A reaction in which Oxygen (O₂) is produced from Mercury Oxide (HgO) would be a decomposition reaction.
2HgO → 2Hg + O₂
If 250g of O₂ is needed to be produced,
then the moles of oxygen needed to be produced = 250g ÷ 32 g/mol
= 7.8125 mol
Now, the mole ratio of Oxygen to Mercury Oxide is 1 : 2
∴ if the moles of oxygen = 7.8125 mol
then the moles of mercury oxide = 7.8125 mol × 2
= 15.625 mol
Thus the number moles of HgO needed to produce 250.0 g of O₂ is 15.625 mol
Answer;
=28.09 amu
Explanation;
In this problem, they did not give us the percentages. However, since we know the number of atoms, we can easily calculate the percentages. For example:
(460 X 100)/500 = 92%
If we do this for all three isotopes,
(460 × 25)/500 = 5 %
(460 × 15) /500 = 3%
-We get 92%, 5%, and 3%. (We'll assume these are absolute numbers for determining our significant figures).
Now the problem is just like the previous one. First convert the percentages into decimals. Then multiply those decimals by the masses and add. Here's the solution:
= (0.92) X (27.98 amu) + (0.05) X (28.98 amu) + (0.03) X (29.97 amu)
= 25.74 amu + 1.449 amu + 0.8991 amu
= 28.09 amu
True
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Chlorine<span> is in group 17 of periodic table, also called the halogens, and is not found as the element in nature only as a compound.
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