a. W = 0 J
b. W = - 308.028 J
<h3>Further explanation</h3>
Given
Nitrogen gas expands in volume from 1.6 L to 5.4 L
Required
The work done
Solution
Isothermal :
W = -P . ΔV
Input the value :
a. At a vacuum, P = 0
So W = 0
b. At pressure = 0.8 atm
W = - 0.8 x ( 5.4 - 1.6)
W = -3.04 L.atm ( 1 L.atm = 101.325 J)
W = - 3.04 x 101.325
W = - 308.028 J
Answer:
The spectator ions is:
and 
Explanation:
The equation of reaction between H₂ SO₄ and KOH is:

Rewriting this equation as ionic;
![[2H^{+} + SO^{2-}_4 + 2K^+ +2OH^- \to 2K^+ SO_4^{2-} + 2H_2O ]](https://tex.z-dn.net/?f=%5B2H%5E%7B%2B%7D%20%2B%20SO%5E%7B2-%7D_4%20%2B%202K%5E%2B%20%2B2OH%5E-%20%5Cto%202K%5E%2B%20%20SO_4%5E%7B2-%7D%20%2B%202H_2O%20%5D)
Spectators ions are ions present on both sides of the ionic equation by the same quantity but do not take part in the net reaction.
Answer:
Explanation:
The air 9% mole% methane have an average molecular weight of:
9%×16,04g/mol + 91%×29g/mol = 27,8g/mol
And a flow of 700000g/h÷27,8g/mol = 25180 mol/h
In the reactor where methane solution and air are mixed:
In = Out
Air balance:
91% air×25180 mol/h + 100% air×X = 95%air×(X+25180)
Where X is the flow rate of air in mol/h = <em>20144 mol air/h</em>
<em></em>
The air in the product gas is
95%×(20144 + 25180) mol/h = 43058 mol air× 21%O₂ = 9042 mol O₂ ×32g/mol = <em>289 kg O₂</em>
43058 mol air×29g/mol <em>1249 kg air</em>
Percent of oxygen is:
=<em>0,231 kg O₂/ kg air</em>
<em></em>
I hope it helps!
False. What you’re describing is a displacement reaction.