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>
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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>
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I hope it helps!
Answer:
4380 mmHg
Explanation:
Boyle's Law can be used to explain the relationship between pressure and volume of an ideal gas. The pressure is inversely related to volume, so if volume decrease the pressure will increase. It can be expressed in the equation as:
P1V1=P2V2
In this question, the first condition is 2L volume and 876 mmHg pressure. Then the system changed into the second condition where the volume is 400ml and the pressure is unknown. The pressure will be:
P1V1= P2V2
876 mmHg * 2L = P2 * 400ml /(1000ml/L)
P2= 876 mmHg * 2L / 0.4L
P2= 4380 mmHg
Yo sup??
we can solve this problem by applying Newton's 2nd law
F*t=Δp
p=momentum
pi=mu=1500*30
pf=mv=m*0=0
Therefore
F*3=1500*30
F=15000 N
Hope this helps.
Answer:
V₂ = 104.76 mL
Explanation:
Given data:
Initial volume = 100.0 mL
Initial temperature = 21°C (21 + 273.15 K = 294.15 K)
Final temperature = 35°C (35 + 273.15 K = 308.15 k)
Final volume = ?
Solution:
Charles Law:
According to this law, The volume of given amount of a gas is directly proportional to its temperature at constant number of moles and pressure.
Mathematical expression:
V₁/T₁ = V₂/T₂
V₁ = Initial volume
T₁ = Initial temperature
V₂ = Final volume
T₂ = Final temperature
Now we will put the values in formula.
V₁/T₁ = V₂/T₂
V₂ = V₁T₂/T₁
V₂ =100.0 mL × 308.15 K / 294.15 K
V₂ = 30815 mL.K /294.15 K
V₂ = 104.76 mL
