The half reactions as they occur at each electrode
is as follows
at the anode Sn(s) =sn^2+(aq) + 2e -
at the cathode 2 ag^+(aq) + 2e - = 2Ag (s)
net cell reaction = Sn (s) + 2Ag^+(aq) = sn^2+ (aq) + 2 Ag (s)
1) Chemical reaction: AgNO₃ + HCl → AgCl + HNO₃.
V(AgNO₃) = 30,0 mL = 0,03 L.
c(AgNO₃) = 0,225 mol/L.
n(AgNO₃) = 0,03 L · 0,225 mol/L.
n(AgNO₃) = 0,00675 mol.
From chemical reaction: n(AgNO₃) : n(HCl) = 1 : 1.
0,00675 mol : n(HCl) = 1 : 1.
n(HCl) = 0,00675 mol.
V(HCl) = n(HCl) ÷ c(HCl).
V(HCl) = 0,00675 mol ÷ 0,130 mol/L.
V(HCl) = 0,0519 L = 51,92 ml.
2) 1) Chemical reaction: AgNO₃ + KCl → AgCl + KNO₃.
V(AgNO₃) = 30,0 mL = 0,03 L.
c(AgNO₃) = 0,225 mol/L.
n(AgNO₃) = 0,03 L · 0,225 mol/L.
n(AgNO₃) = 0,00675 mol.
From chemical reaction: n(AgNO₃) : n(KCl) = 1 : 1.
0,00675 mol : n(KCl) = 1 : 1.
n(KCl) = 0,00675 mol.
m(KCl) = n(KCl) · M(KCl).
m(KCl) = 0,00675 mol · 74,55 g/mol.
m(KCl) = 0,503 g.
n - amount of substance.
M - molar mass.
Answer:
15.98 L
Explanation:
First, you need to find T1, T2, V1 and V2.
T1 = 25 C = 298.15 K (25C + 273.15K)
T2 = 100 C = 373.15 K (100C + 273.15K)
V1 = 20. L
V2 = ? (we are trying to find)
Next, rearrange to fit the formula
V2 = V1 x T1 / T2
Next, fill in with our numbers
V2 = 20. L x 298.15 K / 373.15 K
Do the math and you should get...
15.98 L
- If you need more help or futher explanation please let me know. I would be glad to help!
Answer:
The elements in group 13 and group 15 form a cation with a -3 charge each.
The Mitocondria is the powerhouse of the cells in plants
