Answer :
A car stopped at the top of a hill
Explanation :
Definition of potential energy : energy stored that depends upon the relative position of various parts of a system
The greatest molar amount of metal product is obtained from silver (1) nitrate.
Let us determine the mass of metal obtained in each case.
For La^3+;
La^3+ + 3e ----> La
1 mole of La is deposited by 3F
x moles of La is deposited by 10 F
x = 1 × 10/3
x = 3.33 moles
For Zn^2+;
1 mole Zn^2+ is deposited by 2F
x moles of Zn^2+ is deposited by 10 F
x = 5 F
For Ag^+
1 mole of Ag^+ is deposited by 1 F
x moles of Ag^+ is deposited by 10 F
x = 10 moles
For Ba^2+;
1 mole of Ba^2+ is deposited by 2 F
x moles of Ba^2+ is deposited by 10 F
x = 5 moles
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Q6. 3
Q7. 3
Q8. pH
Q18. 3
Q19. 3
Q20. 4
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CaCO₃ + 2HCl = CaCl₂ + CO₂ + H₂O
n(CaCO₃)=m(CaCO₃)/M(CaCO₃)
n(CaCO₃)=13.00/100.09=0.1299 mol
Δm=13.00+52.65-60.32=5.33 g
m(CO₂)=5.33 g
n(CO₂)=5.33/44.01=0,1211 mol
w=0.1211/0.1299=0,9323 (93.23%)
When the order n of the reaction rate is determined by this formula:
rate = K * [A]^n
when we have the unit of rate M S^-1
and the unit of [A] M
So:
1- for k = 8.79 x 10^2 M S^-1 :
when the unit of K here is M S^-1
by substitution in rate formula:
M S^-1 = M S^-1 * M ^n
∴ M^n = 1
∴M^0 = 1
∴ this reaction rate is zero order reaction
2- when k = 4.46 x 10^-1 and has unit M-1 S-1:
by substitution in rate formula:
M S^-1 = M-1 S-1 * M^n
∴M^2 = M^n
∴n = 2 so this is the second order reaction
3- when K 2.35 x 10^6 and it's unit is S-1 :
So by substitution in rate formula:
M S^-1 = S^-1 * M^n
∴ M^n = M
∴ n = 1
So this is the first order reaction
4- when k = 1.88 x 10^-3 and its unit is M-2 S-1
by substitution in rate formula:
M S^-1 = M^-2 S^-1 * M^n
∴ M^n = M^3
∴ n = 3
∴ this is a third order reaction
from 1 & 2 & 3 & 4
so we can see that the correct answer is (4), K= 1.88 x 10^-3 M-2 S-1 is the highest order reaction.
