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2 years ago

Help me!!! Pleaseeee!!!!

Chemistry

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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Explanation:

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NH₃: 4 moles
O₂: 5 moles
NO: 4 moles
H₂O: 6 moles

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O₂: 5 moles*32 g/mol= 160 g
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$mass of O_{2} =\frac{7.5 g of NH_{3} * 160 g of O_{2} }{68 g of NH_{3} }$

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