120 volt divided by 22 ampere
= 5.4545454545455 ohm (Ω)
P = V × I
= 120 volt × 22 ampere
= 2640 watt (W)
Answer:
0.2 kcal/mol is the value of
for this reaction.
Explanation:
The formula used for is:


where,
= Gibbs free energy for the reaction
= standard Gibbs free energy
R =Universal gas constant
T = temperature
Q = reaction quotient
k = Equilibrium constant
We have :
Reaction quotient of the reaction = Q = 46
Equilibrium constant of reaction = K = 35
Temperature of reaction = T = 25°C = 25 + 273 K = 298 K
R = 1.987 cal/K mol

![=-1.987 cal/K mol\times 298 K\ln [35]+1.987 cal/K mol\times 298K\times \ln [46]](https://tex.z-dn.net/?f=%3D-1.987%20cal%2FK%20mol%5Ctimes%20298%20K%5Cln%20%5B35%5D%2B1.987%20cal%2FK%20mol%5Ctimes%20298K%5Ctimes%20%5Cln%20%5B46%5D)

1 cal = 0.001 kcal
0.2 kcal/mol is the value of
for this reaction.
Answer:The simplify command is used to apply simplification rules to an expression. The simplify routine searches the expression for function calls, square roots, radicals, and powers and invokes the appropriate simplification procedures. For detailed information on the simplify command, see simplify/details.
Explanation:
Answer:
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Explanation:
I am looking forward to be a hero bocah kok tolol of G if
Answer:
The correct solution is "21024 KWh/degree day".
Explanation:
The given query is incomplete. Below is the attachment of complete query is provided.
The given values are:
Indoor design temperature:
= 70°F
Now,
According to the question,
The heat loss annually will be:
= 
= 
Degree days will be:
= 
= 
Hence,
Annual KWh use will be:
= 
On substituting the values, we get
= 
= 