All categories

3 years ago

The symbol for xenon (Xe) would be a part of the noble gas notation for the element antimony. cesium. radium. uranium.

2 answers:

8 0

The noble gas notation is the short or abbreviated form of the electron configuration.

It means that you use the symbol of the previous noble gas as part of the electron configuration of an element.

The gas noble previous to antimony is Kr, so you do not use Xe to write the electron configuration of Sb.

The gas noble previous to radium is Rn, so you do not use Xe to wirte the electron configuration of Ra.

The gas noble previous to uranium is Rn, so you do not use Xe to write the electron configuration of U.

The gas noble previous to cesium is Xe, so you use Xe to write the noble notation for Sb. This is it: Cs: [Xe] 6s.

Answer: cesium

The ga

atroni [7]3 years ago

7 0

the answer is B.

CESIUM.....

You might be interested in

Which of the following reactions would you find in a radioisotope thermal generator?

Mars2501 [29]

The 3rd option

because i’m always right:)

6 0

2 years ago

How many protons, neutrons, and electrons make up an ion of

yKpoI14uk [10]

Answer:

protons=15

electron=15

neutron=16

6 0

3 years ago

Calculate the speed of a marble that rolls 9 cm in 4 seconds

cupoosta [38]

Answer:

2.25

Explanation:

9/4

6 0

3 years ago

Hydrochloride acid +_________ zinc chloride +hydrogen

Klio2033 [76]

Answer:

Hydrochloride acid + Zinc = Zinc Chloride + Hydrogen

Explanation:

When Hydrochloride acid and Zinc react, it results in the formation of Zinc chloride and hydrogen.

<em>Hope I helped</em>

3 0

2 years ago

Draw the alkene formed when 1-heptyne is treated with hbr in the presence of peroxide.

Nimfa-mama [501]

<h2>Heptene formed is -</h2><h2> $CH_3-CH_2-CH_2-CH_2-CH_2-CH=CHBr$ </h2>

Explanation:

The two possibilities when the peroxide is not present

$CH_{3}-CH_{2}-CH_{2}-CH_{2}-CH_{2}-C$ ≡ $CH +HBr$ → $CH_3-CH_2-CH_2-CH_2-CH_2-CBr=CH_{2}$

$CH_3-CH_2-CH_2-CH_2-CH_2-CBr=CH_2$ + HBr → $CH_3-CH_2-CH_2-CH_2-CH_2-CBr_2-CH_3$

In presence peroxide,

$CH_3-CH_2-CH_2-CH_2-CH_2-C$ ≡ $CH$ + HBr → $CH_3-CH_2-CH_2-CH_2-CH_2-CH=CHBr$

When peroxides are present in the reaction mixture, hydrogen bromide adds to the triple bond of heptane with regioselectivity.

This reaction is opposite to that of Markovnikov's rule which says that when asymmetrical alkene reacts with a protic acid HX, then the hydrogen of an acid is attached to the carbon with more in number of hydrogen substituents, and the halide (X) group is attached to the carbon with more in number of substituents of alkyl.
One mole of HBr adds to one mole of 1-heptane.
The structure of heptene formed is -

$CH_3-CH_2-CH_2-CH_2-CH_2-CH=CHBr$

5 0

3 years ago