There are 2071.4662 grams of glucose in 11.5 moles.
Per 1 mole there are 180.15588 grams of glucose. 180.5588 x 11.5 =2076.4262
Answer : The concentration of 
at equilibrium is 0 M.
Solution : Given,
Concentration of = 0.0200 M
Concentration of 
= 1.00 M
The given equilibrium reaction is,
![Fe^{3+}(aq)+3C_2O_4^{2-}(aq)\rightleftharpoons [Fe(C_2O_4)_3]^{3-}(aq)](https://tex.z-dn.net/?f=Fe%5E%7B3%2B%7D%28aq%29%2B3C_2O_4%5E%7B2-%7D%28aq%29%5Crightleftharpoons%20%5BFe%28C_2O_4%29_3%5D%5E%7B3-%7D%28aq%29)
Initially conc. 0.02 1.00 0
At eqm. (0.02-x) (1.00-3x) x
The expression of 
will be,
![K_c=\frac{[[Fe(C_2O_4)_3]^{3-}]}{[C_2O_4^{2-}]^3[Fe^{3+}]}](https://tex.z-dn.net/?f=K_c%3D%5Cfrac%7B%5B%5BFe%28C_2O_4%29_3%5D%5E%7B3-%7D%5D%7D%7B%5BC_2O_4%5E%7B2-%7D%5D%5E3%5BFe%5E%7B3%2B%7D%5D%7D)
By solving the term, we get:
Concentration of at equilibrium = 0.02 - x = 0.02 - 0.02 = 0 M
Therefore, the concentration of at equilibrium is 0 M.
So diffusion is inversely proportional to mass !
so as mass of the particle increases, diffusion decreases !
Answer:
[H_3 O^+] = 1.0 ×10^-13
Explanation:
If we multiply the left side we get, 1e - 13. We add 13 to the right side while subtracting the remaining value from the left side (H3O+) than combine like terms. As you will get pH = 13.00
From the statement of Hess' law, the enthalpy of the reaction A---> C is +90 kJ
<h3>What is Hess' law?</h3>
Hess' law of constant heat summation states that for a multistep reaction, the standard enthalpy of reaction is always constant and is independent of the pathway or intermediate routes taken.
From Hess' law, the enthalpy change for the reaction A ----> C is calculated as follows:
A---> C = A ---> B + B ---> C
ΔH of A---> C = 30 kJ + 60 kJ
ΔH = 90 kJ
Therefore, the enthalpy of the reaction A---> C is +90 kJ
The above reaction A---> C can be shown in the enthalpy diagram below:
A -------------------> C (ΔH = +90 kJ)
\ /
\ / (ΔH = +60 kJ)
(ΔH = +30 J) \ /
> B
Learn more about enthalpy and Hess law at: brainly.com/question/9328637
