topscore GC #56

This forum made possible through the generous support of SDN members, donors, and sponsors. Thank you.
Josh779

Josh779

Full Member
10+ Year Member
Joined
Dec 20, 2010
Messages
145
Reaction score
0
Points
0
  1. Dental Student
Advertisement - Members don't see this ad
What is the temperature change of a 10g sample with specific heat capacity of 20cal/gC that has absorbed 100 cal?

A. 0.5 C
B. 2 C
C. 20 C
D. 30 C
E. 40 C

Solution: 100cal=10g x 20 cal/gC x dT ;dT=0.5C

so i tried using q=mcdT but wasn't sure where the 100 cal was suppose to be used. Do we set it equal to 100cal because that was already the change in energy?
 
Sort by date Sort by votes
Josh779 said:
What is the temperature change of a 10g sample with specific heat capacity of 20cal/gC that has absorbed 100 cal?

A. 0.5 C
B. 2 C
C. 20 C
D. 30 C
E. 40 C

Solution: 100cal=10g x 20 cal/gC x dT ;dT=0.5C

so i tried using q=mcdT but wasn't sure where the 100 cal was suppose to be used. Do we set it equal to 100cal because that was already the change in energy?
Click to expand...

yes, you use q = mcdT
They are telling us that it uses 100 calories; so set it equal to 100 and solve of dT...👍
 
Upvote 0 Downvote
What volume of HCL was added if 20ml of 1M NaOH is titrated with 1M HCl to produce a pH=2

A. 10.2 ml
B. 20.2 ml
C 30.4 ml
D. 35.5 ml
E. None of these

Solution Both acids and bases are 1M therefore, 20mL of HCL will neutralize 20 mL of NaOH. This produces 40ML of .5M NaCl (This i understand)...but this part confuses me. (1M)(xmL)=(.01M)(40+xmL) ; x=.4+.01x ; .99x=.4 ; x=.4 mL ; 40.4 is the final amount, therefore 40.4-20mL=20.4mL of acid added.

Okay so where did the .01M in bold come from?
I calculated (20mL)x(1L/1000mL)x(1mole/1L)=.020 moles, which should give me .02M or is it halved because we doubled the concentration so it becomes .01M? need some help! thanks.
 
Upvote 0 Downvote
Question was asked before. This is how it was solved:

Given:
pH = 2
[H+] = 10^(-2)
[H+] = 0.01 M
-> This is where the 0.01 M came from

Since HCl + NaOH -> H2O + NaCl, it's a neutralization reaction.

20 mL 1M HCl + 20 mL 1M NaOH = 40 mL of water

If we keep adding HCl, the pH goes down as HCl will dissociate into H+ and Cl-.

How much extra HCl do we add?
Given:
Volume ~ 40 mL
Concentration = 0.01 M
moles = ?

c = moles/volume
moles = (0.040L)(0.01M)
moles = 0.0004 mol

We know the concentration of HCl is 1M.
So the amount of 1M HCl needed to add to the solution is..

c=moles/volume
volume = moles/concentration
volume = 0.0004 mol/1.0M
volume = 0.0004 L = 0.4 mL


Total Volume of HCl added = neutralizing volume + volume needed to lower pH
= 20 mL + 0.4 mL
= 20.4 mL

Hope this helps...
 
Upvote 0 Downvote
AlbinoPolarBear said:
Question was asked before. This is how it was solved:

Given:
pH = 2
[H+] = 10^(-2)
[H+] = 0.01 M
-> This is where the 0.01 M came from

Since HCl + NaOH -> H2O + NaCl, it's a neutralization reaction.

20 mL 1M HCl + 20 mL 1M NaOH = 40 mL of water

If we keep adding HCl, the pH goes down as HCl will dissociate into H+ and Cl-.

How much extra HCl do we add?
Given:
Volume ~ 40 mL
Concentration = 0.01 M
moles = ?

c = moles/volume
moles = (0.040L)(0.01M)
moles = 0.0004 mol

We know the concentration of HCl is 1M.
So the amount of 1M HCl needed to add to the solution is..

c=moles/volume
volume = moles/concentration
volume = 0.0004 mol/1.0M
volume = 0.0004 L = 0.4 mL


Total Volume of HCl added = neutralizing volume + volume needed to lower pH
= 20 mL + 0.4 mL
= 20.4 mL

Hope this helps...
Click to expand...
Thanks! I was looking at similar problems people posted but this made everything click. This may seem like a dumb question, but do we HAVE to convert to liters when figuring out the # of moles given the volume and molarity?
 
Upvote 0 Downvote
Josh779 said:
Thanks! I was looking at similar problems people posted but this made everything click. This may seem like a dumb question, but do we HAVE to convert to liters when figuring out the # of moles given the volume and molarity?
Click to expand...

well molarity is mol/L so, yes, you do have to convert to L at one point or another otherwise the units won't cancel out to give you mol. if you keep it in mL, you can still do the problem, but understand that you're getting mmol, not mol, out of the calculation. if you can keep track, that's fine.

M x mL = mmol

...but easiest thing would be to always convert to L.
 
Upvote 0 Downvote
Ok cool, thats what I thought. I have just done a few dilution problems where it wasn't necessary even though we were using the M1V1=M2V2 equation.
 
Upvote 0 Downvote
Josh779 said:
Ok cool, thats what I thought. I have just done a few dilution problems where it wasn't necessary even though we were using the M1V1=M2V2 equation.
Click to expand...

yeah in M1V1 = M2V2, you just have to use the same units on both sides and everything cancels because it's the same on both sides, so it's fine. but if you want to get a certain quantity out, in this case mol, then make sure your units are kosher
 
Upvote 0 Downvote
rockclock said:
yeah in M1V1 = M2V2, you just have to use the same units on both sides and everything cancels because it's the same on both sides, so it's fine. but if you want to get a certain quantity out, in this case mol, then make sure your units are kosher
Click to expand...
Oohh that's where I was getting mixed up. Thanks!
 
Upvote 0 Downvote
You must log in or register to reply here.

Similar threads

C
Advice for GC DAT Bootcamp
Replies
0
Views
859
CalcioRoma2003
C
M
2025 DAT Breakdown (560 AA)
Replies
1
Views
2K
Chicago_predental
C
username16790135
DAT Breakdown (540 AA/560 TS)
Replies
0
Views
1K
username16790135
username16790135
T
DAT breakdown and study tips (520AA/560TS)
Replies
0
Views
2K
thankful_turkey1
T
F
Dat breakdown 23AA, 23TS, 21PAT, 25 OC, 23 Bio, 21 GC, 23 QR, 23 RC
Replies
1
Views
2K
DAT Destroyer
DAT Destroyer
Top Bottom