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Semenov [28]
2 years ago
7

4. Which of the following statements explains the cause of lanthanide contraction?

Chemistry
1 answer:
yawa3891 [41]2 years ago
8 0

Answer:

B. PROTONS EXHIBIT STRONGER PULL ON OUTER f ORBITALS

Explanation:

Lanthanide contraction is the greater than normal decrease in the ionic radius of the lanthanide series from atomic number 57 to atomic number 71. This decrease is rather not expected of the ionic radii of these elements and they result in the greater decrease in the subsequent series of the lanthanides from the atomic number 72. The cause of which is as a result of the poor shielding effects of the nuclear charge around the electrons of the f orbitals. So therefore, protons are strongly pulled out of the 4f orbital and as a result of the poor shielding effect which causes the electrons of the 6s orbitals to be drawn more closer to the nucleus and hence resulting in a smaller atomic radii. It is worthy to note that the shielding effects of the inner electrons decreasing from s orbital to the f orbital; that is s > p > d > f. So from the decrease in the shielding effects from s to the f orbitals, lanthanide contraction results from the inability of the orbitals far away from s like the 4f orbiatls to shield the outermost shells of the lanthanide elements. So the cause of lanthanide contraction is the action of the protons which strongly pull the electrons of the f orbitals because of the poor shielding effects due to the distance of this orbital from the nucleus.

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Give the product for the reaction of 1-butene with methanol in the presence of acid. The mechanism is the same as the mechanism
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Answer:

The final result is <u>2-methoxybutane. </u>

<u />

Explanation:

1-butene has a carbon-carbon double bound between C1 and C2.

When it will react with methanol, in the presence of an acid, the result will be an ether.

C4H8 + CH3OH → C5H12O

An acid-catalyzed ether synthesis from alkenes is limited by carbocation stability.

In the first step, the double bound will disappear. The C2 atom will be a C+ atom, this becaus it has only 3 bounds and not 4.

This C+ -atom will atract the O- atom to form an ether. The CH3 of methanol will bind on the C3 atom, this is the most stable position.

2-methoxybutane will be formed. It has a structural formula of C5H12O

1-methoxybutane will not be formed, because it's less stable.

1-butanol will be formed when water is added to 1-butene. The mechanism has the same principle but not the same product.

1-ethoxybutane and 2-ethoxybutane have a structural formula of C6H14O, this will not be the final result.

The final result is <u>2-methoxybutane. </u>

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8 0
2 years ago
Latent heat of vaporization is used to (1 Point) (a) overcome the forces of attraction between molecules in solid-state. (b) inc
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A student assistant is cleaning up after a chemistry laboratory exercise and finds three one-liter bottles containing alcohol so
Svetllana [295]

The approximate alcohol content is 210 ml.

Explanation:

It can be deduced from the question that each bottle is of 1000ml or 1 litre.

The first bottle is one half full means it has 500 ml of solution and it has 20% alcohol in it. So volume of alcohol in the solution is

20/100*500

=100 ml

The first bottle is one fifth full, so the volume of mixture is 1/5th of 1000ml

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30/100*200

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The third bottle is one tenth full so its volume is 1/10*1000

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50/100*100

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The alcohol content obtained from all these 3 litres is:

100+60+50

= 210 ml of alchohol is obtained from 800 ml of mixture.

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2 years ago
Question 21 (2 points)
m_a_m_a [10]

Answer:

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Explanation:

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Consider the reaction. X ( g ) + Y ( g ) − ⇀ ↽ − Z ( g ) K p = 1.00 at 300 K In which direction will the net reaction proceed fo
marta [7]

Answer:

Explanation:

We have in this question the equilibrium

X ( g ) + Y ( g ) ⇆  Z ( g )

With the equilibrium contant Kp = pZ/(pX x pY)

The moment we change the concentration of Y, we are changing effectively the partial pressure of Y since pressure and concentration are directly proportional

pV = nRT ⇒ p = nRT/V and n/V is molarity.

Therefore we can calculate the reaction quotient Q

Qp = pZ/(pX x pY) = 1/ 1  x 0.5 atm = 2

Since Qp is greater than Kp the system proceeds from right to left.

We could also arrive to the same conclusion by applying LeChatelier´s principle which states that any disturbance in the equilibrium, the system will react in such a way to counteract the change to restore the equilibrium. Therefore, by having reduced the pressure of Y the system will react favoring the reactants side increasing some of the y pressure until restoring the equilibrium Kp = 1.

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2 years ago
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