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lys-0071 [83]
3 years ago
5

How much 0.985 M KOH is needed to to titrate 50.00 ml of 2.32 M HNO3?

Chemistry
1 answer:
Ksenya-84 [330]3 years ago
3 0

Answer:

The answer to your question is: 118 ml of KOH

Explanation:

Data

volume KOH = ?

[KOH] = 0.985 M

volume HNO3 = 50 ml = 0.05 l

[HNO3] = 2.32 M

Reaction

                KOH + HNO3 ⇒  KNO3 + H2O

Formula

Molarity = moles / volume

Process

moles of HNO3 = 2.32 x 0.05 l

                          = 0.116

From the equation we know that the proportion KOH : HNO3 = 1 : 1

then,                    

                        1 mol of KOH ----------------  1 mol of HNO3

                        x                     ---------------    0.116 mol of HNO3

                        x = (0.116 x 1)/ 1

                        x = 0.116 moles of KOH

Now, calculate volume of KOH

Volume of KOH = moles of KOH / molarity

Volume of KOH = 0.116 / 0.985

                          = 0.118 l of KOH or 118 ml

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find the volume of an item using a scale and waterA crown made of alloy of gold and silver has a volume of 60cm3and mass of 1050
KiRa [710]

Answer:

The mass of gold in crown is 960.61 g.And  the volume of gold is 49.77 cm^3.

Explanation:

Density is defined as mass of the substance present in unit volume of the substance.

Mass of gold in an alloy of crown= m

Mass of silver in an alloy of crown = M

Volume  gold in an alloy of crown= v

Volume of silver in an alloy of crown = V

Density of gold = 19.3 g/cm^3

Density of silver = 10.5 g/cm^3

Volume of the crown = 60 cm^3

60 cm^3=v+V

60 cm^3=\frac{m}{19.3 g/cm^3}+\frac{M}{10.5 g/cm^3}..(1)

Mass of eh crown = 1050 g

1050g =m+M..(2)

Solving equation (1) and (2):

m = 960.61 g

M = 89.39 g

Volume of the gold = v = \frac{960.61 g}{19.3 g/cm^3}=49.77 cm^3

Volume of silver = V = 60 cm^3-v=60 cm63-49.77 cm63=10.23 cm^3

The mass of gold in crown is 960.61 g.And  the volume of gold is 49.77 cm^3.

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What change in volume results if 50 mL of gas is cooled from 30 C to 4 C
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The relation between the volume of the gas and the temperature is established by Charles's law. With a decrease in the temperature, the volume decreases by 45.7 mL. Thus, option c is correct.

<h3>What is Charle's law?</h3>

Charle's law states the direct relation present between the temperature and the volume of the gas. The law is given as:

V₁ ÷ T₁ = V₂ ÷ T₂

Given,

V₁ = 50 mL

T₁ = 303.15 K

T₂ = 277.15 K

Substituting the value the final volume is calculated as:

50 ÷ 303.15 = V₂ ÷ 277.15

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