Все написано в скобках правильно
The complete ionization of KBr into its constituents
is:<span>
<span>KBr (s) --->
K+ (aq) + Br- (aq)</span></span>
<span>
During electrolysis, oxidation takes place at the anode electrode. This means
that an ion is stripped off its electron hence becoming more positive:
<span>2 Br- (aq) --->
Br2 (g) + 2e- </span></span>
We can see that Bromine gas Br2 is evolved at the anode.
<span>
<span>Meanwhile at the cathode, the reduction reaction occurs.
Which means that the electron from the anode electrode is used to make an ion
more negative:
<span>2K+ (aq) + 2e- ---> 2K (s) </span></span>
Hence, through reduction, solid potassium is deposited on the
plate.</span>
Half reactions:
<span>Anode: 2 Br- (aq) --->
Br2 (g) + 2e- </span>
<span>Cathode: 2K+ (aq) + 2e-
---> 2K (s) </span>
Answer:
The heavier wagon rolls 1/2 as fast as the lighter wagon.
Explanation:
When the compressed spring that joins them is released then the force acts on both wagons will be of equal magnitude but in the opposite direction. However as the mass of one wagon is twice that of other, so the acceleration will become half of the heavier wagon in comparison with lighter one.
Answer:
(4) A = 3 A, A₂ = 11 A
(5) 7 A
Explanation:
(4)
From the diagram,
A = 3+6+2
A = 11 A
V = A₂R
A₂ = V/R₂............ Equation 1
Given: V = 12 V, R₂ = 4 Ω
Substitute these values into equation 1
A₂ = 12/4
A₂ = 3 A
(5) Applying,
V = IR'
I = V/R'............ Equation 1
Where V = Voltage, I = cuurent, R' = total resistance.
But,
1/R' = (1/3)+(1/4)
1/R' = (3+4)/12
1/R' = 7/12
R' = 12/7 Ω
Given: V = 12 V
Substitute these values into equation 1
I = 12/(12/7)
I = 7 A
Therefore
A = 7 A
