In a closed system, heat should be conserved which means that the heat produced in the calorimeter is equal to the heat released by the combustion reaction. We calculate as follows:
Heat of the combustion reaction = mC(T2-T1)
= 1 (1.50) (41-21)
= 30 kJ
Answer:
1.386 KJ
Explanation:
From the question given above, the following data were obtained:
Mass (M) of copper = 45 g
Initial temperature (T1) = 20.0°C
Final temperature (T2) = 100.0°C
Heat absorbed (Q) =..?
Next, we shall determine the change in temperature. This can be obtained as follow:
Initial temperature (T1) = 20.0°C
Final temperature (T2) = 100.0°C
Change in temperature (ΔT) =?
ΔT = T2 – T1
ΔT = 100 – 20
ΔT = 80 °C
Next, we shall determine the heat absorbed by the sample of copper as follow:
Mass (M) of copper = 45 g
Change in temperature (ΔT) = 80 °C
Specific heat capacity (C) of copper = 0.385 J/gºC
Heat absorbed (Q) =..?
Q = MCΔT
Q = 45 × 0.385 × 80
Q = 1386 J
Finally, we shall convert 1386 J to KJ. This can be obtained as follow:
1000 J = 1 KJ
Therefore,
1386 J = 1386 J × 1 KJ /1000 J
1386 J = 1.386 KJ
Thus, the heat absorbed by the sample of the sample of copper is 1.386 KJ.
Hi!
I think the oxidation state of all the atoms should change. :)
Hope this helps
Answer:
Ok so the answer for 9 is
x/6=4
x=24
Explanation:
Solve for x by simplifying both sides of the equation, then isolating the variable.
The activity series goes top to bottom, most active to least active elements, going: Li, K, Ba, Sr, Ca, Na, Mg, Mn, Zn, Fe, Cd, Co, Ni, Sn, Pb, H, Cu, Ag, Hg, Au.
Thus, your list of metals would go from most reactive to least reactive: Li, K, Mg, Zn, Fe, Cu, Au