Chemical Forums
Chemistry Forums for Students => High School Chemistry Forum => Topic started by: mr cool on February 05, 2009, 02:08:25 PM
-
I can tell this worksheet is going to be a nightmare because im stuck on number one....
What mass of ice was converted from -20.00*C to liquid water at 78.00*C by 7560 calories?
i have gotten this far..
(.5 cal/g)(M)(0*C - 20.00*C) + (80 cal/g)(M) + (1 cal/g)(M)(78.00-0)= 7560 calories
| | |
v v v
10g(M) + 80g(M) + 78.00g(M) = 7560 calories
is this correct up to this point?
i have no clue what to do next, i know it involves algebra though :P
-
Dear reezyfbaby;
Your first equation line is correct!
But units in your last equation are wrong!, - they must be: cal/g * (Mg) = cal !
On the left you have written: a * (M g) + b *( Mg) + c * (Mg) = 7560 cal.
Can you do the little algebra on the left side and finally substitute correctly backward: a = 10 cal/g, and so on?
I hope to have been of help to you.
Good Luck!
ARGOS++
-
so....
10 cal/g (M) + 80 cal/g (M) + 78.00 cal/g (M) = 7560 ?
-
Dear reezyfbaby;
Not exactly! - But already much better!
Can you count the left side together and keep the units, also for your M?
Good Luck!
ARGOS++
-
Dear reezyfbaby;
Not exactly! - But already much better!
Can you count the left side together and keep the units, also for your M?
Good Luck!
ARGOS++
are you asking me to distribute the 10 cal/g into the (M) ?
-
-
168.00 cal/g (M g)?
-
Dear reezyfbaby;
Exact!
If you now complete the equation you get: 168.0 cal/grams * M grams = 7560 cal.
Now you can solve for M grams.
Good Luck!
ARGOS++
-
would i take
168.0 cal/grams * M grams = 7560 cal
and divide it by 7560 cal to get rid of the cal and then be left with grams?
i got 45.00 g when i divided the two.
just going out on a limb here, algebra is not my forte either.
-
Dear reezyfbaby;
Correct!
Good Luck!
ARGOS++
-
i think i got this next one right so far....
A 150.00 g piece of -50.00*C ice is exposed to enough energy (85.00 kJ) to convert it to liquid water at what temperature?
my work....
(this first part is taking 85.00 kJ into calories...)
85.00 kJ * 100j / 1 kJ * 1 cal / 2.092 = 4063 (i used 2.092 because my chem teacher said 1 cal = 4.1845 J and since we start at ice i split it in half.)
(.5 cal/g)(150.00g)(0*C - -50.00*C) + (80 cal/g)(150.00g) + (1 cal/g)(150.00g)(F - 0*C) = 4063 calories... (the F is my place holder for the final temp)
3750 cal/g + 12000 cal/g + 150.00 cal/g(F - 0*C) = 4063 calories...
15750 cal/g + 150.00 cal/g * F = 4063 calories.
do i add 15750 to 150.00 and get 15900 and divide it by 4063?
-
Why would you divide by 2 just because you are starting with ice?
-
-
-
-
-
-
-
First:[/b] don't i use half of 4.184 because, i start in ice?
No you don't. Were you taught that, or did you read that somewhere?
Also, I really do not recommend using an online conversion tool unless you can convert the units by hand first. Almost all work in chemistry requires being able to convert units, or use dimensional analysis (which is what you can use to convert units). If you are not comfortable with those techniques it is going to make your life only harder later on (even later in the semester).
-
no im pretty good at dimensional analysis it's just that i was frustrated so i just took the easy way out :(
-
-
-
Dear reezyfbaby;
Sorry, - Yes!
Please do it only stepwise! Please start with First: .
Good Luck!
ARGOS++
-
Dear reezyfbaby;
Sorry, - Yes!
Please do it only stepwise! Please start with First: .
Good Luck!
ARGOS++
oh you want me to do #1 first ? oki...
85000 kj / 4.1845 = 20313
P.S. Here is another problem i was trying in the meantime...
What is the final temperature of a 50.00g sample of ice at -10.00*C if it is exposed to 59.75 J of energy and remains ice?
(.5 cal/g)(50.00g)(F- -10.00*C) = 250 cal
25 cal (F - -10*C) =250 cal
25 cal(F) + -250 cal = 250 cal
-225 cal = 250cal
250cal / -225 cal = 0.9 *C
?
-
-
-
-
4563 left?
-
-
-
Dear reezyfbaby;
Bravo!, - You did it!
Please think once again how we did it!
For your new Question: You mixed Joules up with calories!
Good Luck!
ARGOS++
-
Dear reezyfbaby;
Bravo!, - You did it!
Please think once again how we did it!
For your new Question: You mixed Joules up with calories!
Good Luck!
ARGOS++
where did i mix them up? i don't see joules anywhere in that problem.
-
Dear reezyfbaby;
Can you identify: 59.75 J in your question/problem? J stands for Joules!
Good Luck!
ARGOS++
-
Dear reezyfbaby;
Can you identify: 59.75 J in your question/problem? J stands for Joules!
Good Luck!
ARGOS++
yea but i converted 59.75 joules into 250 calories. i know that J stands for Joules.
-
Dear reezyfbaby;
Remember!: For that you have to divide it, - not multiply!
Good Luck!
ARGOS++
-
Dear reezyfbaby;
Remember!: For that you have to divide it, - not multiply!
Good Luck!
ARGOS++
oh, that would explain it lol :)
so this is my first step?
(.5 cal/g)(50.00g)(F - -10.00*C) = 14.28 cal?
-
-
-
-
-
-
w/e i gave up on that problem because it frustrated me and i started the last one....
26.46 KJ of energy is released from 10.00 grams of steam as it condenses from 125.00*C. What is the final temerature of the condensed water?
(.5 cal/g *C)(10.00g)(100-125) + (-540 cal/g)(10.00g)+(1.0 cal/g)(10.00g)(F-100*C) = 6323 calories
-125 cal + 5400 cal + 10 cal/g(F-100*C)=6323 calories
(10.00 cal/g)(Fo - 100*C) = 1038 calories
10.00 cal/g Fo
1038 calories
------------- = 103.8 *C
10.00 cal/g Fo
-
-