Answer:

Explanation:
Hello.
In this case, since the acid is monoprotic and the KOH has one hydroxyl ion only, we can see that at the equivalence point the moles of both of them are the same:

Thus, since we are given 1.70 g of the acid, we compute the moles of acid that were titrated:

Which equal the moles of KOH. In such a way, since the molarity is defined as moles over liters (M=n/V), the liters are moles over molarity (V=n/M), thus, the resulting volume is:

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I believe it’s the third option
Chemically combined to make a new pure substance
Molar mass Argon = 39.948 g/mol
1 mol ------ 39.948 g
mol ----- 20.0 g
mol = 20.0 * 1 / 39.948
= 0.5006 moles
1 mol --------------------- 22.4 L ( at STP )
0.5006 moles ------------- L
L = 0.5006 * 22.4
= 11.21 L
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Answer:
Its a 50% chance of it happeing
Explanation:
Answer:
This is a coal combustion process and we will assume
Inlet coal amount = 100kg
It means that there are
15kg of H2O, 2kg of Sulphur and 83kg of Carbon
Now to find the mole fraction of SO2(g) in the exhaust?
Molar mass of S = 32kg/kmol
Initial moles n of S = 2/32 = 0.0625kmols
Reaction: S + O₂ = SO₂
That is 1 mole of S reacts with 1 mole of O₂ to give 1 mole of SO₂
Then, it means for 0.0625 kmoles of S, we will have 0.0625 kmole of SO2 coming out of the exhaust
The mole fraction of SO2(g) in the exhaust=0.0625kmols
Explanation: