The answer is D.) +0.34 V
Answer:
ΔH°rxn = -827.5 kJ
Explanation:
Let's consider the following balanced equation.
2 PbS(s) + 3 O₂(g) → 2 PbO(s) + 2 SO₂(g)
We can calculate the standard enthalpy of reaction (ΔH°rxn) from the standard enthalpies of formation (ΔH°f) using the following expression.
ΔH°rxn = [2 mol × ΔH°f(PbO(s)) + 2 mol × ΔH°f(SO₂(g)
)] - [2 mol × ΔH°f(PbS(s)) + 3 mol × ΔH°f(O₂(g)
)]
ΔH°rxn = [2 mol × ΔH°f(PbO(s)) + 2 mol × ΔH°f(SO₂(g)
)] - [2 mol × ΔH°f(PbS(s)) + 3 mol × ΔH°f(O₂(g)
)]
ΔH°rxn = [2 mol × (-217.32 kJ/mol) + 2 mol × (-296.83)] - [2 mol × (-100.4) + 3 mol × 0 kJ/mol]
ΔH°rxn = -827.5 kJ
The half-life of the carbon-14 isotope is used in dating fossils in a process called radiocarbon dating.
hope this is adequate.
<span>3 NO2 + H2O -------->. 2 HNO3. + NO
3(46g)------------------------> 2 ( 63g) HNO3
? kg-------------------------5.89 x10^3kg HNO3
Mass of NO2. = 5.89x10^3 x 138/ 2(63) = 6.45 x10^3 kg</span>
Copper is a good conductor of heat and electricity so it will not fry easily
