All categories

4 years ago

2Al + 3MnSO4 ↔ Al2(SO4)3 + Mn

1 answer:

5 0

In accordance with the Le Chatelier's Principle, when the concentration of the aluminum increases then the equilibrium would shift to the right. It will produce more products since there are more reactants available and also collision is more frequent.

You might be interested in

which diagram best illustrates the ion-molecule attractions that occur when the ions of NaCl(s) are added to water

tatuchka [14]

Diagram is on the picture below.
Answer is: 1).
Sodium chloride is ionic compound and in the water dissociate in sodium cation (positive charge) and chloride anion (negative charge). Water is polar compound, oxagan has negative charge and hydrogen charge. Positive interact witn negative charge and negative with positive charge.

8 0

4 years ago

A metallic bond can exist between atoms that have low ionization energies and

zavuch27 [327]

Answer: The elements that have the lowest electronegativity are the VIII A elements or noble gases. These elements have a theoretical electronegativity of zero. These elements are stable in their electron configuration there is not force moving the noble gases to gain any electrons.

Explanation:

7 0

3 years ago

QUESTION 57

Tema [17]

57 - True

58 - True

59 - True

60 - Fe and O

61 - Sr and F

62 - <span>both ionic and covalent bonds

Hope this helps!</span>

6 0

4 years ago

Read 2 more answers

Name two methods scientists use to obtain empirical evidence

adell [148]

Through hypothesis and experiments

4 0

4 years ago

A 0.20 M solution of a weak acid has a pH of 5.40. What is the Ka for the acid?

sweet [91]

Answer:

The Ka for this weak acid is 7,92 * 10^-11

Explanation:

First of all, let's think the equation

HA + H2O <------> H3O+ + A-

When we add water to a weak acid, it dissociates in an equilibrium to generate the corresponding anion and the hydronium cation (the acid form of water)

How do you calculate Ka?? Ka is the acid equilibrium constant.

( [H3O+] . [A-] ) / [HA] where all the concentrations are in equilibrium.

We don't have the concentration in equilibrium but we have the initial concentration. So...

HA + H2O <------> H3O+ + A-

initial- 0.2 M I don't have H3O+, either A-

reaction - an specific amount reacted (X)

in equilibrium (0,2 - X) <-----> X + X

And now, how's the formula for Ka

( [H3O+] . [A-] ) / [HA] = Ka

(X . X) / (0.2-X)

X^2 / (0.2-X) = Ka

Look, that we don't have X as the [H3O+] but we know the pH, so we can know the [H3O+] indeed.

10^-pH = [H3O+]

10^-5,40 = 3,98 * 10^-6

Let's go back to Ka

( [H3O+] . [A-] ) / [HA] = Ka

(3,98 * 10^-6)^2 / (0.2 - 3,98 * 10^-6) = Ka

(3,98 * 10^-6 is an small number, soooo small that we can approximate to 0)

If we have in order 10^-6, 10^-5 we can consider that.

So now, we have

(3,98 * 10^-6)^2 / (0.2) = Ka

1,58 * 10^-11 / (0.2) = Ka = 7,92* 10^-11

5 0

3 years ago