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mihalych1998 [28]
3 years ago
14

What is the half-reaction that occurs at the cathode during the electrolysis of molten potassium bromide?

Physics
1 answer:
polet [3.4K]3 years ago
7 0

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>

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Answer:

Explanation:

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initial velocity, u = 20 m/s

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y=2+20Sin5 t-4.9t^{2}

As it hits the ground in time t, so put y = 0

0=2+1.74 t-4.9t^{2}

4.9t^{2}-1.74t-2=0

t= \frac{1.74\pm\sqrt{1.74^{2}+4\times\2\times4.9}}{9.8}

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The ball travels a horizontal distance x in time t

X = 20 Cos5 x t

X =  16.76 m

As this distance is more than the distance of net, so it clears the net.

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Let the vertical distance traveled by the ball in time t' is y'.

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y'=y_{0}+uSin\theta t'-1/2 gt'^{2}

y'=2+20Sin5 t-4.9\times0.35^{2}

y' = 2.008 m

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It clears the net by a vertical distance of 2.008 - 1 = 1.008 m and horizontal distance 16.76 - 7 = 9.76 m

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3 years ago
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50200 J of heat are removed from
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Correct Answer:

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4 years ago
Light waves are not mechanical waves. The Sun transmits light waves to Earth through __________ _________. *
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Electromagnetic waves?
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Two objects have a force of gravity of 100 N. If the distance between both objects decreases by a factor of 7, while the masses
Shkiper50 [21]

The gravitational force <em>F</em> between two masses <em>M</em> and <em>m</em> a distance <em>r</em> apart is

<em>F</em> = <em>G M m</em> / <em>r</em> ²

Decrease the distance by a factor of 7 by replacing <em>r</em> with <em>r</em> / 7, and decrease both masses by a factor of 8 by replacing <em>M</em> and <em>m</em> with <em>M</em> / 8 and <em>m</em> / 8, respectively. Then the new force <em>F*</em> is

<em>F*</em> = <em>G </em>(<em>M</em> / 8) (<em>m</em> / 8) / (<em>r</em> / 7)²

<em>F*</em> = (1/64 × <em>G M m</em>) / (1/49 × <em>r</em> ²)

<em>F*</em> = 49/64 × <em>G M m</em> / <em>r</em> ²

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