Solid potassium fluoride decomposes into fluorine gas and solid potassium .
1 answer:
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The allowable combination for the atomic orbital is n=3, l=1 or 2, =+1.
Three quantum numbers specify an atomic orbital:
- The principal quantum number, n, which is a positive integer, describes the relative size of the orbital and its distance from the nucleus.
- l is the angular momentum quantum number that is related to the shape of the orbital; l is an integer from 0 to n-1 (so n limits l ),
- is the magnetic quantum number that prescribes the three-dimensional shape of the orbital around the nucleus;
values are integers from -l to =l(l limits ml)
For n = 3, l can have three values: 0, 1, and 2. Hence, the l value must be lower than 3. Because of = +1, the l value must be higher than 0. Two l values are consistent with n and
values:
l= 1 or 2
Therefore, the allowable condition is n=7, l=1 or 2, =+3.
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<u>Answer:</u> The mass of iron in the ore is 10.9 g
<u>Explanation:</u>
We are given:
Mass of iron (III) oxide = 15.6 g
We know that:
Molar mass of Iron (III) oxide = 159.69 g/mol
Molar mass of iron atom = 55.85 g/mol
As, all the iron in the ore is converted to iron (III) oxide. So, the mass of iron in iron (III) oxide will be equal to the mass of iron present in the ore.
To calculate the mass of iron in given mass of iron (III) oxide, we apply unitary method:
In 159.69 g of iron (III) oxide, mass of iron present is
So, in 15.6 g of iron (III) oxide, mass of iron present will be =
Hence, the mass of iron in the ore is 10.9 g