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>
Time period = 1 / frequency
Time period = 1 / 250 th of a second
U use the train to play hope i helped please thank me
Answer:
340 seconds = 5.667 minutes
Explanation:
As we know, S = v t or t = S / v (S = 51 x 10^9 m and v = 3 x 10^8 ms^-1)
So, t = 51 x 10^9 / 3 x 10^8 = 17 x 10^1 = 170 s
For a RTT estimation, the time span will be doubled of one way propagation for transmission and receive delay.
The over all round trip time will be = 170 x 2 = 340 seconds = 5.667 minutes
Call the capacitance C.
<span>Note the energy in a capacitor with voltage V is E =½CV². </span>
<span>Initial energy = ½C(12)² = 72C </span>
<span>40% of energy is delivered, so 60% remains.in the capacitor. </span>
<span>Remaining energy = (60/100) x 72C =43.2C </span>
<span>If the final potential difference is X, the energy stored is ½CX² </span>
<span>½CX² = 43.2C </span>
<span>X² = 2 x 43.2 = 86.4 </span>
<span>X = 9.3V</span>
