Help :D

[bloop]

Hi please help me with the below questions. Thank you.

[annn1e]

ok, for the first question, i got A
first off,i found how many mols of butanol there is using mols=mass/molar bass
mols is 0.00518,
ratio given for mols/heat of combustion is 1:2670kj, however as only 35% is used (65%lost), the ration is 1:93.45kj
if one mol gives off 93.45kj, 0.00518 mols gives off 0.484071kj
this is your delta H and you can work backwards from here by subbing everything into mcat formula
0.484071= 0.25 x 4.19 x (x-23.5)
when i found x that rounded off to 24 so im assuming thats the answer? praying that im right

[MisterNeo]

Hi please help me with the below questions. Thank you.

The answer is B.
-You find the moles of butanol.
-Multiply by molar heat for the kJ.
-Multiply by 0.35 since 65% was lost.
-Convert to joules.
-Divide by 250 and 4.18.
-Add 23.5
The answer is 24.13 degrees Celsius.
I'm not sure about the second question since Tc-99m decays into Tc-99 in 6 hours, which takes over 200000 years to decay into Ru-99.

[annn1e]

oh it is B for the first one, convert 250 to 0.25 and the answer is 28 whoops...

[Shadowxo]

oh it is B for the first one, convert 250 to 0.25 and the answer is 28 whoops...

Hi
If you're referring to converting the mass from 250g to 0.250kg, remember that in chemistry you (pretty much always) use grams instead of kilograms. The big trick with this question is having only 35% of the energy going into the water and having the molar heat in kJ while E=mc∆T uses J.
n(Butanol) = m/M = 0.384/74.0= 0.00519 mol
Energy released = 0.00519 * 2670 kJ = 13.9 kJ= 1.39*10⁴ J
Energy absorbed by water = 1.39*10⁴*0.35 =4.85* 10³ as it absorbs 35% of it
E=mc∆T
4.85* 10³=250*4.18*∆T
∆T = 4.64ºC
T = T_initial + ∆T = 23.5+ 4.64 = 28.1ºC