Mercury (ii) oxide is made up of mercury and oxygen. The total mass of mercury (ii) oxide is 14.2 g, after decomposition 13.2 g of mercury were formed, therefore the mass of oxygen 1 g (14.2 g -13.2 g).
Percentage of oxygen = (1/14.2)×100 = 7.04%
Percentage of mercury = (13.2/14.2) × 100 = 92.96%
Therefore, percentage composition of the compound, oxygen is 7.04% and mercury is 92.96%.
Answer:
because they need only one electron
The compound that is formed when a copper cation and a bromine anion reacts would be Copper (II) bromide. It will have a chemical formula CuBr2. This reaction is a synthesis reaction where two substances reacts to form one product.
Answer:
Can you tell me the question in the comments on this answer or like how you do it then ill answer you in the comments under this answer
Answer:
The answer to the question is
The pressure of carbon dioxide after equilibrium is reached the second time is 0.27 atm rounded to 2 significant digits
Explanation:
To solve the question, we note that the mole ratio of the constituent is proportional to their partial pressure
At the first trial the mixture contains
3.6 atm CO
1.2 atm H₂O (g)
Total pressure = 3.6+1.2= 4.8 atm
which gives
3.36 atm CO
0.96 atm H₂O (g)
0.24 atm H₂ (g)
That is
CO+H₂O→CO(g)+H₂ (g)
therefore the mixture contained
0.24 atm CO₂ and the total pressure =
3.36+0.96+0.24+0.24 = 4.8 atm
when an extra 1.8 atm of CO is added we get Increase in the mole fraction of CO we have one mole of CO produces one mole of H₂
At equilibrium we have 0.24*0.24/(3.36*0.96) = 0.017857
adding 1.8 atm CO gives 4.46 atm hence we have
(0.24+x)(0.24+x)/(4.46-x)(0.96-x) = 0.017857
which gives x = 0.031 atm or x = -0.6183 atm
Dealing with only the positive values we have the pressure of carbon dioxide = 0.24+0.03 = 0.27 atm
