Iodate has a charge of -1, this means its chemical symbol
must be IO3(-1) while Lanthanum is La(+3). Therefore making the compound
La(IO3)3.
However sulfate has a chemical symbol of SO4(-2), it has
charge of -2, therefore the formula for lanthanum sulfate is:
<span>La2(SO4)<span>3</span></span>
Mg (s) + HCl (aq) → MgCl₂(s) + H₂(g)
Looking at the equation :
We have 1 Mg at the left hand side and 1 Mg as well on the right hand side.
So that is balanced.
We have 1 H at the left hand side and 2 H on the right hand side.
So that is not balanced. Same for Chlorine. Cl.
We add 2 to the HCl on the left hand side and that balances it.
Mg(s) + 2HCl(aq) → MgCl₂(s) + H₂(g)
Answer:
0.464 L
Explanation:
Molarity (M) = number moles (n) ÷ volume (V)
According to the information given in this question:
number of moles (n) = 4.36 moles
Molarity = 9.4M
Volume = ?
Using M = n/V
9.4 = 4.36/V
9.4V = 4.36
V = 4.36/9.4
V = 0.464 L
Hence, 0.464L of water are needed the volume of water.
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!
