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3 years ago

What’s the electron configuration of an N-3 ion

Chemistry

1 answer:

Alla [95]3 years ago

8 0

Answer:

1s2 2s2 2p6

Explanation:

The nitride ion(N^3-) is formed when nitrogen gains three electrons. Nitrogen possesses seven electrons in its orbitals and ordinarily has the electronic configuration; 1s2 2s2 2p3. However,being in group 15, nitrogen can accept three electrons to form the nitride ion and complete its octet of electrons. When this happens, three electrons are added to the nitrogen atom and the electronic configuration is now the same as that of Neon, its closest noble gas which is 1s2 2s2 2p6. Hence the answer given above.

Elements can accept or donate electrons in order to complete their octet structure in accordance to the octet rule which states that atoms and ions must possess eight electrons in their outermost shell in order to attain chemical stability. The reason for ion formation and chemical reaction is in order for species to attain the octet structure.

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The equilibrium 2NO(g)+Cl2(g)⇌2NOCl(g) is established at 500 K. An equilibrium mixture of the three gases has partial pressures

QveST [7]

Answer:

For A: The $K_p$ for the given reaction is $4.0\times 10^1$

For B: The $K_c$ for the given reaction is 1642.

Explanation:

The given chemical reaction follows:

$2NO(g)+Cl_2(g)\rightleftharpoons 2NOCl(g)$

For A:

The expression of $K_p$ for the above reaction follows:

$K_p=\frac{(p_{NOCl})^2}{(p_{NO})^2\times p_{Cl_2}}$

We are given:

$p_{NOCl}=0.24 atm\\p_{NO}=9.10\times 10^{-2}atm=0.0910atm\\p_{Cl_2}=0.174atm$

Putting values in above equation, we get:

$K_p=\frac{(0.24)^2}{(0.0910)^2\times 0.174}\\\\K_p=4.0\times 10^1$

Hence, the $K_p$ for the given reaction is $4.0\times 10^1$

For B:

Relation of $K_p$ with $K_c$ is given by the formula:

$K_p=K_c(RT)^{\Delta ng}$

where,

$K_p$ = equilibrium constant in terms of partial pressure = $4.0\times 10^1$

$K_c$ = equilibrium constant in terms of concentration = ?

R = Gas constant = $0.0821\text{ L atm }mol^{-1}K^{-1}$

T = temperature = 500 K

$\Delta ng$ = change in number of moles of gas particles = $n_{products}-n_{reactants}=2-3=-1$

Putting values in above equation, we get:

$4.0\times 10^1=K_c\times (0.0821\times 500)^{-1}\\\\K_c=\frac{4.0\times 10^1}{(0.0821\times 500)^{-1})}=1642$

Hence, the $K_c$ for the given reaction is 1642.

7 0

2 years ago

Identify the phase changes in the figure that are labeled a

BigorU [14]

Isndidndjncidbdjbfidhjdbidhxidhfidhjfjxidjdidjdj

6 0

2 years ago

The fuel used in many disposable lighters is liquid butane, C4H10. Butane has a molecular weight of 58.1 grams in one mole. How

forsale [732]

Doesnt the number of carbon atoms stay the same.
Though the weight of carbon in 1.5g is 1.24g.

This is because the RAM of C4 is 48.

The RFM of C4H10 is 58. Therefore, 48/58 is carbon in butane.

48/58 x 1.5 = 1.24g

7 0

3 years ago

At a particular temperature, K 3.39 for the reaction SO2(9) + NO2(9) SOs(9) +NO(9) If all four gases had initial concentrations

Korvikt [17]

Answer:

The equilibrium concentrations are:

[SO2]=[NO2] = 0.563 M

[SO3]=[NO] = 1.04 M

Explanation:

Given:

Equilibrium constant K = 3.39

[SO2] = [NO2] = [SO3] = [NO] = 0.800 M

To determine:

The equilibrium concentrations of the above gases

Calculation:

Set-up an ICE table for the given reaction

$SO2(g) + NO2(g)\rightleftharpoons SO3(g) + NO(g)$

I 0.800 0.800 0.800 0.800

C -x -x +x +x

E (0.800-x) (0.800-x) (0.800+x) (0.800+x)

The equilibrium constant is given as:

$Keq = \frac{[SO3][NO]}{[SO2][NO2]}=\frac{(0.800+x)^{2}}{(0.800-x)^{2}}$

$3.39=\frac{(0.800+x)^{2}}{(0.800-x)^{2}}$

x = 0.2368 M

[SO2]=[NO2] = 0.800 -x = 0.800 - 0.2368 = 0.5632 M

[SO3]=[NO] = 0.800 +x = 0.800 + 0.2368 = 1.037 M

4 0

2 years ago

What must differ beween the atom of two different elements?

kodGreya [7K]

The number of protons must differ between two different elements

5 0

3 years ago

Read 2 more answers