All categories

3 years ago

What do you add to a base to weaken it

Chemistry

1 answer:

lesantik [10]3 years ago

6 0

Answer:

you dont do anything because then you are supporting it

if u add something

Explanation:

You might be interested in

PLEASE HELP AS SOON AS POSSIBLE

lakkis [162]

1. Atoms
2 neutral, proton
3.

7 0

3 years ago

What is the hydrogen ion concentration of 0.450M solution of acetic acid (HC2H3O2), if the ionization constant(Ka) is 2.19 x 10-

WITCHER [35]

Answer:

3.02 Mm

Explanation:

3 0

2 years ago

Calculate the change in pH when 71.0 mL of a 0.760 M solution of NaOH is added to 1.00 L of a solution that is 1.0O M in sodium

Eddi Din [679]

Explanation:

It is known that $pK_{a}$ value of acetic acid is 4.74. And, relation between pH and $pK_{a}$ is as follows.

pH = pK_{a} + log $\frac{[CH_{3}COOH]}{[CH_{3}COONa]}$

= 4.74 + log $\frac{1.00}{1.00}$

So, number of moles of NaOH = Volume × Molarity

= 71.0 ml × 0.760 M

= 0.05396 mol

Also, moles of $CH_{3}COOH$ = moles of $CH_{3}COONa$

= Molarity × Volume

= 1.00 M × 1.00 L

= 1.00 mol

Hence, addition of sodium acetate in NaOH will lead to the formation of acetic acid as follows.

$CH_{3}COONa + NaOH \rightarrow CH_{3}COOH$

Initial : 1.00 mol 1.00 mol

NaoH addition: 0.05396 mol

Equilibrium : (1 - 0.05396 mol) 0 (1.00 + 0.05396 mol)

= 0.94604 mol = 1.05396 mol

As, pH = pK_{a} + log $\frac{[CH_{3}COONa]}{[CH_{3}COOH]}$

= 4.74 + log $\frac{0.94604}{1.05396}$

= 4.69

Therefore, change in pH will be calculated as follows.

pH = 4.74 - 4.69

= 0.05

Thus, we can conclude that change in pH of the given solution is 0.05.

8 0

2 years ago

What was one main point of daltons atomic theory

bagirrra123 [75]

The main points of Dalton's atomic theory are: Everything is composed of atoms, which are the indivisible building blocks of matter and cannot be destroyed. All atoms of an element are identical. The atoms of different elements vary in size and mass.

4 0

3 years ago

Fish breathe the dissolved air in water through their gills. Assuming the partial pressures of oxygen and nitrogen in air to be

amid [387]

Answer:

X(O₂) = 0.323

X(N₂) = 0.677

Explanation:

We have the partial pressures of oxygen (O₂) and nitrogen (N₂):

P(O₂) = 0.20 atm

P(N₂) = 0.80 atm

In order to solve the problem, you need the solubilities of each gas in water at 298 K. We can consider 1.3 x 10⁻³ mol/(L atm) for oxygen (O₂) and 6.8 x 10⁻⁴mol/(L atm) for nitrogen (N₂) from the bibliography.

s(O₂) = 1.3 x 10⁻³ mol/(L atm)

s(N₂) = 6.8 x 10⁻⁴mol/(L atm)

So, we calculate the concentration (C) of each gas as the product of its partial pressure (P) and the solubility (s):

C(O₂) = P(O₂) x s(O₂) = 0.20 atm x 1.3 x 10⁻³ mol/(L atm) = 2.6 x 10⁻⁴mol/L

C(N₂) = P(N₂) x s(N₂) = 0.80 atm x 6.8 x 10⁻⁴mol/(L atm) = 5.44 x 10⁻⁴ mol/L

In 1 liter of water, we have the following number of moles (n):

n(O₂) = 2.6 x 10⁻⁴ mol

n(N₂) = 5.44 x 10⁻⁴ mol

Thus, the total number of moles (nt) is calculated as the sum of the number of moles of the gases in the mixture:

nt = n(O₂) + n(N₂) = 2.6 x 10⁻⁴ mol + 5.44 x 10⁻⁴ mol = 8.04 x 10⁻⁴ mol

Finally, the mole fraction of each gas is calculated as the ratio between the number of moles of each gas and the total number of moles:

X(O₂) = n(O₂)/nt = 2.6 x 10⁻⁴ mol/(8.04 x 10⁻⁴ mol) = 0.323

X(N₂) = n(N₂)/nt = 5.44 x 10⁻⁴ mol/(8.04 x 10⁻⁴ mol) = 0.677

5 0

2 years ago