Answer:
-209 kJ
Explanation:
I did the math. You're welcome ;)
Answer:
403 mL
Explanation:
First, I will assume that the mole is 1, because you are not specifing this.
Now, with the innitial data, we need to get the pressure:
T = 65+273 = 338 K
V = 500 / 1000 = 0.5 L
Now if:
PV = nRT
Then:
P = nRT/V and V = nRT/P
Let's calculate the P:
P = 1 * 0.082 * 338 / 0.5 = 55.432 atm
The standard temperature is 0° C or 273 K so, the volume is:
V = 1 * 0.082 * 273 / 55.432
V = 0.40384 L or simply 403.84 mL
Hey there!:
8) ΔTb = i*Kb*m
m is molality
Since same number of mol is added to same amount of water in both cases
m will be same for both
is 1 for glucose since it is covalent compound
is 4 of Al(NO3)3 as it breaks into 1 Al₃⁺ and 3 NO₃⁻
So, ΔTb will be 4 times in aluminum nitrate case
So, boiling point will change by 4ºC
9) use Q = m* L
L = heat of vaporization so:
T1=T2=100ºC
5.40 * 1000 => 5400 cal/g
Q = 5400 / 540
Q = 10 grams
Hope that thlps!
Answer:
el primero es Li, y el segundo es ... oh, está cortado
Explanation:
