It would be C, because Ionic bonds have to deal with valence electrons ( the outer shell ones)
Using the relative atomic weights of both copper and sulfur ie copper = 63.55 and sulfur is 32.06 so 63.55+32.06=95.56 total mass and so of this, copper = 63.55/95.56=66.4%. So to get 10 grams of copper, use the formula 10g=66.4%xCuS so CuS=10/0.664=15.06 grams of CuS.
Answer:
7000 kg*m/s E
Explanation:
Momentum formula: p=mv
m=200kg
v=35 m/s East
p=(200kg)(35m/s E)
m=7000 kg*m/s E
If you want to simplify it further, m=7*10^3 kg*m/s E
We cannot solve this problem without using empirical data. These reactions have already been experimented by scientists. The standard Gibb's free energy, ΔG°, (occurring in standard temperature of 298 Kelvin) are already reported in various literature. These are the known ΔG° for the appropriate reactions.
<span>glucose-1-phosphate⟶glucose-6-phosphate ΔG∘=−7.28 kJ/mol
fructose-6-phosphate⟶glucose-6-phosphate ΔG∘=−1.67 kJ/mol
</span>
Therefore, the reaction is a two-step process wherein glucose-6-phosphate is the intermediate product.
glucose-1-phosphate⟶glucose-6-phosphate⟶fructose-6-phosphate
In this case, you simply add the ΔG°. However, since we need the reverse of the second reaction to end up with the terminal product, fructose-6-phosphate, you'll have to take the opposite sign of ΔG°.
ΔG°,total = −7.28 kJ/mol + 1.67 kJ/mol = -5.61 kJ/mol
Then, the equation to relate ΔG° to the equilibrium constant K is
ΔG° = -RTlnK, where R is the gas constant equal to 0.008317 kJ/mol-K.
-5.61 kJ./mol = -(0.008317 kJ/mol-K)(298 K)(lnK)
lnK = 2.2635
K = e^2.2635
K = 9.62
