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kenny6666 [7]
3 years ago
6

Which group in the the periodic table contains elements with the valence electron configuration of ns2 npl1?

Chemistry
1 answer:
marissa [1.9K]3 years ago
3 0
3A (also known as group 13) all the elements have 3 electrons in their outer most shell thats why theyll be divided into the ns2 and the np1
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Which element would you expect to be more metallic?<br> (a) S or Cl (b) In or Al (c) As or Br
Natalija [7]

Explanation:

When we move across a period from left to right then there will occur an increase in electronegativity and also there will occur an increase in non-metallic character of the elements.

As sulfur (S) is a group 16 element and chlorine (Cl) is a group 17 element. Hence, sulfur (S) is more metallic in nature than chlorine.

This means that chlorine (S) is less metallic than chlorine (Cl).

Both indium (I) and aluminium (Al) are group 13 elements. And, when we move down a group then there occur an increase in non-metallic character of the elements. As indium belongs to group 13 and period 5 whereas aluminium belongs to group 13 and period 3.

Therefore, aluminium (Al) is more metallic than indium (In).

Arsenic (Ar) is a group 15 element and bromine (Br) is a group 17 element. Therefore, arsenic is more metallic than bromine.

4 0
4 years ago
Which type of star has the lowest density?
julsineya [31]
The white dwarf has the lowest density of the stars listed. Hope this helps and have a nice day!
8 0
4 years ago
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The carbon-carbon double bond in ethene is ________ and ________ than the carbon-carbon triple bond in ethyne.
lbvjy [14]

Answer:

weaker and longer

Explanation:

Since there are 3 bonds in ethyne in comparision with the 2 bonds of ethyne between carbon atoms, they are attracted more to each other → the bond gets shorter . And since there are one more bond that supports the union → the bond gets stronger

thus the carbon-carbon double bond in ethene is weaker and longer than the carbon-carbon triple bond in ethyne

3 0
3 years ago
Consider the following multistep reaction:
chubhunter [2.5K]

The question is incomplete, here is the complete question:

Consider the following multistep reaction:

C+D⇌CD (fast)

CD+D→CD₂ (slow)

CD₂+D→CD₃ (fast)

C+3D→CD₃

Based on this mechanism, determine the rate law for the overall reaction.

<u>Answer:</u> The rate law for the reaction is \text{Rate}=k'[C][D]^2

<u>Explanation:</u>

Rate law is the expression which is used to express the rate of the reaction in terms of the molar concentration of reactants where each term is raised to the power their stoichiometric coefficient respectively from a balanced chemical equation.

In a mechanism of the reaction, the slow step in the mechanism determines the rate of the reaction.

For the given chemical reaction:

C+3D\rightarrow CD_3

The intermediate reaction of the mechanism follows:

<u>Step 1:</u>  C+D\rightleftharpoons CD;\text{ (fast)}

<u>Step 2:</u>  CD+D\rightarrow CD_2;\text{(slow)}

<u>Step 3:</u>  CD_2+D\rightarrow CD_3;\text{(fast)}

As, step 2 is the slow step. It is the rate determining step

Rate law for the reaction follows:

\text{Rate}=k[CD][D]           ......(1)

As, [CD] is not appearing as a reactant in the overall reaction. So, we apply steady state approximation in it.

Applying steady state approximation for CD from step 1, we get:

K=\frac{[CD]}{[C][D]}

[CD]=K[C][D]

Putting the value of [CD] in equation 1, we get:

\text{Rate}=k.K[C][D]^2\\\\\text{Rate}=k'[C][D]^2  

Hence, the rate law for the reaction is \text{Rate}=k'[C][D]^2

5 0
3 years ago
A student is given the following information about an unknown solution:
Harlamova29_29 [7]

Answer:

answer is strong base hope it helps

3 0
3 years ago
Read 2 more answers
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