The reaction is
Zn(s) + 2MnO₂(s) + H₂O(l) ⟶ Zn(OH)₂(s) + Mn₂O₃(s)
The reactants are Zn(s) and MnO₂(s) while Zn(OH)₂(s) and Mn₂O₃(s) products.
Let's see the oxidation state of each compound.
sum of the o.s. of the each element of compound = charge of the compound
O.s of O = -2
O.s of OH⁻ = -1
O.s of Zn(s) = 0
O.s of Mn in MnO₂(s) = x
x + (-2) * 2 = 0
x = +4
O.s of Zn in Zn(OH)₂(s) = a
a + (-1) * 2 = 0
a = +2
O.s of Mn in Mn₂O₃(s) = b
2*b + (-2) * 3 = b
2b = 6
b = +3
Since the O.s is reduced in Mn from +4 to +3, the reduced species is Mn₂O₃(s).
Answer:
Five significant figures.
Explanation:
The given measurement have five significant figures 52301.
All non-zero digits are consider significant figures like 1, 2, 3, 4, 5, 6, 7, 8, 9.
Leading zeros are not consider as a significant figures. e.g. 0.03 in this number only one significant figure present which is 3.
Zero between the non zero digits are consider significant like 104 consist of three significant figures.
The zeros at the right side e.g 2400 are also significant. There are four significant figures are present.
-130KJ is the standard heat of formation of CuO.
Explanation:
The standard heat of formation or enthalpy change can be calculated by using the formula:
standard heat of formation of reaction = standard enthalpy of formation of product - sum of enthalpy of product formation
Data given:
Cu2O(s) ---> CuO(s) + Cu(s) ∆H° = 11.3 kJ
2 Cu2O(s) + O2(g) ---> 4 CuO(s) ∆H° = -287.9 kJ
CuO + Cu ⇒ Cu2O (-11.3 KJ) ( Formation of Cu2O)
When 1 mole Cu20 undergoes combustion 1/2 moles of oxygen is consumed.
Cu20 + 1/2 02 ⇒ 2CuO (I/2 of 238.7 KJ) or 119.35 KJ
So standard heat of formation of formation of Cu0 as:
Cu + 1/2 02 ⇒ CuO
putting the values in the equation
ΔHf = ΔH1 + ΔH2 (ΔH1 + ΔH2 enthalapy of reactants)
heat of formation = -11.3 + (-119.35)
= - 130.65kJ
-130.65 KJ is the heat of formation of CuO in the given reaction.
