Answer: E=∆H*n= -40.6kj
Explanation:
V(CO) =15L=0.015M³
P=11200Pa
T=85C=358.15K
PV=nRT
n=(112000×0.015)/(8.314×358.15)
n(Co)= 0.564mol
V(Co)= 18.5L = 0.0185m³
P=744torr=98191.84Pa
T= 75C = 388.15k
PV=nRT
n= (99191.84×0.0185)/(8.314×348.15)
n(H2) = 0.634mol
n(CH30H) =1/2n(H2)=1/2×0.634mol
=0.317mol
∆H =∆Hf{CH3OH}-∆Hf(Co)
∆H=-238.6-(-110.5)
∆H = 128.1kj
E=∆H×n=-40.6kj.
Answer:
5.37 L
Explanation:
To solve this problem we need to use the PV=nRT equation.
First we <u>calculate the amount of CO₂</u>, using the initial given conditions for P, V and T:
- P = 785 mmHg ⇒ 785/760 = 1.03 atm
- T = 18 °C ⇒ 18 + 273.16 = 291.16 K
1.03 atm * 4.80 L = n * 0.082 atm·L·mol⁻¹·K⁻¹ * 291.16 K
We <u>solve for n</u>:
Then we use that value of n for another PV=nRT equation, where T=37 °C (310.16K) and P = 745 mmHg (0.98 atm).
- 0.98 atm * V = 0.207 mol * 0.082 atm·L·mol⁻¹·K⁻¹ * 310.16 K
And we <u>solve for V</u>:
Answer:
Volume = mass/density
Rearrange the equation for Mass:
Mass = Volume x Density
That is one way you can do it
Conceptually, just look at the units, the wood block's density is 0.6<u>g/cm^3</u> while the volume is 2.2 <u>cm^3</u>
So if density is every gram per centimeter cubed, and the volume is at centimeter cubed, the logical thing to do would be to multiply the density by the volume to get the total mass.
0.6g/cm^3 x 2.2cm^3 = 1.32g
<u>Therefore the mass of the block of wood is 1.32g </u>
ΔG > 0
is always true for the freezing of water.
Explanation:
- The freezing of water is only spontaneous when the temperature is fairly small. Over 273 K, the higher value of TΔS causes the sign of ΔG to be positive, and there is no freezing point.
- The entropy decreases as water freezes. This does not infringe the Thermodynamics second law. The second law doesn't suggest entropy will never diminish anywhere.
- Entropy will decline elsewhere, provided it increases by at least as much elsewhere.
