<span>HNO2 =====> H+ + NO2-
</span>I<span>nitial concentration</span> = 0.311
<span>C = -x,x,x </span>
<span>E = 0.311-x,x,x
</span>KNO2 ====>K+ + NO2-
<span>Initial concentration = 0.189 </span>
<span>C= -0.189,0.189,0.189 </span>
E = 0,0.189,0.189
Energy is stored in chemical bonds during photosynthesis.
During photosynthesis, the radiant energy from the sun is converted to chemical energy in carbohydrates.
Inorganic materials in the form of carbon dioxide and oxygen combine to form carbohydrates in the presence of radiant energy according to the equation below:
The energy is thus, stored in chemical bonds in the carbohydrate and this is what is oxidized during respiration to release the locked energy.
More on photosynthesis can be found here: brainly.com/question/1388366
<u>Answer:</u> The value of equilibrium constant for the given reaction is 56.61
<u>Explanation:</u>
We are given:
Initial moles of iodine gas = 0.100 moles
Initial moles of hydrogen gas = 0.100 moles
Volume of container = 1.00 L
Molarity of the solution is calculated by the equation:
Equilibrium concentration of iodine gas = 0.0210 M
The chemical equation for the reaction of iodine gas and hydrogen gas follows:
<u>Initial:</u> 0.1 0.1
<u>At eqllm:</u> 0.1-x 0.1-x 2x
Evaluating the value of 'x'
The expression of 
for above equation follows:
![K_c=\frac{[HI]^2}{[H_2][I_2]}](https://tex.z-dn.net/?f=K_c%3D%5Cfrac%7B%5BHI%5D%5E2%7D%7B%5BH_2%5D%5BI_2%5D%7D)
![[HI]_{eq}=2x=(2\times 0.079)=0.158M](https://tex.z-dn.net/?f=%5BHI%5D_%7Beq%7D%3D2x%3D%282%5Ctimes%200.079%29%3D0.158M)
![[H_2]_{eq}=(0.1-x)=(0.1-0.079)=0.0210M](https://tex.z-dn.net/?f=%5BH_2%5D_%7Beq%7D%3D%280.1-x%29%3D%280.1-0.079%29%3D0.0210M)
![[I_2]_{eq}=0.0210M](https://tex.z-dn.net/?f=%5BI_2%5D_%7Beq%7D%3D0.0210M)
Putting values in above expression, we get:
Hence, the value of equilibrium constant for the given reaction is 56.61
Answer:
F2 is the limiting reactant
27.6 grams of NaF is produced.
Explanation:
Balance the equation first.
2Na+ F2 ---> 2NaF
To find the limiting reactant, solve for how much NaF can be produced with Na and F2
12.5g F2 x (1 mole F2/ 38.00 grams F2)x (2 mole NaF/ 1 mole F2)
=0.658 moles NaF
16.2g Na x (1 mole Na/ 22.99 grams Na)x (2 mole NaF/ 2 mole Na)
=0.705 moles NaF
Since F2 produced the least NaF, F2 is the limiting reactant.
Now, to find how much NaF there is, use the moles solved above with F2 as the limiting reactant.
0.658 moles NaF x (41.99 grams NaF/ 1 mole NaF)= 27.6 moles NaF
27.6 moles of NaF would be theoretically produced.
Answer:
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