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:
i would say 10, so the anser is A.
Explanation:
because there are the same number of protons and electrons, therefore for a regular O, you are supposed to have only 8 protons, but it is charged, thus, whatever the charge is will be taken into consideration into how much the proton and electron doe it have. Thus, for this case, it has 10, because the charge is negative and you have 8 electron plus 2 = 10.
Answer:
Explanation:
We can use the Noyes-Whitney equation to calculate the rate of dissolution.
Data:
D = 1.75 × 10⁻⁷ cm²s⁻¹
A = 2.5 × 10³ cm²
Cₛ = 0.35 mg/mL
C = 2.1 × 10⁻⁴ mg/mL
d = 1.25 µm
Calculations:
Cₛ - C = (0.35 - 2.1 × 10⁻⁴) mg·cm⁻³ = 0.350 mg·cm⁻³
d = 1.25 µm = 1.25 × 10⁻⁶ m = 1.25 × 10⁻⁴ cm
As in math or
?
/.........
Answer: 406 hours
Explanation:
where Q= quantity of electricity in coloumbs
I = current in amperes = 39.5 A
t= time in seconds = ?
The deposition of copper at cathode is represented by:
Coloumb of electricity deposits 1 mole of copper
i.e. 63.5 g of copper is deposited by = 193000 Coloumb
Thus 19.0 kg or 19000 g of copper is deposited by = 
Coloumb
(1hour=3600s)
Thus it will take 406 hours to plate 19.0 kg of copper onto the cathode if the current passed through the cell is held constant at 39.5 A
