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25. 00 ml of a buffer solution contains 0. 500 m hclo and 0. 380 m naclo. if 50. 00 ml of water is added to the buffer, what are

geniusboy [140]

The new concentrations of $HClO$ and $NaClO$ are 0.25M and 19M

Calculation of number of moles of each component,

Molarity of $HClO$ = number of moles/volume in lit = 0. 500 M

Number of moles = molarity of $HClO$ × volume in lit = 0. 500 M× 0.025 L

Number of moles of $HClO$ = 0.0125 mole

Molarity of $NaClO$ = number of moles/volume in lit = 0. 38 M

Number of moles = molarity of $NaClO$ × volume in lit = 0. 38 M× 0.025 L

Number of moles of $NaClO$ = 0.95 mole

Calculation of new concentration at volume 50 ml ( 0.05L)

Molarity of $HClO$ = number of moles/volume in lit = 0.0125 mole/0.05L

Molarity of $HClO$ = 0.25M

Molarity of $NaClO$ = number of moles/volume in lit = 0.95mole/0.05L

Molarity of $NaClO$ = 19 M

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5 0

2 years ago

In a solution the solute is the substance present in the greatest amount true or false

kati45 [8]

Solute is something that is being dissolved { ex : sugar , salt}

Solvent is something that has ability to dissolve things { ex : water}

False because the solvent is present in larger amounts...

3 0

3 years ago

Read 2 more answers

A manganese electrode was oxidized electrically. If the mass of the electrode decreased by 225 mg during the passage of 1580 cou

qwelly [4]

Answer:

Oxidation state of Mn = +4

Explanation:

Atomic mass of Mn = 55g/mol

From Faraday's law of electrolysis,

Electrochemical equivalent = $\frac{mass}{charge}$

i.e Z = $\frac{m}{Q}$ = $\frac{0.225g}{1580C}$ = 0.0001424 g/C

But Equivalent weight, E = atomic mass ÷ valency = Z × 96,485

⇒ $\frac{55}{valency}$ = 0.0001424 × 96,485

∴ Valency of Mn = +4

8 0

3 years ago

a) What substances are present in an aqueous buffer composed of HC2H3O2 and C2H3O2 - ?b) What happens when LiOH is added to a bu

Alex17521 [72]

Answer:

a) HC₂H₃O₂, C₂H₃O₂⁻, H₃O⁺, H₂O, OH⁻

b) HC₂H₃O₂ + LiOH ⇄ H₂O + LiC₂H₃O₂

c) C₂H₃O₂⁻ + HBr ⇄ HC₂H₃O₂ + Br⁻

Explanation:

a) In a HC₂H₃O₂/C₂H₃O₂⁻ buffer system, the following reactions take place:

HC₂H₃O₂ + H₂O ⇄ C₂H₃O₂⁻ + H₃O⁺

C₂H₃O₂⁻ + H₂O ⇄ HC₂H₃O₂ + OH⁻

Thus, the species present are: HC₂H₃O₂, C₂H₃O₂⁻, H₃O⁺, H₂O, OH⁻.

b) When LiOH is added to the buffer system, it is partially neutralized according to the following equation.

HC₂H₃O₂ + LiOH ⇄ H₂O + LiC₂H₃O₂

c) When HBr is added to the buffer system, it is partially neutralized according to the following equation.

C₂H₃O₂⁻ + HBr ⇄ HC₂H₃O₂ + Br⁻

3 0

3 years ago

Consider the reaction. PCl5(g)↽−−⇀PCl3(g)+Cl2(g) K=0.042 The concentrations of the products at equilibrium are [PCl3]=0.18 M and

tatyana61 [14]

Answer: The equilibrium concentration of $PCl_5$ is 1.285 M.

Explanation:

The chemical equation for the decomposition of phosphorus pentachloride follows:

$PCl_5(g)\rightleftharpoons PCl_3(g)+Cl_2(g)$

The expression for equilibrium constant is given as:

$K_c=\frac{[PCl_3][Cl_2]}{[PCl_5]}$

We are given:

$K_c=0.042$

$[PCl_3]=0.18M$

$[Cl_2]=0.30M$

The concentration of solid substances are taken to be 1. Thus, they do not appear in the equilibrium constant expression.

Putting values in above equation, we get:

$0.042=\frac{0.18\times 0.30}{[PCl_5]}$

$[PCl_5]=1.285$

Hence, the equilibrium concentration of $PCl_5$ is 1.285 M.

6 0

3 years ago