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myrzilka [38]
3 years ago
14

Use the changes in oxidation numbers to identify which atoms are oxidized and which are reduced in each reaction.

Chemistry
1 answer:
krok68 [10]3 years ago
5 0
You didn't include much information as to what you were referring too, however I answered a similar question recently to a friend so I'm hoping this is the same question in complete as to yours, if not I apologize.

a.) 2Na(s) + Cl2(g) -----> 2NaCl(s) 
Na --> Na+1 is oxidized 
Cl2 --> Cl- is reduced 

b.) 2HNO3(aq) + 6HI(aq) ----> 2NO(g) + 3I2(s) + 4H2O(l) 
N+5 --> N+2 is reduced 
I-1 --> I2 is oxidized 
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The intensity of illumination at any point from a light source is proportional to the square of the reciprocal of the distance b
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the stronger light 5.5 m apart from the total illumination​

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From the problem's statement , the following equation can be deducted:

I= k/r²

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denoting 1 as the stronger light and 2 as the weaker light

I₁= k/r₁²

I₂= k/r₂²

dividing both equations

I₂/I₁ = r₁²/r₂²=(r₁/r₂)²

solving for r₁

r₁ = r₂ * √(I₂/I₁)

since we are on the line between the two light​ sources , the distance from the light source to the weaker light is he distance from the light source to the stronger light + distance between the lights . Thus

r₂ = r₁ + d

then

r₁ = (r₁ + d)* √(I₂/I₁)

r₁ = r₁*√(I₂/I₁) + d*√(I₂/I₁)

r₁*(1-√(I₂/I₁)) =  d*√(I₂/I₁)

r₁ = d*√(I₂/I₁)/(1-√(I₂/I₁))  =

r₁ = d/[√(I₁/I₂)-1)]

since the stronger light is 9 times more intense than the weaker

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then since d=11 m

r₁ = d/[√(I₁/I₂)-1)] = 11 m / (3-1) = 5.5 m

r₁ = 5.5 m

therefore the stronger light 5.5 m apart from the total illumination​

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