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3 years ago

what volume of a 0.155 M calcium hydroxide solution is required to neutralize 28.8 mL of a 0.106 M nitric acid?

Chemistry

1 answer:

Hitman42 [59]3 years ago

7 0

Answer:

9.85mL

Explanation:

First, let us write a balanced equation for the reaction. This is illustrated below:

Ca(OH)2 + 2HNO3 —> Ca(NO3)2 + 2H2O

From the balanced equation above,

nA (mole of the acid) = 2

nB (mole of the base) = 1

Data obtained from the question include:

Vb (volume of the base) =?

Mb (Molarity of base) = 0.155 M

Va (volume of the acid) = 28.8 mL

Ma (Molarity of acid) = 0.106 M

Using MaVa/MbVb = nA/nB, the volume of calcium hydroxide (i.e the base) can be obtain as follow:

MaVa/MbVb = nA/nB

0.106 x 28.8 / 0.155 x Vb = 2/1

Cross multiply to express in linear form as shown below:

2 x 0.155 x Vb = 0.106 x 28.8

Divide both side by 2 x 0.155

Vb = (0.106 x 28.8) / (2 x 0.155)

Vb = 9.85mL

Therefore, the Volume of calcium hydroxide is 9.85mL

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Answer:

0.862 J/gºC

Explanation:

The following data were obtained from the question:

Mass of metal (Mₘ) = 50 g

Initial temperature of metal (Tₘ) = 100 °C

Mass of water (Mᵥᵥ) = 400 g

Initial temperature of water (Tᵥᵥ) = 20 °C

Equilibrium temperature (Tₑ) = 22 °C

Specific heat capacity of water (Cᵥᵥ) = 4.2 J/gºC

Specific heat capacity of metal (Cₘ) =?

The specific heat capacity of the metal can be obtained as follow:

Heat lost by metal = MₘCₘ(Tₘ – Tₑ)

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Heat gained by water = MᵥᵥCᵥᵥ(Tₑ – Tᵥᵥ)

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= 3360 J

Heat lost by metal = Heat gained by water

3900 × Cₘ = 3360

Divide both side by 3900

Cₘ = 3360 / 3900

Cₘ = 0.862 J/gºC

Therefore, the specific heat capacity of the metal is 0.862 J/gºC

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3 years ago

what volume of a 0.155 M calcium hydroxide solution is required to neutralize 28.8 mL of a 0.106 M nitric acid?​

what volume of a 0.155 M calcium hydroxide solution is required to neutralize 28.8 mL of a 0.106 M nitric acid?