Chemistry help

aretardedorange · 02-15-2011, 09:54 PM

I need to get these by midnight tonight but I have no idea how to do them.

Problem 1-

Problem 2-

skinnycow · 02-15-2011, 10:14 PM

are you working with Q=mcAT?

BrokenVert · 02-15-2011, 10:17 PM

Quote:

Originally Posted by skinnycow

are you working with Q=mcAT?

sounds like it. Just remember to do your phase changes. Thats a diff formula.

skinnycow · 02-15-2011, 10:17 PM

1) 63360 degrees C

2) 114153.62 degrees C

I think that's right, last time i took chem was in hs lol

BrokenVert · 02-15-2011, 10:18 PM

Quote:

Originally Posted by skinnycow

1) 63360 degrees C

2) 114153.62 degrees C

The answers would be in Joules.

skinnycow · 02-15-2011, 10:21 PM

Quote:

Originally Posted by BrokenVert

The answers would be in Joules.

oops sorry i meant J

TiAg335i · 02-15-2011, 10:23 PM

Quote:

Originally Posted by skinnycow

1) 63360 degrees C

2) 114153.62 degrees C

I think that's right, last time i took chem was in hs lol

geez. i must be a dumbass. i don't remember a thing in high school. i'm sure i took chemistry somewhere in those 4 years. hahaha

i did end up being a biochemistry major and graduating with honors out of college though...

but i still would have to do some mental digging to do those problems, even tho they look simple as pie

aretardedorange · 02-15-2011, 10:24 PM

Yes that formula. But I reboot my computer and it gave me new problems. If you could do these two it would be much appreciated.

Inspired · 02-15-2011, 10:30 PM

-1.2x10^5 for problem 2?
(1200g)(0.901J/gC)(21-133C)

skinnycow · 02-15-2011, 10:30 PM

give me a sec lol

aretardedorange · 02-15-2011, 10:39 PM

Quote:

Originally Posted by twinturbo335

-1.2x10^5 for problem 2?
(1200g)(0.901J/gC)(21-133C)

THANKS!

skinnycow · 02-15-2011, 10:41 PM

1) 0.1222
2) 142718.4

the first one's tricky

on the first one, since the metal is in solution, you'll end up having the same temperature (between water and the metal) which means T-final for the water is also 61.1, so you throw that into the Q = mc(Tf-Ti) equation and you get the energy change by the water, Q = 329.28 since you have Q, which will fit into the Q=mcT on the metal equation, you can put everything in and find c 329.28 = (55.89)(c)(61.1 - 12.9) rearrange the equation and solve for c

Inspired · 02-15-2011, 10:47 PM

Quote:

Originally Posted by skinnycow

1) 63360 degrees C

2) 114153.62 degrees C

I think that's right, last time i took chem was in hs lol

Isn't it -63360? It is t final minus t initial. 21-186

aretardedorange · 02-15-2011, 10:49 PM

Quote:

Originally Posted by twinturbo335

-1.2x10^5 for problem 2?
(1200g)(0.901J/gC)(21-133C)

This one is wrong. What you had before was correct.

aretardedorange · 02-15-2011, 10:49 PM

Quote:

Originally Posted by skinnycow

1) 0.1222
2) 142718.4

the first one's tricky

on the first one, since the metal is in solution, you'll end up having the same temperature (between water and the metal) which means T-final for the water is also 61.1, so you throw that into the Q = mc(Tf-Ti) equation and you get the energy change by the water, Q = 329.28 since you have Q, which will fit into the Q=mcT on the metal equation, you can put everything in and find c 329.28 = (55.89)(c)(61.1 - 12.9) rearrange the equation and solve for c

1. Is incorrect

. Thanks anyway though. Much appreciated!

skinnycow · 02-15-2011, 10:49 PM

Quote:

Originally Posted by twinturbo335

Isn't it -63360? It is t final minus t initial. 21-186

yeah it is

i didn't put the negative cause the question was worded funny 'how much is removed' but yeah it's technically negative

skinnycow · 02-15-2011, 10:51 PM

Quote:

Originally Posted by aretardedorange

1. Is incorrect

. Thanks anyway though. Much appreciated!

awww i'm sorry i was trying to do it as quickly as i could and couldn't really look it through

Inspired · 02-15-2011, 10:51 PM

Quote:

Originally Posted by aretardedorange

This one is wrong. What you had before was correct.

The first one I didn't put to 2 significant figures. thought it would be better than having a whole bunch of #s

aretardedorange · 02-15-2011, 10:53 PM

Quote:

Originally Posted by skinnycow

awww i'm sorry i was trying to do it as quickly as i could and couldn't really look it through

No problem

. I kind of got the hang of these problems (kind of) being the key word lol. Will attempt some of these tomorrow.

