Answer:
V=0.68L
Explanation:
For this question we can use
V1/T1 = V2/T2
where
V1 (initial volume )= 0.75 L
T1 (initial temperature in Kelvin)= 303.15
V2( final volume)= ?
T2 (final temperature in Kelvin)= 273.15
Now we must rearrange the equation to make V2 the subject
V2= (V1/T1) ×T2
V2=(0.75/303.15) ×273.15
V2=0.67577931717
V2= 0.68L
Answer:
D 2,2
Explanation:
We can see that there are 2 chlorines on the reactant side so there has to be a 2 on the product side
Now we have Na + Cl2 --> 2NaCl
The problem now is that there are 2 sodiums on the product side so add a 2 to the Na on the reactant side
2Na + Cl2 --> 2NaCl
Now it's balanced!
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!
Answer:
So, you're dealing with a sample of cobalt-60. You know that cobalt-60 has a nuclear half-life of
5.30
years, and are interested in finding how many grams of the sample would remain after
1.00
year and
10.0
years, respectively.
A radioactive isotope's half-life tells you how much time is needed for an initial sample to be halved.
If you start with an initial sample
A
0
, then you can say that you will be left with
A
0
2
→
after one half-life passes;
A
0
2
⋅
1
2
=
A
0
4
→
after two half-lives pass;
A
0
4
⋅
1
2
=
A
0
8
→
after three half-lives pass;
A
0
8
⋅
1
2
=
A
0
16
→
after four half-lives pass;
⋮
Explanation:
now i know the answer
