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Andreyy89
2 years ago
13

N2 molecules absorb ultraviolet light but not visible light. I2 molecules absorb both visible and ultraviolet light. Which of th

e following statements explains the observations?
More energy is required to make N2 molecules vibrate than is required to make I2 molecules vibrate.
A

More energy is required to remove an electron from an I2 molecule than is required to remove an electron from a N2 molecule.
B

Visible light does not produce transitions between electronic energy levels in the N2 molecule but does produce transitions in the I2 molecule.
C

The molecular mass of I2 is greater than the molecular mass of N2.
Chemistry
1 answer:
Umnica [9.8K]2 years ago
4 0

C. Visible light does not produce transitions between electronic energy levels in the N₂ molecule but does produce transitions in the I₂ molecule.

Explanation:

Abortion of light by molecules will produce electronic transitions from a ground level to a higher level equal to the energy of absorbed light.

In the case of nitrogen (N₂) the allowed electronic transitions are between electronic energy levels with a energy difference equal to the energy of photons of the ultraviolet light. Nitrogen will not absorb from the visible range so it is colorless.

Now iodine (I₂) have the allowed electronic transitions between electronic energy levels with a energy difference equal to the energy of photons of the visible light. As a consequence Iodine vapors have a violet color.

Learn more about:

electronic transitions

brainly.com/question/11328705

#learnwithBrainly

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Answer: d. More than 6.5 grams of copper (II) is formed, and some copper chloride is left in the reaction mixture.

Explanation: 2Al+3CuCl_2\rightarrow 2AlCl_3+3Cu

As can be seen from the chemical equation, 2 moles of aluminium react with 3 moles of copper chloride.

According to mole concept, 1 mole of every substance weighs equal to its molar mass.

Aluminium is the limiting reagent as it limits the formation of product and copper chloride is the excess reagent as (14-7.5)=6.5 g is left as such.

Thus 54 g of of aluminium react with 270 g of copper chloride.

1.50 g of aluminium react with=\frac{270}{54}\times 1.50=7.5 gof copper chloride.

3 moles of copper chloride gives 3 moles of copper.

7.5 g of copper chloride gives 7.5 g of copper.

8 0
2 years ago
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Determine ΔH for the reaction CaCO3 → CaO + CO2 given these data: 2 Ca + 2 C + 3 O2 → 2 CaCO3 ΔH = −2,414 kJ C + O2 → CO2 ΔH = −
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Answer:

The ΔH for the reaction is -456.5 KJ

Explanation:

Here we want to determine ΔH for the reaction;

Mathematically;

ΔH = ΔH(product) - ΔH(reactant)

In the case of the first reaction;

ΔH = ΔH(CaO) + ΔH(CO2) - ΔH(CaCO3)  ...........................(*)

From the other reactions, we can get the respective ΔH for the individual molecule in the reaction

In second reaction;

Kindly note that for elements, molecule of gases, ΔH = 0

What this means is that throughout the solution;

ΔH(Ca)  = 0 KJ

ΔH(O2) = 0 KJ

ΔH(C) = 0 KJ

Thus, in writing the equation for the subsequent chemical reactions, we shall need to write and equate the overall ΔH for the reaction to that of the product alone

So in the second reaction

ΔH = 2ΔH(CaCO3)

Thus;

-2414/2 = ΔH(CaCO3)

ΔH(CaCO3) = -1,207  KJ

Moving to the third reaction, we have;

ΔH = ΔH(CO2)

Hence ΔH(CO2) = -393.5 KJ

For the last reaction;

ΔH = ΔH(CaO)

Hence ΔH(CaO) = -1270 KJ

Going back to equation *

ΔH = ΔH(CaO) + ΔH(CO2) - ΔH(CaCO3)

Using the values of the ΔH  of the respective molecules given above,

ΔH  = -1270 + (-393.5) - (-1207)

ΔH  = -456.5 KJ

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A fox spots a rabbit in a field. The fox begins to chase the rabbit, and the rabbit runs away. Which statement best describes th
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Answer:

See explanation below

Explanation:

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In the second step, reacts with the lindlar catalyst to do a reduction, and form a double bond between carbon 1 and 2. In essence, compound A is similar to compound B.

Finally B reacts with water in acid and makes a addition reaction, and form an alcohol.

The whole process can be seen in the picture below.

Hope this helps

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