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3 years ago

The atomic mass of magnesium is the weighted average of the atomic masses of

Chemistry

1 answer:

Debora [2.8K]3 years ago

6 0

Answer:

2. All the naturally occurring isotopes of Mg.

Explanation:

You want to know the atomic mass of the magnesium you use in the lab. That’s “natural” magnesium. So, you must use the weighted average of all the naturally occurring isotopes in natural Mg.

1. and 3. are <em>wrong</em>. You won’t get the correct mass for natural Mg if you use only the artificial isotopes for your calculation.

4. is <em>wrong</em>. You must use all the naturally occurring isotopes. The two most abundant isotopes of Mg account for only 90 % of the atoms. If you ignore the other 10 %, your calculation will be wrong.

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The rays produced in a cathode tube are

wel

Answer:

Electrons

Explanation:

Cathode rays carry electronic currents through the tube. Electrons were first discovered as the constituents of cathode rays. J.J. Thomson used the cathode ray tube to determine that atoms had small negatively charged particles inside of them, which he called “electrons.”

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3 years ago

Read 2 more answers

The boiling points of different substances are given in the table below.

Oksi-84 [34.3K]

Answer:

Water

Explanation :-

Higher the intermolecular forces between the liquid particles, higher its boiling point.

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3 years ago

A powdered drink mix is stirred into a glass of water to form a solution. If any more is added it will not dissolve

Brut [27]

SaturTED, hope this helps, m8

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3 years ago

Noble gas compounds like KrF, XeCl, and XeBr are used in excimer lasers. Draw an approximate molecular orbital diagram appropria

Aneli [31]

Answer:

Here's what I get.

Explanation:

The MO diagrams of KrBr, XeCl, and XeBr are shown below.

They are similar, except for the numbering of the valence shell orbitals.

Also, I have drawn the s and p orbitals at the same energy levels for both atoms in the compounds. That is obviously not the case.

However, the MO diagrams are approximately correct.

The ground state electron configuration of KrF is

$(1\sigma_{g})^{2}\, (1\sigma_{u}^{*})^{2} \, (2\sigma_{g})^{2} \, (2\sigma_{u}^{*})^{2} \, (3\sigma_{g})^{2} \, (1\pi_{u})^{4} \, (1\pi_{g}^{*})^{4} \, (3\sigma_{g}^{*})^{1}$

KrF⁺ will have one less electron than KrF.

You remove the antibonding electron from the highest energy orbital, so the bond order increases.

The KrF bond will be stronger.

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3 years ago

Were you in ms.secrihrst class in dutchtown middle.

IrinaVladis [17]

NO, why though this is a simple question lol!

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3 years ago