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Nata [24]
3 years ago
10

Refraction occurs when the speed of light changes as it travels between mediums. T F

Chemistry
1 answer:
Mariana [72]3 years ago
5 0

The answer is true

Refraction is an effect that occurs when a light wave, incident at an angle away from the normal, passes a boundary from one medium into another in which there is a change in velocity of the light.The wavelength decreases as the light enters the medium and the light wave changes direction.

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Answer:

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The ion n3− has _____ protons and _____ electrons.enter your answers as integers separated by a comma.
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The ion N³⁻ is called the azide ion. In its neutral state, it occurs as the element Nitrogen. The atomic number of Nitrogen is 7. When it turns into an anion (negatively charged ion), it gains 3 more electrons. That's why its net charge becomes -3. It means that the protons is still 7, but the electrons are now 10. 

Overall charge = +7 + -10 = -3
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Bromine reacts with nitric oxide to form nitrosyl bromide as shown in this reaction: br2(g) + 2 no(g) → 2 nobr(g) a possible mec
Svetach [21]

The overall reaction is given by:

Br_{2}(g) + 2 NO(g) \rightarrow  2 NOBr(g)

The fast step reaction is given as:

NO(g) + Br_{2}(g) \rightleftharpoons NOBr_{2}(g) (fast; k_{eq}= \frac{k_{1}}{k_{-1}})

The slow step reaction is given as:

NOBr_{2}(g) + NO(g) \rightarrow  2 NOBr(g) (slow step k_{2})

Now, the expression for the rate of reaction of fast reaction is:

r_{1}=k_{1}[NO][Br_{2}]-k_{-1}[NOBr_{2}]

The expression for the rate of reaction of slow reaction is:

r_{2}=k_{2}[NOBr_{2}] [NO]

Slow step is the rate determining step. Thus, the overall rate of formation is the rate of formation of slow reaction as [NOBr_{2}] takes place in this reaction.

The expression of rate of formation is:

\frac{d(NOBr)}{dt}=r_{2}

= k_{2}[NOBr_{2}][NO]    (1)

Now, consider that the fast step is always is in equilibrium. Therefore, r_{1}=0

k_{1}[NO][Br_{2}]= k_{-1}[NOBr_{2}]

[NOBr_{2}] = \frac{k_{1}}{k_{-1}}[NO][Br_{2}]

Substitute the value of [NOBr_{2}] in equation (1), we get:

\frac{d(NOBr)}{dt}=k_{2}[NOBr_{2}][NO]

=k_{2} \frac{k_{1}}{k_{-1}}[NO][Br_{2}][NO]

= \frac{k_{1}k_{2}}{k_{-1}}[NO]^{2}[Br_{2}]

Thus, rate law of formation of NOBr in terms of reactants is given by \frac{k_{1}k_{2}}{k_{-1}}[NO]^{2}[Br_{2}].









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3 years ago
What are the answers to the blanks to make this equation balanced
Eddi Din [679]

1Al+6NaOH==>2Na_{3}AlO_{3}+3H_{2}


Hope this helps!

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