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3 years ago

15

How would the electron configuration of nitrogen change to make a stable configuration?

2 answers:

kozerog [31]3 years ago

7 0

Answer:

It would gain three electrons.

Explanation:

I just took it

lesantik [10]3 years ago

5 0

Answer: The electronic configuration of nitrogen is 1s2 2s2 2p3. Thus nitrogen has a half-filled p orbital, which is comparatively more stable. Thus the p orbital is the outermost orbital. To achieve a stable gas configuration, nitrogen needs to have a fulfilled p orbital.

Explanation: please mark brainlyest i really nead it

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When the [H+] in a solution is 1.7 × 10−9 M, what is the pOH

Harrizon [31]

The pOH of a solution is related to the concentration of hydroxide ions in a solution. Also, it is related to the pH of the solution since the sum of pOH and pH is equal to 14. We can use this relation to solve this problem. We do as follows:

pH = -log[<span>1.7 × 10−9 M</span>] = 8.8
pOH + pH = 14
pOH + 8.8 = 14
pOH = 5.2

7 0

3 years ago

Read 2 more answers

1. Define a Family.

Nataliya [291]

Answer:

a family is a group consisting of 2 parents and their children living together in a union.

or all desendants of a common ancestor

2. A period is a name given to the horizontal row of the periodic table.

3.

a.mg

b.k

c.Fe

d.cu

e.Al

4.

a.Carbon

b.Chlorine

c.Gold

d.strontium

5.

a.period 1

b.period 4

c.period 5

d.period 5

6.

a.group 6

b.group 2

c.group 7

d.group 8 belong ti 1st transitional elements.

7.

a.F

b.Na

c.Ar

d.V

e.I

4 0

3 years ago

The modern atomic model includes electrons orbiting the nucleus. which of the following parts of dalton's atomic model was dispr

lana66690 [7]

Atom can never be divided into smaller particles is the answer was disproved

7 0

3 years ago

A sample of pure NH4HS is placed in a sealed 2.0-L container and heated to 550 K at which the equilibrium constant is 3.5 x 10-3

Zolol [24]

Answer:

The mass of $NH_{3}$ in the container is 2.074 gram

Explanation:

Given:

Volume of $NH_{4} HS$ $V = 2$ lit

Equilibrium constant $k _{eq} = 3.5 \times 10^{-3}$

The reaction in which $NH_{3}$ is produced

$NH_{4} HS$ ⇄ $NH_{3} + H_{2}S$

Here equal moles of $NH_{3}$ and $H_{2}S$ is formed.

From the formula of equilibrium constant,

$k_{eq} = (NH_{3})(H_{2}S )$

$x^{2} = 3.5 \times 10^{-3}$

$x = 0.061$ M

Above value shows,

$NH_{3} = 0.061$ $\frac{moles}{L}$

So in 2 L no. moles of $NH_{3}$ = $0.061 \times 2 = 0.122$ moles.

So mass of 0.122 mole of $NH_{3}$ is = $0.122 \times 17 = 2.074$ g

Therefore, the mass of $NH_{3}$ in the container is 2.074 gram

8 0

3 years ago

How many molecules are in 41.8 g of sulfuric acid

Anton [14]

Answer

× 10²³ molecules are in 41.8 g of sulfuric acid

Explanation

The first step is to convert 41.8 g of sulfuric acid to moles by dividing the mass of sulfuric acid by its molar mass.

Molar mass of sulfuric acid, H₂SO₄ = 98.079 g/mol

$Mole=\frac{Mass}{Molar\text{ }mass}=\frac{41.8\text{ }g}{98.079\text{ }g\text{/}mol}=0.426187053\text{ }mol$

Finally, convert the moles of sulfuric acid to molecules using Avogadro's number.

Conversion factor: 1 mole of any substance = 6.022 × 10²³ molecules.

Therefore, 0.426187053 moles of sulfuric acid is equal

$\frac{0.426187053\text{ }mol}{1\text{ }mol}\times6.022×10²³\text{ }molecules=2.57\times10^{23}\text{ }molecules$

Thus, 2.57 × 10²³ molecules are in 41.8 g of sulfuric acid.

3 0

1 year ago