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 molarity of KBr in the final solution is 1.42M
Explanation:
We can calculate the molarity of the KBr in the final solution by dividing the total number of moles of KBr in the solution by the final volume of the solution.
We will first calculate the number of moles of KBr in the individual sample before mixing together
In the first sample:
Volume (V) = 35.0 mL
Concentration (C) = 1.00M
Number of moles (n) = C × V
n = (35.0mL × 1.00M)
n= 35.0mmol
For the second sample
V = 60.0 mL
C = 0.600 M
n = (60.0 mL × 0.600 M)
n = 36.0mmol
Therefore, we have (35.0 + 36.0)mmol in the final solution
Number of moles of KBr in final solution (n) = 71.0mmol
Now, to get the molarity of the final solution , we will divide the total number of moles of KBr in the solution by the final volume of the solution after evaporation.
Therefore,
Final volume of solution (V) = 50mL
Number of moles of KBr in final solution (n) = 71.0mmol
From
C = n / V
C= 71.0mmol/50mL
C = 1.42M
Therefore, the molarity of KBr in the final solution is 1.42M
Sodium metal forms at the cathode
Answer: (a) The solubility of CuCl in pure water is 
.
(b) The solubility of CuCl in 0.1 M NaCl is 
.
Explanation:
(a) Chemical equation for the given reaction in pure water is as follows.
Initial: 0 0
Change: +x +x
Equilibm: x x
And, equilibrium expression is as follows.
![K_{sp} = [Cu^{+}][Cl^{-}]](https://tex.z-dn.net/?f=K_%7Bsp%7D%20%3D%20%5BCu%5E%7B%2B%7D%5D%5BCl%5E%7B-%7D%5D)
x =
Hence, the solubility of CuCl in pure water is .
(b) When NaCl is 0.1 M,
,
, 
Net equation: 
= 0.1044
So for,
Initial: 0.1 0
Change: -x +x
Equilibm: 0.1 - x x
Now, the equilibrium expression is as follows.
K' = 
0.1044 = 
x =
Therefore, the solubility of CuCl in 0.1 M NaCl is .
