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2 years ago

14

Help me!!! Pleaseeee!!!!

1 answer:

Sidana [21]2 years ago

5 0

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)

= 50 × Cₘ × (100 – 22)

= 50 × Cₘ × 78

= 3900 × Cₘ

Heat gained by water = MᵥᵥCᵥᵥ(Tₑ – Tᵥᵥ)

= 400 × 4.2 × (22 – 20)

= 400 × 4.2 × 2

= 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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<u>Explanation:</u>

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<u>For magnesium chloride:</u>

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Volume of solution = 664 mL

Putting values in equation 1, we get:

$0.338=\frac{\text{Moles of }MgCl_2\times 1000}{664}\\\\\text{Moles of }MgCl_2=\frac{(0.338mol/L\times 664)}{1000}=0.224mol$

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Putting values in equation 1, we get:

$\text{Concentration of chloride ions}=\frac{0.648mol\times 1000}{961}\\\\\text{Concentration of chloride ions}=0.674M$

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