Chemistry, 08.12.2020 01:00 tiffanibell71

How many nonbonding electrons are on each nitrogen atom
A 0
B 2
C 4
D 6

Answers: 2

Show answers

Answers

The number of valence electron in the outermost shell of a nitrogen atom is a pair of electrons. Hence, the number of non - bonding electrons on a nitrogen atom is 2.

Nitrogen has an atomic number of 7. Hence, 7 electrons in it's shell. Using the Lewis structure with the K, L, M, N configuration in the formation of Ammonia $(NH_{3})$ Nitrogen has 5 Electrons in its valence shell with one each covalently bonded to each of the 3 hydrogen atoms to attain a stable state. Hence, with 3 of the 5 electrons being bonded to one Hydrogen each. Then there are only two of the Nitrogen atoms which aren't paired.

Therefore, there are two non - bonding atoms.

Learn more :link

Answer from: andrewmena05

In ammonia (NH3) the number of non-bonded electrons are 2.

In nitrogen atom there are five electrons in its outermost shell. In ammonia (NH3) three electrons of nitrogen atom covalently bonded with three hydrogen atoms whereas the remaining two are non-bonded electrons.

Due to this non-bonded electron pair, ammonia reacts violently with oxidizing gases such as chlorine, bromine, and other halogens etc so we can conclude that 2 electrons are present in the compound of ammonia.

Learn more: link

Answer from: Quest

plants need sunlight, water, and carbon dioxide to make food. plants use the raw materials of water, nutrients they absorb from soil, and carbon dioxide to make sugars the plant can use. energy from the sun and chlorophyll are also needed.

Answer from: Quest

answer: the value of enthalpy change will be -152.80 kj.

solution:

pressure = 1atm , temperature = $25^oc$ = 298 k $)0^oc=273kelvins$

$o_3+3no\rightarrow 3no_2,\delta h_{reaction}=? [/)</p><p>[tex]\delta h_{no}=90.4kj/mol$

$\delta h_{no_2}=33.85kj/mol$

$\delta h_{o_3}=142.2kj/mol$

$\delta h_{reaction}=\sum \delta h_{products}-\sum\delta h_{reactants}=3\times\delta h_{no_2}-(3\times \delta h_{no}+\delta h_{o_3})$

$\delta h_{reaction}=33.85-(90.4+142.2)=-311.85kj/mol$

given that pressure and temperature are same in both gases , the number of moles of both gases can be determined by the ideal gas equation.

$v_{no}=12.00 l,v_{o_3}=8.50 l$

$pv_{no}=n_{no}rt=1atm \times 12.00 l=n_{no}\times 0.0821 atml/k mol \times 298 k$

$n_{no}=0.49 moles$

$pv_{o_3}=n_{o_3}rt=1atm \times 8.50 l=n_{o_3}\times 0.0821 atml/k mol \times 298 k$

$n_{o_3}=0.34 moles$

according to reaction 3 moles no are reacting with 1 mol of $o_3$ ,then 0.49 mol of no will react with $\frac{1}{3}\times 0.49$ moles of ozone that is 0.16 moles of ozone. since, no is limiting reagent and $o_3$ is an excessive reagent.

enthalpy change for the reaction between the given volumes of the gases is :

$\delta h_{reaction}\times \text{moles of no}=-311.85kj/mol\times 0.49 mol=-152.80kj/mol$

Another question on Chemistry

Chemistry, 22.06.2019 09:30

What is the mass of 0.46 mol of mgcl2

Answers: 3

Answer

Chemistry, 22.06.2019 16:40

Identify the lewis acid in this balanced equation: ag+ + 2nh3 -> ag(nh3)2+a. ag+b. nh3c. ag(nh3)2+

Answers: 1

Answer

Chemistry, 22.06.2019 16:50

Answer asap need by wednesday morning calculate the ph of 0.16m ch3cooh which has ka = 1.74 x 10-5 mol dm-3 best answer will be brainliest

Answers: 3

Answer

Chemistry, 22.06.2019 23:00

Consider the reaction: 2al(s) + fe2o3(s) → al2o3(s) + 2fe(s) the δhf for fe2o3(s) = -824.3 kj/mole. the δhf for al2o3(s) = -1675.7 kj/mole. finish the equation. δhrxn = [(1)( kj/mole) + (2)( kj/mole)] - [(1)( kj/mole) + (2) ( kj/mole)]

Answers: 1

Answer

You know the right answer?

How many nonbonding electrons are on each nitrogen atom
A 0
B 2
C 4
D 6...

Questions

Chemistry, 27.08.2019 00:30

Which soil can't be rolled into balls or clumps...