The empirical formula for this would be FeS
Answer:
B
Explanation:
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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:
1. The length of the wire is halved.
the resistance of a conductor is directly proportional to it's length (L) as R - L. Thus doubling it's length will double it's resistance, while halving it's length would halve the resistance. Also the resistance of a conductor is inversely proportional to it's cross sectional area.
2. The area of cross-section of the conductor in increased.
On increasing the area of cross-section, resistance decreases. Thus is because resistance is inversely proportional to area.
3. The temperature of the conductor is increased.
With increasing temperature, the resistance of the wire increases as collisions within the wire increases and "slow" the flow of current... Since conductors typically display an increased resistivity with temperature increase, they have a positive temperature coefficient.
Answer: option d
Explanation:
Project teams must be proactive. Each member of the team should empowered and each member can share ideas and experience in order to make the WBS more precise.