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erik [133]
1 year ago
12

How many hydrogen bonds can CH2O make to water

Chemistry
1 answer:
VladimirAG [237]1 year ago
5 0
Hydrogen bonds are not like covalent bonds. They are nowhere near as strong and you can't think of them in terms of a definite number like a valence. Polar molecules interact with each other and hydrogen bonds are an example of this where the interaction is especially strong. In your example you could represent it like this: 

<span>H2C=O---------H-OH </span>

<span>But you should remember that the H2O molecule will be exchanging constantly with others in the solvation shell of the formaldehyde molecule and these in turn will be exchanging with other H2O molecules in the bulk solution. </span>

<span>Formaldehyde in aqueous solution is in equilibrium with its hydrate. </span>

<span>H2C=O + H2O <-----------------> H2C(OH)2</span>
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Why is it important to have regular supervision of the weight and measurements in the market
Sergio [31]

Answer:

Supervision of weights and measures promotes accurate measurements of goods and services to ensure that everybody gets a fair trade in the marketplace. Not so coincidentally it also is a deterrent to ensure that traders are being honest in their trade practises.

Explanation:

4 0
1 year ago
The reaction between hydrogen gas and chlorine gas produces hydrogen chloride according to the following equation: H2(g) + Cl2(g
Andreyy89

Answer:

The enthalpy of reaction is -185 kJ

Explanation:

To get the reaction:

 H₂(g) + Cl₂(g) → 2 HCl(g)

you must follow the following steps:

<em>1)  Reactive molecules must break their bonds to obtain their atoms.</em>

H₂(g) → 2 H(g)

Cl₂(g) → 2 Cl(g)

Bond energy (or enthalpy) is the energy required to break one mole of bonds of a gaseous substance. In the case of diatomic molecules with a single bond, it corresponds to the energy necessary to dissociate 1 mole of said substance in the atoms that form it.

Whenever you want to break links you must supply energy, so the link enthalpy will have positive values; while when a mole of bonds is formed energy is released and the bond enthalpy of this process will be negative.

In this case you will then have:

H₂(g) → 2 H(g)           ΔH=436 kJ/mol

Cl₂(g) → 2 Cl(g)         ΔH=243 kJ/mol

So the total energy needed to break all the bonds is:

ΔH=1 mol*436 kJ/mol +1 mol* 243 kJ/mol= 679 kJ

2) The atoms that were obtained in the break of the bonds must be combined to obtain the product.

2 H (g) + 2 Cl (g) → 2 HCl (g)

Being the single bond energy for one mole of 431 kJ H-Cl bonds and considering that two moles of H-Cl bonds are formed, the ΔH is:

ΔH = -2 moles* (432 kJ/mol) = -864 kJ

As mentioned, when a mole of bonds is formed energy is released, the bond enthalpy of this process will be negative.  So the formation of HCl is negative.

Hess's law states that the energy change in an overall chemical reaction is equal to the sum of the energy changes in the individual reactions comprising it. So:

ΔHtotal= -864 kJ + 679 kJ

ΔHtotal= -185 kJ

<u><em>The enthalpy of reaction is -185 kJ</em></u>

3 0
1 year ago
What is the hydronium ion concentration of a solution whose pH is 7.30
Assoli18 [71]
[ H₃O⁺] = 10 ^ - pH

[ H₃O⁺ ] = 10 ^ - 7.30

[ H₃O⁺ ] = 5.011 x 10⁻⁸ M

hope this helps!
6 0
1 year ago
Hector claims that a single sodium ion and a single oxygen ion do not bond together to form a stable binary ionic compound. Is h
levacccp [35]

Answer:

C

Explanation:

Looking at the periodic table, we can see that sodium is in group 1, so a sodium ion would be Na⁺, with a charge of +1. Oxygen is in group 16, so an oxygen ion would be O²⁻, with a charge of -2.

A compound formed only by a single sodium ion and a single oxygen ion would thus have a charge of -1, and in order to have a stable ionic compound its charge must be zero.

8 0
1 year ago
Read 2 more answers
When 9.2 g of frozen N2O4 is added to a 0.50 L reaction vessel and the vessel is heated to 400 K and allowed to come to equilibr
Amanda [17]

<u>Answer:</u> The value of K_c for the given reaction is 1.435

<u>Explanation:</u>

To calculate the molarity of solution, we use the equation:

\text{Molarity of the solution}=\frac{\text{Mass of solute}}{\text{Molar mass of solute}\times \text{Volume of solution (in L)}}

Given mass of N_2O_4 = 9.2 g

Molar mass of N_2O_4 = 92 g/mol

Volume of solution = 0.50 L

Putting values in above equation, we get:

\text{Molarity of solution}=\frac{9.2g}{92g/mol\times 0.50L}\\\\\text{Molarity of solution}=0.20M

For the given chemical equation:

                 N_2O_4(g)\rightleftharpoons 2NO_2(g)

<u>Initial:</u>          0.20

<u>At eqllm:</u>     0.20-x        2x

We are given:

Equilibrium concentration of N_2O_4 = 0.057

Evaluating the value of 'x'

\Rightarrow (0.20-x)=0.057\\\\\Rightarrow x=0.143

The expression of K_c for above equation follows:

K_c=\frac{[NO_2]^2}{[N_2O_4]}

[NO_2]_{eq}=2x=(2\times 0.143)=0.286M

[N_2O_4]_{eq}=0.057M

Putting values in above expression, we get:

K_c=\frac{(0.286)^2}{0.143}\\\\K_c=1.435

Hence, the value of K_c for the given reaction is 1.435

6 0
2 years ago
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