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Anna71 [15]
3 years ago
12

Hello! I am working on some homework and came across this question, which completely confuses me for various reasons. Experiment

al evidence suggests that the nitrogen atom in ammonia, NH3, has four identical orbitals in the shape of a pyramid or tetrahedron. a). Draw an energy-level diagram to show the formation of these hybrid orbitals. (Hint: no electron promotion is required). b). Name the type of hybrid orbitals found in NH3. Of the four hybrid orbitals on the N atom, how many will take part in bonding? Explain.
Chemistry
1 answer:
Mariana [72]3 years ago
8 0
Thats hard as hell just drop out of the class lol
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Help please !!!
castortr0y [4]

Answer:

B

pls mark

Explanation:

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3 years ago
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Please help me I really don’t understand
Leokris [45]
I dont understand life, for example, look at my username.

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When magnesium chloride reacts with water, 5.85 L HCl(g) is produced.
amid [387]

Answer: How many moles of HCl was produced?

⇒ 0.261 moles of HCl

How many moles of MgCl2 reacted?

⇒ 0.131 moles of MgCl2

What mass of MgCl2 reacted?

⇒ 12.4 g MgCl2

Explanation:

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Vinegar 59g, Oil 177g, Brown sugar 10g, and total mass
denis-greek [22]

Answer:

Total mass = 246 g

Explanation:

Given data:

Mass of vinegar = 59 g

Mass of oil = 177 g

Mass of brown sugar = 10 g

Total mass = ?

Solution:

Total mass = masses of [sugar + oil + vinegar]

Total mass = 59 g + 177 g + 10 g

Total mass = 246 g

3 0
3 years ago
A volume of 125 mL of H2O is initially at room temperature (22.00 ∘C). A chilled steel rod at 2.00 ∘C is placed in the water. If
VMariaS [17]

Answer:

41.9 g

Explanation:

We can calculate the heat released by the water and the heat absorbed by the steel rod using the following expression.

Q = c × m × ΔT

where,

c: specific heat capacity

m: mass

ΔT: change in temperature

If we consider the density of water is 1.00 g/mol, the mass of water is 125 g.

According to the law of conservation of energy, the sum of the heat released by the water (Qw) and the heat absorbed by the steel (Qs) is zero.

Qw + Qs = 0

Qw = -Qs

cw × mw × ΔTw = -cs × ms × ΔTs

(4.18 J/g.°C) × 125 g × (21.30°C-22.00°C) = -(0.452J/g.°C) × ms × (21.30°C-2.00°C)

ms = 41.9 g

3 0
3 years ago
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