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jek_recluse [69]
2 years ago
5

True or False When one side of a molecule is electronegative (δ-) and the other side of the

Chemistry
1 answer:
lesya [120]2 years ago
8 0

Answer:

True; When one side of a molecule is electronegative (δ-) and the other side of the

molecule is electropositive (δ+), it is said to have a dipole moment.

Explanation:

A dipole moment exists in a molecule as a result of differences in the electronegativity values between the atoms of the elements involved in the chemical bonding.

When a strogly electronegative atom such as oxygen or chlorine is chemically bonded to a less electronegative or an electropositive atom such as hydrogen, there is an uneven sharing of the electrons involved in the bonding. The more electronegative atoms tends to draw the shared electrons mostly to themselves. This induces a partially negative charge (δ-) on them while leaving the electropositive atoms with a partially positive charge (δ+).

Water is an example of a molecule having a dipole moment. The oxygen atoms are more electronegative than hydrogen and as such draw the shared electrons to themselves more, inducing a partial positive charge (δ+) on the hydrogen atoms while they themselves develop a partial negative charge (δ-).

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3 years ago
Chemical reactions forming ions _____.
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Porcentaje en masa formula y unidad<br><br>​
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2 years ago
What is the volume of ammonia produced at 243 K at a pressure of 1.38 atm by the unbalanced reaction on the left if 5740 moles o
Neporo4naja [7]

Answer:

49671 L is the produced volume of ammonia

Explanation:

We think the reaction of ammonia 's production:

N₂(g) + 3H₂(g)  → 2NH₃ (g)

We have the moles of each reactant so let's determine the limiting reactant:

Ratio is 1:3. 1 mol of nitrogen reacts with 3 moles of H₂

Then, 1720 moles of N₂ will react with (1720 .3) /1 = 5160 moles of H₂

We have 5740 moles of hydrogen, so we have enough hydrogen. This is the excess reagent, so the limiting is the N₂

1 mol of N₂ can produce 2 moles of ammonia

Therefore 1720 moles of N₂ will produce (1720 . 2) /1 = 3440 moles of NH₃

We apply now, the Ideal Gases Law → P . V = n . R .T

V = (n . R . T) /P → V = (3440 mol . 0.082 L.atm/mol.K . 243K) / 1.38 atm

V = 49671 L

We confirm that the nitrogen was the limiting reactant

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Therefore, 5740 moles of H₂ will react with (5740 . 1) /3 = 1913 moles of N₂

It was ok to say, that N₂ was the limiting reactant because we need 1913 moles in the reaction, and we only have 1720 moles

6 0
3 years ago
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