All categories

3 years ago

Explain the periodic trend for ionization energy and why this occurs?

Chemistry

1 answer:

Dahasolnce [82]3 years ago

4 0

Answer:

Ionization energy tends to increase across a period because the nuclear pull is increasing while shielding is not changing because electrons are in the same energy level. Not only that as we move across a period elements are closer to a full octet.

Explanation:

You might be interested in

Calculate the molar mass of the acid.

JulijaS [17]

Citric acid has the molecular formula C6H8O7 so you can add the molar masses of the elements from the periodic table. C has a molar mass of 12.01 g/mol, H has 1.01 g/mol and O has 15.999 g/mol. Now you calculate the total molar mass= (6*12.01 + 8*1.01 + 7*15.999). This yields a molar weight of 192.124 g/mol (anhydrous)

3 0

3 years ago

The radius of an indium atom is 0.163 nm. If indium crystallizes in a face-centered unit cell, what is the length of an edge of

adoni [48]

Answer: The edge length of the unit cell is 0.461 nm

Explanation:

We are given:

Atomic radius of iridium = 0.163 nm

To calculate the edge length, we use the relation between the radius and edge length for FCC lattice:

$a=2\sqrt{2}R$

Putting values in above equation, we get:

$a=2\sqrt{2}\times 0.163=0.461nm$

Hence, the edge length of the unit cell is 0.461 nm

8 0

3 years ago

What does the second law of thermodynamics say? O A. The entropy of the universe is increasing. B. A system at equilibrium has n

gayaneshka [121]

Answer:

A. The entropy of the universe is increasing.(always)

Explanation:

The Second Law of Thermodynamics states that Entropy cannot decrease, because it keeps increasing and increasing and increasing. It will always stay on the increasing side.

Hope this helps!

Please Mark Brainliest!

7 0

3 years ago

Draw the major organic substitution product(s) for (2R,3S)-2-bromo-3-methylpentane reacting with the given nucleophile. Indicate

Andrew [12]

Answer:

(2R,3S)-2-ethoxy-3-methylpentane

and

(2S,3S)-2-ethoxy-3-methylpentane

Explanation:

For this case, we will have $CH_3CH_2O^-$ as nucleophile. Also, this compound is also in excess. So, we will have as solvent $CH_3CH_2OH$ a protic solvent. Therefore the Sn1 reaction would be favored.

The first step would be the carbocation formation followed by the attack of the nucleophile. In this case both isomers would be produced: R and S (see figure).

7 0

2 years ago

A general chemistry student found a chunk of metal in the basement of a friend's house. To figure out what it was, she tried the

mr_godi [17]

Answer:

do you have options for this

6 0

3 years ago

Read 2 more answers