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borishaifa [10]
4 years ago
8

Which molecule contains a triple covalent bond? (1) H2 (3) O2 (2) N2 (4) Cl2

Chemistry
2 answers:
Allushta [10]4 years ago
6 0
N2 because it has 5 valence atoms so it needs 3 to have 8, wich is the octet rule, so three is a triple covalent bond
dexar [7]4 years ago
6 0

<u>Answer:</u> The correct answer is Option 2.

<u>Explanation:</u>

Covalent bond is defined as the b which is formed by the sharing of electrons between the atoms forming a bond.

Non-polar covalent bond is defined as the bond which is formed when there is no difference in electronegativity between the atoms. When atoms of the same element combine, it results in the formation of non-polar covalent bond. <u>For Example:</u> N_2,O_2 etc..

For the given options:

  • <u>Option 1:</u>  H_2

Hydrogen is the 1st element of the periodic table having electronic configuration 1s^1

This element needs 1 electron to attain stable electronic configuration and will form single covalent bond with another hydrogen atom.

  • <u>Option 2:</u>  N_2

Nitrogen is the 7th element of the periodic table having electronic configuration 1s^22s^22p^3

This element needs 3 electrons to attain stable electronic configuration and will form triple covalent bond with another nitrogen atom.

  • <u>Option 3:</u>  O_2

Oxygen is the 8th element of the periodic table having electronic configuration 1s^22s^22p^4

This element needs 2 electrons to attain stable electronic configuration and will form double covalent bond with another oxygen atom.

  • <u>Option 4:</u>  Cl_2

Chlorine is the 17th element of the periodic table having electronic configuration 1s^22s^22p^63s^23p^5

This element needs 1 electron to attain stable electronic configuration and will form single covalent bond with another chlorine atom.

Hence, the correct answer is Option 2.

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No. Of particles in H2SO4 in which 8grams oxygen is present????
steposvetlana [31]

Answer:

7?

Explanation:

Its somewhat hard to comprehend the question, but if the way I read it was right, its 7.

6 0
3 years ago
If 0.50 mol/L of butane is added to the original equilibrium mixture and the system shifts to a new equilibrium position, what i
Ostrovityanka [42]

Consider the isomerization of butane with equilibrium constant is 2.5 .The system is originally at equilibrium with :

[butane]=1.0 M , [isobutane]=2.5 M

If 0.50 mol/L of butane is added to the original equilibrium mixture and the system shifts to a new equilibrium position, what is the equilibrium concentration of each gas?

Answer:

The equilibrium concentration of each gas:

[Butane] = 1.14 M

[isobutane] = 2.86 M

Explanation:

Butane  ⇄  Isobutane

At equilibrium

1.0 M               2.5 M

After addition of 0.50 M of butane:

(1.0 + 0.50) M               -

After equilibrium reestablishes:

(1.50-x)M            (2.5+x)

The equilibrium expression will wriiten as:

K_c=\frac{[Isobutane]}{[Butane]}

2.5=\frac{2.5+x}{(1.50-x)}

x = 0.36 M

The equilibrium concentration of each gas:

[Butane]= (1.50-x) = 1.50 M - 0.36M = 1.14 M

[isobutane]= (2.5+x) = 2.50 M + 0.36 M = 2.86 M

3 0
3 years ago
What would ne the acceleration in a body moving wit uniform velocity and why​
meriva

Answer: The derivative of a constant term is always 0. So the acceleration of the body would be zero. Hence, the acceleration of a body moving with uniform velocity will always be zero.

Hope this helps :) :)

3 0
3 years ago
Read 2 more answers
Two equilibrium reactions of nitrogen with oxygen, with their corresponding equilibrium constants (Kc) at a certain temperature,
7nadin3 [17]

Answer:

Kc = 1.54e - 31 / 2.61e - 24

Explanation:

1 )   N_{2}(gas) + O_{2}(gas)\rightarrow 2NO(gas)  ; Kc = 1.54e - 31

2)   N_{2}(gas) + 1/2O_{2}(gas)\rightarrow N_{2}O(gas)  ; Kc = 2.16e - 24

   upon reversing  ( 2 )  equation

     N_{2}O(gas)\rightarrow N_{2}(gas) + 1/2O_{2}(gas)   Kc = 1/2.16e - 24  

    now adding 1 and reversed equation (2)

       N_{2}(gas) + O_{2}(gas)\rightarrow 2NO(gas)

      N_{2}O(gas)\rightarrow N_{2}(gas) + 1/2O_{2}(gas)

   we get ,

                  N_{2}O(gas) + 1/2O_{2}(gas)\rightarrow 2NO(gas)  Kc = 1.54e-31 × 1/2.61e - 24

       equilibrium constant of equation (3) is -

            Kc = 1.54e - 31 / 2.61e - 24

3 0
3 years ago
A sample of urea contains 1.121 g of N, 0.161 g of H, 0.48 g of C, and 0.64 g of O. What is the empirical formula of urea?
dybincka [34]

Answer:

The empirical formula is N2 CH4 O

Explanation:

5 0
3 years ago
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