Ch 12.1 Help
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Ch 12.1 Help
Question 1
1) A temperature is quoted as 60.16 oF. What is it in oC (see sheet 3)?
Answer: 15.644444444444442
How to solve:
Tc = (5/9)*(Tf32)
Tc = temperature in degrees Celsius
Tf = temperature in degrees Fahrenheit
Question 2
2) A temperature difference is quoted as 22.27 oC. What is the temperature difference quoted in oK (see sheet 3) ? Indicate with a positive (negative) sign whether the relation between DT in oC and oK is (not) the same for DT in oC and oF.
Answer: 22.27
How to solve:
Change in temperature will be the same in degrees Kelvin.
Question 3
3) A substance with a density of 0.5031 g/cm3 occupies a volume of 2.905 cm3. What is the number of Atomic Mass Units (see sheet 6) ? Indicate with a negative (positive) sign whether the chemical composition of the substance has (does not have) to be known.
Answer: 8.80425e+23
How to solve:
Question 4
4) What is the mass of 2.828 moles of H2O in kg (see sheet 6,8') ? Indicate with a negative (positive) sign whether the mass of the same number of hydrogen moles and oxygen moles is (not) the same.
Answer: 0.050903999
How to solve:
Mass of H2O in 1 mole = 18g = .018kg
Set up a proportion then solve for the mass
(mass of substance/moles of substance) = (molecular mass of substance/one mole)
(mass/2.828 moles) = (.018kg/1 mole)
mass of 2.828 moles = 0.050903999
1) A temperature is quoted as 60.16 oF. What is it in oC (see sheet 3)?
Answer: 15.644444444444442
How to solve:
Tc = (5/9)*(Tf32)
Tc = temperature in degrees Celsius
Tf = temperature in degrees Fahrenheit
Question 2
2) A temperature difference is quoted as 22.27 oC. What is the temperature difference quoted in oK (see sheet 3) ? Indicate with a positive (negative) sign whether the relation between DT in oC and oK is (not) the same for DT in oC and oF.
Answer: 22.27
How to solve:
Change in temperature will be the same in degrees Kelvin.
Question 3
3) A substance with a density of 0.5031 g/cm3 occupies a volume of 2.905 cm3. What is the number of Atomic Mass Units (see sheet 6) ? Indicate with a negative (positive) sign whether the chemical composition of the substance has (does not have) to be known.
Answer: 8.80425e+23
How to solve:
Question 4
4) What is the mass of 2.828 moles of H2O in kg (see sheet 6,8') ? Indicate with a negative (positive) sign whether the mass of the same number of hydrogen moles and oxygen moles is (not) the same.
Answer: 0.050903999
How to solve:
Mass of H2O in 1 mole = 18g = .018kg
Set up a proportion then solve for the mass
(mass of substance/moles of substance) = (molecular mass of substance/one mole)
(mass/2.828 moles) = (.018kg/1 mole)
mass of 2.828 moles = 0.050903999
Daisy Guest
ques 3
So for three, I just used the numbers from the given question above and I get an answer pretty close to the one provided, but I don't even know if this is the way the notes say how to do it because I didn't look at them, i just used chemistry knowledge.
So i think you'd have to find the mass first
D= m/v , so D*v = mass
You should get a mass of about 1.4615
Then you do 1.4615 grams x (6.02214e23) = 8.801e23
The third decimal place is off so I'm not sure exactly what to do but maybe someone else can work off of this to find the right way?
And my guess is that it would be positive since I didn't even know the composition while attempting this. Any input?
So i think you'd have to find the mass first
D= m/v , so D*v = mass
You should get a mass of about 1.4615
Then you do 1.4615 grams x (6.02214e23) = 8.801e23
The third decimal place is off so I'm not sure exactly what to do but maybe someone else can work off of this to find the right way?
And my guess is that it would be positive since I didn't even know the composition while attempting this. Any input?
Kathleen Guest
question 3
for this question u=m/(1.66e27 kg).
so find mass in kg, which is m=(rho*Volume)/1000
so find mass in kg, which is m=(rho*Volume)/1000
guest145 Guest
Q3
...and does anyone know how we are supposed to enter that answer on MapleTA for the quiz. Im not sure if im getting it wrong because im not entering in my ans right...thank
meg Guest
Re: Ch 12.1 Help
Yeah I'm not getting question 3 right either. I've tried it in #.###e+# format, and #.###e# format
Kathleen Guest
Question 3
Okay, so Question 3...
Do the formula that is provided above  Density x Volume/1000 = X
Then do X/(1.66e27) = your answer
I don't know why it is 1.66e27, maybe guest145 could explain where they got that?
And you have to actually put your whole answer in...so this was my question:
A substance with a density of 1.1953 g/cm3 occupies a volume of 4.184 cm3. What is the number of Atomic Mass Units
1.1953x4.1484/1000 = .0050011352
.0050011352/ (1.66e27) = 3.012732048e24
The answer I actually put into Maple TA that was graded as correct was: 3012732048000000000000000 , which means you have to put all the numbers from your calculator plus any additional zeroes to add up to 24 places after the decimal point
Do the formula that is provided above  Density x Volume/1000 = X
Then do X/(1.66e27) = your answer
I don't know why it is 1.66e27, maybe guest145 could explain where they got that?
And you have to actually put your whole answer in...so this was my question:
A substance with a density of 1.1953 g/cm3 occupies a volume of 4.184 cm3. What is the number of Atomic Mass Units
1.1953x4.1484/1000 = .0050011352
.0050011352/ (1.66e27) = 3.012732048e24
The answer I actually put into Maple TA that was graded as correct was: 3012732048000000000000000 , which means you have to put all the numbers from your calculator plus any additional zeroes to add up to 24 places after the decimal point
Kathleen Guest
question 3
well, of course i can explain. The answer is indeed in slide 6, in which 1 u=1.66 e27kg, and we all know that u is the atomic mass unit!
guest145 Guest
Re: Ch 12.1 Help
lol thank you, I figured that but I haven't actually printed out or even looked at the slides for these quizzes yet
k Guest
Re: Ch 12.1 Help
can anyone who got question 3 correct post what you did because i have tried using the equation posted and cant get the right answer. thanks
Question Guest
#3
as an old chem lab Prof once said,"Think about your units."
you are multiplying g/cm^3 and cm^3. cm^3 cancel out, you are only left with grams. Grams must be converted then to kg by multiplying by 0.001 or (as posted above) divide by 1000. Entering your answer by moving the decimals and having all of those zero's is accepted by MapleTA, so count carefully before your enter your answer!
you are multiplying g/cm^3 and cm^3. cm^3 cancel out, you are only left with grams. Grams must be converted then to kg by multiplying by 0.001 or (as posted above) divide by 1000. Entering your answer by moving the decimals and having all of those zero's is accepted by MapleTA, so count carefully before your enter your answer!
super mo Guest
#3
if u = 1.66*10^27kg and we know the mass, why wouldnt we do (1.66*10^27)*(mass in kg)
Above it says mass/(1.66*10^27)
Above it says mass/(1.66*10^27)
guest 23 Guest
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