In 1st orbit 2
2nd 8
3rd 10
f orbital has 16
Answer:
The mass of 0.280 mole sample of sodium hydroxide NaOH is 11.2 grams.
Explanation:
To know the mass in grams of 0.280 moles of sample of sodium hydroxide NaOH, you must know the molar mass of the compound, that is, the mass of one mole of a substance, which can be an element or a compound.
So you know:
- Na: 23 g/mole
- O: 16 g/mole
- H: 1 g/mole
So, the molar mass of NaOH is:
NaOH= 23 g/mole + 16 g/mole+ 1 g/mole= 40 g/mole
Then the following rule of three can be applied: if in 1 mole of sodium hydroxide there are 40 grams, in 0.280 moles how much mass is there?
mass= 11.2 grams
<u><em>The mass of 0.280 mole sample of sodium hydroxide NaOH is 11.2 grams.</em></u>
Answer:
BaBr2 (aq) + H2SO4 (aq) → BaSO4 (s) + 2 HBr (aq)
Explanation:
This is a precipitation reaction: BaSO4 is the formed precipitate.
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.
Yes, you're right the answer is 0,02 moles.
