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: The atomic weight of any atom can be found by multiplying the abundance of an isotope of an element by the atomic mass of the element and then adding the results together. This equation can be used with elements with two or more isotopes: Carbon-12: 0.9889 x 12.0000 = 11.8668. Carbon-13: 0.0111 x 13.0034 = 0.1443.
Explanation:
441 g CaCO₃ would have to be decomposed to produce 247 g of CaO
<h3>Further explanation</h3>
Reaction
Decomposition of CaCO₃
CaCO₃ ⇒ CaO + CO₂
mass CaO = 247 g
mol of CaO(MW=56 g/mol) :
From equation, mol ratio CaCO₃ : CaO = 1 : 1, so mol CaO :
mass CaCO₃(MW=100 g/mol) :
