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Artyom0805 [142]
3 years ago
15

A 23.5g aluminum block is warmed to 65.9°C and plunged into an insulated beaker containing 55.0g water initially at 22.3°C. The

aluminum and the water are allowed to come to thermal equilibrium.
Assuming that no heat is lost, what is the final temperature of the water and aluminum?
Chemistry
1 answer:
atroni [7]3 years ago
4 0

Answer:

25.97oC

Explanation:

Heat lost by aluminum = heat gained by water

M(Al) x C(Al) x [ Temp(Al) – Temp(Al+H2O) ] = M(H2O) x C(H2O) x [ Temp(Al+H2O) – Temp(H2O) ]

Where M(Al) = 23.5g, C(Al) = specific heat capacity of aluminum = 0.900J/goC, Temp(Al) = 65.9oC, Temp(Al+H2O)= temperature of water and aluminum at equilibrium = ?, M(H2O) = 55.0g, C(H2O)= specific heat capacity of liquid water = 4.186J/goC

Let Temp(Al+H2O) = X

23.5 x 0.900 x (65.9-X) = 55.0 x 4.186 x (X-22.3)

21.15(65.9-X) = 230.23(X-22.3)

1393.785 - 21.15X = 230.23X – 5134.129

230.23X + 21.15X = 1393.785 + 5134.129

251.38X = 6527.909

X = 6527.909/251.38

X = 25.97oC

So, the final temperature of the water and aluminum is = 25.97oC

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Answer:

Average rate of reaction expressed in moles H₂ consumed per liter per second = 0.0025 M/s = 0.0025 mol/L.s

Explanation:

The complete, correct Question, is presented in the attached image to this answer.

The average rate of reaction in terms of the reactant is defined as the total amount of reactant consumed over a period of time divided by total period of time.

Mathematically,

Average rate of reaction = (change in concentration of H₂ over a period of time) ÷ (total period of time)

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The minus sign is there because the concentration of reactants reduce with time.

change in concentration of H₂ = -ΔC

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We could solve for the average rate of reaction expressed in moles Cl₂ consumed per liter per second

change in concentration of Cl₂ = -ΔC

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Hope this Helps!!!

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