Ask question

Ask question

All categories

2 years ago

6

What is the bond energy per mole for breaking all the bonds of oxygen, O 2 ?

1 answer:

vaieri [72.5K]2 years ago

4 0

There is only one O=O bond in an O2 molecule, so the energy ΔHO2 required for breakingall the bonds in a mole of O2 molecules is 498 kJ/mol O2 molecules. Calculate the bond energy per mole for forming all the bonds of water molecules, H2O

You might be interested in

Like all equilibrium constants, Kw varies somewhat with temperature. Given that Kw is 3.31 × 10−13 at some temperature, compute

Artyom0805 [142]

Since Kw= [H⁺][OH⁻], and the concentration of both substances are the same, the equation is now Kw=[H⁺]²
So,
3.31x10⁻¹³ = [H⁺]²
Take the square root= 5.75x10⁻⁷
Then take the negative log to find the pH:
-log(5.75x10⁻⁷) = 6.25

5 0

2 years ago

A 1.50-liter sample of dry air in a cylinder exerts a pressure of 3.00 atmospheres at a temperature of 25°C. Without changing th

Aneli [31]

Hope this helps you.

3 0

2 years ago

Read 2 more answers

1. What is the oxidation number for the silver ion in tarnish?

spayn [35]

Tarnish is Ag2S-silver sulfide and the oxidation state of silver is +1

7 0

1 year ago

If an equal number of moles of reactants are used, do the following equilibrium mixtures contain primarily reactants or products

puteri [66]

<u>Answer:</u>

<u>For a:</u> The equilibrium mixture contains primarily reactants.

<u>For b:</u> The equilibrium mixture contains primarily products.

<u>Explanation:</u>

There are 3 conditions:

When $K_{eq}>1$ ; the reaction is product favored.
When $K_{eq}$ ; the reaction is reactant favored.
When $K_{e}=1$ ; the reaction is in equilibrium.

For the given chemical reactions:

<u>For a:</u>

The chemical equation follows:

$HCN(aq.)+H_2O(l)\rightleftharpoons CN^-(aq.)+H_3O^+(aq.);K_{eq}=6.2\times 10^{-10}$

The expression of $K_{eq}$ for above reaction follows:

$K_{eq}=\frac{[CN^-][H_3O^+]}{[HCN][H_2O]}=6.2\times 10^{-10}$

As, $K_{eq}$ , the reaction will be favored on the reactant side.

Hence, the equilibrium mixture contains primarily reactants.

<u>For b:</u>

The chemical equation follows:

$H_2(g)+Cl_2(g)\rightleftharpoons 2HCl(g);K_{eq}=2.51\times 10^{4}$

The expression of $K_{eq}$ for above reaction follows:

$K_{eq}=\frac{[HCl]^2}{[H_2][Cl_2]}=2.51\times 10^{4}$

As, $K_{eq}>1$ , the reaction will be favored on the product side.

Hence, the equilibrium mixture contains primarily products.

4 0

2 years ago

What is the overall charge of the compound frbr

sammy [17]

Answer:

Zero

Explanation:

FrBr is an ionic compound .

Fr is in Group 1. Br is in Group 17.

The charges on the ions are +1 and -1, respectively.

The compound consists of Fr⁺Br⁻ ions.

However, there are equal numbers of + and - charges, so

The overall charge of the compound is zero.

5 0

2 years ago