No one can do that by him or her self
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).
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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.
Answer - Inter-molecular attractions
Explanation-
As we know everything around us is made up of matter that means everything has molecules as their basic structure. The state of anything is decided by the spaces between the molecules.
The state of the objects that have strong inter-molecular attractions a solid and gradually the lesser will be in state of liquid and gas. The attraction between the molecules is overcome only when a certain amount of energy is provided from outside.
Now ,
C + O2 → CO2
According to above equation, 1 mole of carbon reacts with one mole of oxygen to produce one mole of carbon dioxide.Thus this implies that 12 g of carbon reacts with 32 g of O2 to produce 44 g of CO2.
No of moles = mass of the substance/molecular mass of the substance.
In this case 1.2 g of carbon reacts with "x "g of O2 to produce 4.4 g of CO2.
No of moles of carbon in this case = 1.2÷ 12 = 0.1 moles.
No of moles of carbon dioxide formed = 4.4÷44 =0.1 moles
Thus already discussed above, 1 mole of carbon reacts with 1 mole of oxygen to produce 1 mole of carbon dioxide. Hence to produce 0.1 mole of CO2 ,0.1 mole of carbon needs to react with 0.1 mole of oxygen.
Also number of moles of O2 = mass of O2÷ molar mass of O2
Substituting number of moles of O2 as 0.1 we get
mass of O2(x) = Number of moles of O2 × Molar mass of O2
Mass of O2 (x) = 0.1 × 32= 3.2 g
Thus mass of 3.2 g O2 reacts with 1.2 g of CO2 to produce 4.4 g of CO2.
The answer is [Pb^2+][Cl-]^2
