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
Answer:
<em>Mg </em>(<em>s</em>) + 2<em>HCI2 </em>(<em>aq</em>) → <em>MgCI2 </em>(<em>aq</em>) + <em>H2 </em>(<em>g</em>)
I think this is the correct answer I not a 100% sure if it is correct.
Explanation:
Guessing
Answer:
The correct answer is - D. Freezing point depression.
Explanation:
When rock salt is spread over snow-covered icy roads, it generates a liquid layer over it by melting from the surface thereby lowering or depression in the freezing point below the ice.
Therefore, due to this liquid layer comes into the contact with the ice present on the road and causes other ice to melts. This keeps on decreasing the volume of the ice on the road therefore, rock salts spread on the roads during a snowstorm.
Oxidation is when the overall charge (or oxidation number) increases. The only way to increase an oxidation number is to lose an electron, thereby making the negative charges less. The correct answer is C.