Thanks everyone that helped! Once again much appreciated!

Inspired · 02-15-2011, 11:04 PM

Quote:

Originally Posted by aretardedorange

Yes that formula. But I reboot my computer and it gave me new problems. If you could do these two it would be much appreciated.

A 55.89 g sample of metal changes its temperature from 12.9 °C to 61.1 °C when it absorbed energy from a 78.7 g sample of water whose initial temperature was 66.2 °C. What is the specific heat of the metal in J/g·°C?

A skillet made of Aluminum (Al) weighing 1.2 kg is heated on a stove to a temperature of 133 oC. If the skillet is removed from the stove and allowed to cool to room temperature (21oC), how much heat must be removed from the skillet. The specific heat of Aluminum is 0.901 J/(g.oC)

is Q1 .0316?

C=Q/M(deltaT)

First I tried to solve for TF for the 2nd part.
(78.7g)(4.184)(Tf-66.2)

=85J

(55.89)(C)(61.1-12.9)

C=2693.9

85J/2693.9

.0316J/G [C]

skinnycow · 02-15-2011, 11:09 PM

Quote:

Originally Posted by twinturbo335

is Q1 .0316?

C=Q/M(deltaT)

First I tried to solve for TF for the 2nd part.
(78.7g)(4.184)(Tf-66.2)

=85J

(55.89)(C)(61.1-12.9)

C=2693.9

85J/2693.9

.0316J/G [C]

booo it bugs me when i get something wrong!!!!

Inspired · 02-15-2011, 11:10 PM

Quote:

Originally Posted by skinnycow

booo it bugs me when i get something wrong!!!!

I'm not sure BTW. Kind of rusty.....

02-15-2011, 09:54 PM	#1
aretardedorange Banned 177 Rep 609 Posts Drives: E60 Join Date: Nov 2009 Location: Empire State iTrader: (6)	Chemistry help I need to get these by midnight tonight but I have no idea how to do them. Problem 1- Problem 2- Last edited by aretardedorange; 02-15-2011 at 10:59 PM..
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02-15-2011, 10:14 PM	#2
skinnycow condoms prevent minivans 244 Rep 223 Posts Drives: Batmobile Join Date: Nov 2010 Location: Kitchen iTrader: (0)	are you working with Q=mcAT? __________________ I have strong opinions about things that don't matter.
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02-15-2011, 10:17 PM	#4
skinnycow condoms prevent minivans 244 Rep 223 Posts Drives: Batmobile Join Date: Nov 2010 Location: Kitchen iTrader: (0)	1) 63360 degrees C 2) 114153.62 degrees C I think that's right, last time i took chem was in hs lol __________________ I have strong opinions about things that don't matter.
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02-15-2011, 10:24 PM	#8
aretardedorange Banned 177 Rep 609 Posts Drives: E60 Join Date: Nov 2009 Location: Empire State iTrader: (6)	Yes that formula. But I reboot my computer and it gave me new problems. If you could do these two it would be much appreciated. Last edited by aretardedorange; 02-15-2011 at 10:59 PM..
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02-15-2011, 10:30 PM	#9
Inspired Colonel 156 Rep 2,368 Posts Drives: 335i Join Date: Dec 2007 Location: California iTrader: (2)	-1.2x10^5 for problem 2? (1200g)(0.901J/gC)(21-133C) Last edited by Inspired; 02-15-2011 at 10:36 PM..
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02-15-2011, 10:30 PM	#10
skinnycow condoms prevent minivans 244 Rep 223 Posts Drives: Batmobile Join Date: Nov 2010 Location: Kitchen iTrader: (0)	give me a sec lol __________________ I have strong opinions about things that don't matter.
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02-15-2011, 10:41 PM	#12
skinnycow condoms prevent minivans 244 Rep 223 Posts Drives: Batmobile Join Date: Nov 2010 Location: Kitchen iTrader: (0)	1) 0.1222 2) 142718.4 the first one's tricky on the first one, since the metal is in solution, you'll end up having the same temperature (between water and the metal) which means T-final for the water is also 61.1, so you throw that into the Q = mc(Tf-Ti) equation and you get the energy change by the water, Q = 329.28 since you have Q, which will fit into the Q=mcT on the metal equation, you can put everything in and find c 329.28 = (55.89)(c)(61.1 - 12.9) rearrange the equation and solve for c __________________ I have strong opinions about things that don't matter.
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