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
Answer:
Drag force is directly proportional to squared velocity
Explanation:
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Hope I helped! ( Smiles )
Answer:
C3H6 + Br2 → C3H6Br2
Explanation:
The reaction in which C3H6Br2 (1,2-Dibromopropane) is created is:
We can see that the only difference between the product (C3H6Br2) and the known reactant (C3H6) of the reaction is two bromine atoms (Br2). Br2 is diatomic bromine - a molecule we get after combining two bromine atoms. This compound is a red-brown liquid at room temperature, which means that that is the liquid described in your question.
