Answer:
I think it will option D hope it helps
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:
35.9 ml
Explanation:
Start with the balanced equation:
3CuCl2(aq)+2Na3PO4(aq)→Cu3(PO4)2(s)+6NaCl(aq)
This tells us that 3 moles of CuCI2 react with 2 moles Na3PO4-
∴ 1 mole CuCl2 will react with 2/3 moles Na3PO4
We know that concentration = moles/volume i.e:
c= n/v
∴n=c×v
∴nCuCl2=0.107×91.01000=9.737×10−3
I divided by 1000 to convert ml to L
∴nNa3PO4=9.737×10−3×23=6.491×10−3
v=nc=6.491×10−30.181=35.86×10−3L
∴v=35.86ml
Answer:
Thick deposits of sediments carried off of the shelf
Explanation:
The continental rise is a thick deposit of sediments that accumulate between the continental slope and the abyssal plain
A is wrong. The area of land that drops toward the deep ocean basins is the continental slope.
B is wrong. The 75-mile shallow flat area just off coastlines is the continental shelf.
D is wrong. The surf area along coastlines is the surf zone
.
A catalyst will speed up the activation energy and therefore speed up the reaction. The products will form fast because of this.
