pH of buffer can be calculated as:
pH=pKa+log[salt]/[Acid]
As ka = 4.58 x 10-4
Concentration of [Salt] that is NO2(-1)=0.380M
Concentration of [Acid] that is HNO2=0.500M
So, pH= -log(4.58*10^-4)+log((0.380)/0.500))
=3.21
So pH of solution will be 3.21
Ca-Cl is an ionic bond. We know this because the difference between their electronegativities is 2.16.The electronegativity tells us which atom will attract the electron more than the other. Hence, chlorine attracts the electron a lot more than the calcium.A difference of more than 1.6 (or 1.7 depending on the source) implies that the electrons are so unevenly shared, that the bond is ionic, rather than polar covalent.
B. horizontally Thats the answer
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.
