0365g * 1 mol / 122.12 g

Answer 18

yes, so how many moles were burned?

Answer 19

.025 mol

Answer 20

Is that right?

Answer 21

yes, then multiply it by the enthalpy of combustion, what will you get? (include the unit)

Answer 22

Is the -3226.7 that?

Answer 23

3226.7 kJ/mol

Answer 24

0.025 mol x (3226.7 kJ/mol) = ?

Answer 25

Is the -3226.7 that?

Answer 26

80.67

Answer 27

and the unit for that is?

Answer 28

80.67 kJ

Answer 29

Then do I divide that by the change of oC which would be 1.83 right? So 80.67kJ/1.83oC=44.08 kJ/oC

Answer 30

no not yet....

you still have to find the heat energy of water (in the calorimeter) using this formula: \(\sf q=m C \Delta T \)

where m is the amt of water around the bomb: 2250 g
C is the specific heat capacity of water: 4.18 J/g C
and \(\sf \Delta T \) is the temperature rise of water and the calorimeter: 24.67°C-22.84°C

Answer 31

q=2250g*4.18*1.83

Answer 32

yeahh so what will you get? (please include the proper unit)

Answer 33

q=17211.15 J oC

Answer 34

That's the right units since grams cancel out, right?

Answer 35

it should only be in Joules, remember the Celsius unit from the specific capacity and the temperature will also get cancelled out.

Now, convert it to kJ , just to make sure we have the proper unit when we subtract it from the heat energy released from the combustion of the acid that you got before.

Answer 36

17.21115 kJ

Answer 37

Would the heat energy release be the change in temperature?

Answer 38

what do you mean? heat energy released is not equal to the change in temperature. they are related but not equal.

Answer 39

So am I needing the 80.67kJ?

Answer 40

yes. Now let's find the heat energy in the bomb calorimeter by using the ffg values:

heat released from the combustion of benzoic acid: 80.67 kJ
heat energy of water (in the calorimeter): 17.21115 kJ

Just simply subtract those two :)

Answer 41

80.67-17.21115=63.46kJ

Answer 42

right, now for the final step :)

to get the heat capacity: heat energy in the bomb calorimeter / temperature change

Answer 43

63.46/1.83=34.68 kJ/oC

Answer 44

yes , it seems right to me

Answer 45

What about sig figs? Should it be 3 so 34.7 kJ/oC?

Answer 46

I'm not really sure about the rules that you are follwing about the sig figs because here, we depend it on the lowest sig fig from the given values on the question...what did your teacher say about it though?

Answer 47

The lowest deal too I believe.

Answer 48

The 3 sig figs was right. Do you have time to help with the other one?

Answer 49

Sorry, I'm afraid not, I need to sleep now, I still have to get up early tomorrow. it is already 12 am here. Just post your question and probably other users can also help you :) goodluck!

Answer 50

Alright. Thanks for helping out!