<span>Answer:
Moles Ca(NO3)2 = 100 x 0.250 / 1000 = 0.025
Ca(NO3)2 >> Ca2+ + 2NO3-
Moles NO3- = 2 x 0.025 = 0.05
Moles HNO3 = 400 x 0.100 / 1000 = 0.04
Total moles = 0.05 + 0.04 = 0.09
Total volume = 500 ml = 0.500 L
M = 0.09 / 0.500 = 0.18</span>
When a female animal and a male see each other the male would show off and they woulf matr and make a baby then the babys will grow up and make more
Answer:
a. Heat is often absorbed during reactions.
c. Heat is often released during chemical reactions.
Explanation:
Thermochemistry -
The study of thermochemistry involve the change in the amount of heat , during any physical or chemical process , is referred to as thermodynamics .
The focus of thermochemistry is on the changing amount of energy in the form of heat , on the system with respect to the surroundings .
The process like boiling , melting , sublimation , may require energy or releases energy , and hence are studied under thermochemistry .
Hence , from the given question ,
The correct options are - a , c.
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
<u>Answer:</u> The equilibrium concentration of HCl is 
<u>Explanation:</u>
We are given:
Moles of
= 0.564 moles
Volume of vessel = 1.00 L
Molarity is calculated by using the equation:

Molarity of 
The given chemical equation follows:

<u>Initial:</u> 0.564
<u>At eqllm:</u> 0.564-x x x
The expression of
for above equation follows:
![K_c=[NH_3][HCl]](https://tex.z-dn.net/?f=K_c%3D%5BNH_3%5D%5BHCl%5D)
The concentration of pure solid and pure liquid is taken as 1.
We are given:

Putting values in above equation, we get:

Negative sign is neglected because concentration cannot be negative.
So, ![[HCl]=2.26\times 10^{-3}M](https://tex.z-dn.net/?f=%5BHCl%5D%3D2.26%5Ctimes%2010%5E%7B-3%7DM)
Hence, the equilibrium concentration of HCl is 