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Westkost [7]
3 years ago
15

For H 2 O, the H-O-H bond angle is 104.5 o and the measure dipole moment is 1.85 Debye. What is the magnitude of the O-H bond di

pole moment for H 2 O? Estimate the bond angle in H 2 S given that the H-S bond dipole moment is 0.67 Debye and the resultant molecular dipole moment is 0.95 Debye.
Chemistry
1 answer:
viva [34]3 years ago
6 0

Answer:

Dipole moment of H₂O: 2.26 Debye

Bond angle in H₂S: 89.7°

Explanation:

The total dipole moment(μr) for H₂O can be calculated by the sum of the dipole moments of each bond of O-H. Because the dipole moment is a vector, the sum of these vectors can be calculated by the cosine law.

μr² = (μO-H)² + (μO-H)² + 2*(μO-H)*(μO-H)*cos(104.5°)

μr² = 1.85² + 1.85² + 2*1.85*1.85*cos(104.5°)

μr² = 3.4225 + 3.4225 + 6.845*(-0.2504)

μr² = 5.13115

μr = √5.13115

μr = 2.26 Debye.

For H₂S:

0.95² = 0.67² + 0.67² + 2*0.67*0.67*cosθ

0.9025 = 0.4489 + 0.4489 + 0.8978cosθ

0.8978cosθ = 0.0047

cosθ = 0.00523

θ = arcos0.00523

θ = 89.7°

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Answer:

0.56M of acetate ions

Explanation:

Given parameters:

Mass of Ba(C₂H₃O₂)₂  = 69g

Volume of water  = 970mL  = 0.97dm³

Molar mass of Ba(C₂H₃O₂)₂ = 255.415g/mol

Unknown:

Concentration of acetate ion in the final solution = ?

Solution:

Let us represent the dissociation;

                Ba(C₂H₃O₂)₂  = Ba²⁺   + 2C₂H₃OO⁻

We see that 1M of will produce 2M of acetate ions

Now, let us find the molarity of the barium acetate;

            Molarity  = \frac{number of moles }{volume}

                Number of moles of Ba(C₂H₃O₂)₂ = \frac{mass}{molar mass}

                Number of moles  = \frac{69}{255.415}  = 0.27moles

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since 1M of Ba(C₂H₃O₂)₂ will produce 2M of acetate ions

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The reaction between ethyne (acetylene, C 2 H 2 ) and hydrogen. The product is ethane (C 2 H 6 ). Which is the limiting reactant
serious [3.7K]

Answer:

Three possible cases:

- If amount are equal for each reactant (for example 1 mol each), the limiting is the hydrogen and the excess reagent is the acetylene.

- When moles of H₂ are greater than C₂H₂

The acetylene is the limiting reagent so the H₂ is the excess

-  When moles of C₂H₂ are greater than H₂

For this case, H₂ is the limiting reactant and the excess is the C₂H₂

Explanation:

First of all we determine the reaction:

Reactants, acetylene and hydrogen

Products are ethane

Then, the balanced reaction is: C₂H₂ + 2H₂ → C₂H₆

1 mol of acetylene reacts with 2 moles of hydrogen ir order to produce 1 mol of ethane.

If amount are equal for each reactant, the limiting is the hydrogen,

For example, 1 mol each

For 1 mol of acetylene I need 2 moles of H₂. I've only got 1 mol, so I do not have enough H₂. The excess reagent is the acetylene.

- When moles of H₂ are greater than C₂H₂

For example, 3 moles of H₂ and 0.5 mol of C₂H₂

2moles of H₂ need 1 mol of C₂H₂ for the reaction

Then 3 moles of H₂ will need (3 . 1) / 2 = 1.5 moles

We have 0.5 moles, so the acetylene is the limiting reagent, again.

- When moles of C₂H₂ are greater than H₂

For example 1 mol of C₂H₂ and 0.001 moles of H₂

If I have 1 mol of C₂H₂, I definetly need the double of moles of hydrogen, so in this case, H₂ is the limiting reactant and the excess is the C₂H₂

If we have 1 mol of H₂ and 0.5 mol of C₂H₂, notice that moles of acetylene are lower than hydrogen

1 mol of C₂H₂ needs 2 moles of H₂

So 0.5 moles of C₂H₂ will need 1 mol of H₂ (it's ok because we have 1 mol)

2 moles of H₂ need 1 mol of C₂H₂ for reaction

Then, 1 mol of H₂ will need 0.5 moles of C₂H₂ (it's ok because we have that amount)

In this case, there is no excess neither limiting. That's why we can choose any of them to determine the moles (or mass) for the product

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Learn more: brainly.com/question/6505878

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