Answer:
ΔH for formation of 197g Fe⁰ = 1.503 x 10³ Kj => Answer choice 'B'
Explanation:
Given Fe₂O₃(s) + 2Al⁰(s) => Al₂O₃(s) + 2Fe⁰(s) + 852Kj
197g Fe⁰ = (197g/55.85g/mol) = 3.527 mol Fe⁰(s)
From balanced standard equation 2 moles Fe⁰(s) => 852Kj, then ...
3.527 mole yield (a higher mole value) => (3.527/2) x 852Kj = 1,503Kj (a higher enthalpy value).
______
NOTE => If 2 moles Fe gives 852Kj (exo) as specified in equation, then a <u>higher energy value</u> would result if the moles of Fe⁰(s) is <u>higher than 2 moles</u>. The ratio of 3.638/2 will increase the listed equation heat value to a larger number because 197g Fe⁰(s) contains more than 2 moles of Fe⁰(s) => 3.527 mole Fe(s) in 197g. Had the problem asked for the heat loss from <u>less than two moles Fe⁰(s)</u> - say 100g Fe⁰(s) (=1.79mole Fe⁰(s)) - then one would use the fractional ratio (1.79/2) to reduce the enthalpy value less than 852Kj.
Gases in which the molecules that make it up naturally consist of two atoms of the same type.
Ionization energy is the measure of the extend to which the nucleus attracts the outermost electron
if ionization energy us high than force of attraction Is high so it is not easy to remove and vice versa .
hope you understand.....
Hydrochloric acid ionisation is as follows;
HCl ---> H⁺ + Cl⁻
HCl is a strong base so there's complete dissociation of acid to H⁺ ions
The number of HCl moles is equivalent to number of H⁺ ions present
1 L of solution contains - 11.6 moles of H⁺ ions
In 35 ml number of moles - 11.6 mol/L / 1000 ml x 35 ml = 0.406 mol
This number of moles are dissolved in 500 ml
therefore molarity = 0.406 mol /500 ml x 1000 ml = 0.812 M
Explanation:
the answer will be 98.4 kJ
