Answer:
3%
Explanation:
Substract the actual error from the final and multiply by 100
Assuming that you’re looking for the concentration of water in the solution, then it would be 0.028 M.
You would have to use the formula:
c1v1 = c2v2, where c =concentration and
v = volume
C1 = ?
V1 = 250 mL
C2 = 0.2 M
V2 = 35 mL
C1 x 250 mL = 0.2 M x 35 mL
C1 = (0.2 M x 35 mL) / 250 mL
C1 = 0.028 M of water added to 35mL of 0.2M HCl
Therefore, there is 0.028 M of water added to 35mL of 0.2M HCl
Answer:
0.085 moles of N₂O₅ are needed
Explanation:
Given data:
Mass of NO₂ produces = 7.90 g
Moles of N₂O₅ needed = ?
Solution:
2N₂O₅ → 4NO₂ + O₂
Number of moles of NO₂ produced :
Number of moles = mass/ molar mass
Number of moles = 7.90 g/ 46 g/mol
Number of moles = 0.17 mol
now we will compare the moles of NO₂ with N₂O₅.
NO₂ : N₂O₅
4 : 2
0.17 : 2/4×0.17 = 0.085 mol
Thus, 0.085 moles of N₂O₅ are needed.
Answer:
Electrolysis, I hope this helped!
