It’s on September 12th :)
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:
0.1 M
Explanation:
Molarity = number of moles / litres of solution.
4 g of calcium bromide = 0.02 mol
(found by dividing 4 g by the atomic mass of CaBr2, which is 199.886)
200 mL of solution = 0.2 litres
Molarity = 0.02 mol / 0.2 L = 0.1 M
Answer:
B
Explanation:
Reactants are the compounds which reaction produce the products. In general terms this can be expressed symbolically as follows:
reactants -> products
Other phenomena like heat are omitted because are not always present, that is, only compounds are included. Therefore, in this reaction the reactants are C3H8 (propane) and O2 ( oxygen) and the products are CO2 (carbon dioxide) and H2O (water)
Answer:
- <u><em>1.7 × 10³ kg of ore.</em></u>
Explanation:
Call X the amount of aluminum ore mined to produce 1.0 × 10³ kg the aluminum metal.
Then, taking into account the yield of the reaction (82 % = 0.82) and the percent of aluminun in the ore (71% = 0.71), you can write the following equation:
- X × 71% × 82% = 1.0 × 10³ kg
↑ ↑ ↑ ↑
(mass of ore) (% of Al in the ore) (yield) ( Al metal to obtain)
You must just simplify, solve and compute:
- X = 1,000 / (0.71 × 0.82) = 1,000 / 0.5822 = 1,717.6 Kg
Round to two significant figures; 1,700 kg = 1.7 × 10³ kg of ore ← answer.