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galina1969 [7]
3 years ago
7

29. Ammonia is formed with one nitrogen and three hydrogen atoms. What is the electron configuration of nitrogen? How

Chemistry
1 answer:
masya89 [10]3 years ago
5 0

Answer:

<u>One lone-Pair is present in Ammonia</u>

<u></u>

Explanation:

The number of valence electron in N = 5

The number of Valence electron in H = 1

The formula of ammonia = NH3

Total valence electron in ammonia molecule = 5 +3(1) = 5+3 = 8

The lewis structure suggest that :

Nitrogen completes its octet by sharing the electron pair with 3 hydrogen atoms.

3 electron of Nitrogen are involved in sharing with Hydrogen

So,<u><em> remaining two electron are left non-bonded</em></u> . Hence they exist as lone- pair

So, there is only 1 lone pair in the ammonia molecule .

The shape of NH3 is bent according to VSEPR theory . This is so because the presence of 1 lone pair causes more repulsion and occupy more space.

Thus the lone pair is changing the shape of the ammonia molecule . It also increase the dipole moment of the molecule , which gives polarity to it.

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An exothermic reaction occurs spontaneously ,why?​
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Answer:

See below  

Step-by-step explanation:

An exothermic reaction tends to occur spontaneously because the products are more stable than the reactants.

Nature tries to get to the lowest energy state.

4 0
3 years ago
An atom of gold has a mass of 3.271 X 10-22 g. How many atoms of gold are in 5.00 g of gold? (Give your answer in scientific not
Kobotan [32]

Answer:

1.53 × 10²² atoms Ag

Explanation:

Step 1: Define conversions

3.271 × 10⁻²² g = 1 atom

Step 2: Use Dimensional Analysis

5.00 \hspace{3} g \hspace{3} Ag(\frac{1 \hspace{3} atom \hspace{3} Ag}{3.271(10)^{-22} \hspace{3} g \hspace{3} Ag} ) = 1.52858 × 10²² atoms Ag

Step 3: Simplify

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1.52858 × 10²² atoms Ag ≈ 1.53 × 10²² atoms Ag

5 0
2 years ago
What volume of Co2 (carbon (iv) oxide)
hram777 [196]

Answer:

2.1056L or 2105.6mL

Explanation:

We'll begin by calculating the number of mole in 10g of Na2CO3. This can be obtained as follow:

Molar mass of Na2CO3 = (23x2) + 12 + (16x3) = 106g/mol

Mass of Na2CO3 = 10g

Mole of Na2CO3 =.?

Mole = mass /molar mass

Mole of Na2CO3 = 10/106

Mole of Na2CO3 = 0.094 mole

Next, we shall determine the number of mole CO2 produced by the reaction of 0.094 mole of Na2CO3. This is illustrated below:

Na2CO3 + 2HCl —> 2NaCl + H2O + CO2

From the balanced equation above,

1 mole of Na2CO3 reacted to produce 1 mole of CO2.

Therefore, 0.094 mole of Na2CO3 will also react to 0.094 mole of CO2.

Next, we shall determine the volume occupied by 0.094 mole of CO2 at STP. This is illustrated below:

1 mole of a gas occupy 22.4L at STP. This implies that 1 mole CO2 occupies 22.4L at STP.

Now, if 1 mole of CO2 occupy 22.4L at STP, then, 0.094 mole of CO2 will occupy = 0.094 x 22.4 = 2.1056L

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7 0
3 years ago
When will a seed begin to germinate?
Vlada [557]
When it has sunlight and water
4 0
3 years ago
Read 2 more answers
Give all possible ml values for orbitals that have each of the following: (a) l = 2; (b) n = 1; (c) n = 4, l = 3.
Olin [163]

Answer:

(a) ml = 0, ±1, ±2

(b) ml = 0

(c) ml = 0, ±1, ±2, ±3, ±4

Explanation:

The rules for electron quantum numbers are:

1. Shell number, 1 ≤ n

2. Subshell number, 0 ≤ l ≤ n − 1

3. Orbital energy shift, -l ≤ ml ≤ l

4. Spin, either -1/2 or +1/2

So in our exercise,

(a) l = 2; equivalent with with sublevel <em>d</em>

-l ≤ ml ≤ l, ml = 0, ±1, ±2, equivalent with dxy, dxz, dyz, dx2-y2, dz2

(b) n = 1;

n = 1, only 01 level

l = 0, equivalent with sublevel <em>s</em>

ml = 0

(c) n = 4, l = 3.

l = 3, equivalent with sublevel <em>f</em>

ml = 0, ±1, ±2, ±3, ±4

7 0
3 years ago
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