Explanation:
<h2><u>Steps </u><u>:</u></h2>
- <u>Move </u><u>decimal</u><u> </u><u>from</u><u> </u><u>left </u><u>to </u><u>right</u><u> </u><u>=</u><u>0</u><u> </u><u>0</u><u>0</u><u>0</u><u>0</u><u>0</u><u>0</u><u>2</u><u>4</u><u>0</u><u>.</u><u>0</u>
- <u>Then </u><u>count </u><u>the</u><u> </u><u>numbers</u><u> </u><u>before</u><u> </u><u>decimal </u><u>and </u><u>w</u><u>rite </u><u>it </u><u>like</u><u> </u><u>this </u><u>=</u><u>2</u><u>4</u><u>0</u><u>.</u><u>0</u><u>x</u><u>1</u><u>0</u><u> </u><u>power-</u><u>9</u><u> </u>
- <u>That's</u><u> </u><u>all </u>
<u>hope</u><u> it</u><u> </u><u>help</u>
<h2><u>#</u><u>H</u><u>o</u><u>p</u><u>e</u></h2>
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The current passing through a circuit consisting of a battery of 12 V and resistor of 2 ohms is 6 Ampere
.
Explanation:
- Assume the wires are ideal with zero resistance.
- The current passing through the circuit will be
I = V/R = 12/2 = 6.000 A.
Energy is released in the reaction
Explanation:
In a given where the energy of the products is greater than that of the reactants, we can infer that energy is released in the reaction.
This indicates that the reaction is an exothermic or exergonic reaction.
These reaction types are accompanied by release of energy.
- In an exothermic change energy is released to the surroundings.
- The surrounding becomes hotter at the end of the change.
- This applies in exergonic reaction which leaves a reaction having more energy than it originally started with.
Learn more:
Exothermic process brainly.com/question/10567109
#learnwithBrainly
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>