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Setler79 [48]
2 years ago
9

How many moles of nacl are needed to make 5.2l of a 2m solution

Chemistry
1 answer:
son4ous [18]2 years ago
7 0

Answer:

10.4 moles

Explanation:

    n / 5.2   =   2  M

     n = 10.4  moles

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If a buffer solution is 0.220 M in a weak acid ( Ka=7.4×10−5) and 0.540 M in its conjugate base, what is the pH?
valkas [14]

Answer: the pH of the solution is 4.52

Explanation:

Consider the weak acid as Ha, it is dissociated as expressed below

HA     H⁺  +  A⁻

the Henderson -Haselbach equation can be expressed as;

pH = pKa + log( [A⁻] / [HA])

the weak acid is dissociated into H⁺ and A⁻ ions in the solution.

now the conjugate base of the weak acid HA is

HA(aq) {weak acid}     H⁺(aq)  +  A⁻(aq) {conjugate base}

so now we calculate the value of Kₐ as well as pH value by substituting the values of the concentrations into the equation;

pKₐ = -logKₐ

pKₐ = -log ( 7.4×10⁻⁵ )

pKₐ = 4.13

now thw pH is

pH = pKₐ  + log( [A⁻] / [HA])

pH = 4.13 + log( [0.540] / [0.220])

pH = 4.13 + 0.3899

pH = 4.5199 = 4.52

Therefore the pH of the solution is 4.52

6 0
3 years ago
Question 12 of 25
NISA [10]

Answer:

B. Equal amounts of all gases have the same volume at the same

conditions

Explanation:

Amedo Avogadro found the relationship between volume of a gas and the number of molecules contained in the volume.

The law states that "equal volumes of all gases at the same temperature and pressure contains equal number of molecules or moles".

The law describes the behavior of gases when involve in chemical reactions. It enables one to change over at will in any statement about gases from volumes to molecules and vice versa.

So, the right option is B which implies that equal amounts of all gases have the same volume at the same conditions.

8 0
2 years ago
calculate the water potential of a solution of 0.15m sucrose. the solution is at standard temperature.
Mrac [35]

Answer:

The water potential of a solution of 0.15 M sucrose solution is -3.406 bar.

Explanation:

Water potential = Pressure potential + solute potential

P_w=P_p+P_s

P_w=P_p+(-iCRT)

We have :

C = 0.15 M, T = 273.15 K

i = 1

The water potential of a solution of 0.15 m sucrose= P_w

P_p=0 bar (At standard temperature)

P_s=-iCRT=-\times 1\times 8.314\times 10^{-2}bar L/mol K\times 273.15 K=-3.406 bar

P_w=0 bar+(-3.406 ) bar

The water potential of a solution of 0.15 M sucrose solution is -3.406 bar.

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2 years ago
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