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kari74 [83]
3 years ago
12

Which of these chemical formulas represents both an element and a molecule?

Chemistry
2 answers:
inna [77]3 years ago
8 0
O2.

Because it is only made up of one type of element it is an element. Howevern there is more than one oxygen atom because oxygen cannot exist as a single atom, therefore making it a molecule as well.
anastassius [24]3 years ago
6 0

Answer:

O2

Explanation:

The other answer on here has a wonderful explanation! :D I just wanted to confirm. :)

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Mercury(II) oxide (HgO) decomposes to form mercury (Hg) and oxygen (O2). The balanced chemical equation is shown below. 2HgO Rig
BigorU [14]

Taking into account the reaction stoichiometry, the correct answer is the third option: 15.63 moles of HgO are needed to produce 250 g of O₂.

In first place, the balanced reaction is:

2 HgO  → 2 Hg + O₂

By reaction stoichiometry (that is, the relationship between the amount of reagents and products in a chemical reaction), the following amounts of moles of each compound participate in the reaction:

  • HgO: 2 moles
  • Hg: 2 moles
  • O₂: 1 moles  

The molar mass of the compounds is:

  • HgO: 216.59 g/mole
  • Hg: 200.59 g/mole
  • O₂: 32 g/mole  

Then, by reaction stoichiometry, the following mass quantities of each compound participate in the reaction:  

  • HgO: 2 moles× 216.59 g/mole= 433.18 grams
  • Hg: 2 moles× 200.59 g/mole= 401.18 grams
  • O₂: 1 mole× 32 g/mole= 32 grams

Then the following rule of three can be applied: if by reaction stoichiometry 32 grams of O₂ are produced by 2 moles of HgO, 250 grams of O₂ are produced from how many moles of HgO?

moles of HgO=\frac{250 grams of O_{2} x2 moles of HgO}{32grams of O_{2}}

<u><em>moles of HgO= 15.625 moles≅ 15.63 moles</em></u>

Finally, the correct answer is the third option: 15.63 moles of HgO are needed to produce 250 g of O₂.

Learn more about reaction stoichiometry:

  • <u>brainly.com/question/24741074 </u>
  • <u>brainly.com/question/24653699 </u>
  • <u>brainly.com/question/23871710</u>
  • <u>brainly.com/question/3588546</u>
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