20.06 g of Hg and 1.6 g of O₂
<u>Explanation:</u>
To Find:
Number of Mercury and oxygen that can be obtained from 21.7 g of HgO
First we have to write the balanced equation for the decomposition reaction of Mercury(II) oxide as,
2 HgO (s) → 2Hg(l) + O₂ (g)
21.7 g of HgO = 
= 0.1 mol of HgO.
As per the above equation, we can find the mole ratio between HgO and Hg is 1: 1 and that of HgO and oxygen is 2:1 .
So amount of Hg produced = 0.1 mol × 200.59 g / mol ( molar mass of Hg)
= 20.06 g of Hg
Amount of oxygen produced = 0.05 mol × 32 g/ mol = 1.6 g of O₂
Thus it is clear that 20.06 g of Hg and 1.6 g of O₂ is obtained from 21.7 g of HgO
The volume of Sulfur dioxide-SO₂ at STP : = 102.144 L
<h3>Further explanation</h3>
Given
4.56 mol Sulfur dioxide-SO₂
Required
The volume
Solution
Conditions at T 0 ° C and P 1 atm are stated by STP (Standard Temperature and Pressure). At STP, Vm is 22.4 liters/mol.
so for 4.56 mol :
= 4.56 x 22.4 L
= 102.144 L
Answer:
8 moles
Explanation:
When we are asked to convert from grams of a substance into moles, we have to use the substance's molar mass.
Meaning that for this problem, we'll <em>use the molar mass of hydrogen peroxide</em> (H₂O₂), as follows:
There are 8 moles in 272 grams of hydrogen peroxide.
Answer:
everyone would die
Explanation:
if we did not know about it we would not do anything about it
Answer:
a) Addition reaction, is your answer
