<span>If you look at the chlorine box, with the symbol Cl, you see the atomic mass is equal to 35.453 atomic mass units. This is the weighted average mass of chlorine, including its isotopes, as found in nature. This also means that one mole of chlorine atoms has a mass of 35.453 grams.</span>
Answer:
18,01528
Explanation:
<span>0.925 grams if using hydrochloric acid in the reaction.
0.462 grams if using sulfuric acid in the reaction.
0.000 grams if using nitric acid in the reaction.
Assuming you're using HCl or a similar acid for this reaction, the equation for the reaction is:
Zn + 2 HCl ==> ZnCl2 + H2
So each mole of zinc used, produces 1 mole of hydrogen gas, or 2 moles of hydrogen atoms. So we need to look up the atomic weights of both zinc and hydrogen.
Atomic weight zinc = 65.38
Atomic weight hydrogen = 1.00794
Moles zinc = 30.0 g / 65.38 g/mol = 0.458855919 mol
Since we produce 2 moles of hydrogen atoms per mole of zinc, multiply by 2 and the atomic weight of hydrogen to get the mass of hydrogen produced. So
0.458855919 * 2 * 1.00794 = 0.92499847 grams.
Rounding to 3 significant figures gives 0.925 grams.
To show the assumption of the acid used, the balanced equation for sulfuric acid would be
Zn2 + H2SO4 ==> Zn(SO4)2 + H2
Which means that for every mole of zinc used, 1 mole of hydrogen gas is generated (half that produced via hydrochloric acid).
If nitric acid were used, the reaction is
4Zn + 10HNO3 ==> 4Zn(NO3)2 + N2O + 5H2O
Which means that NO hydrogen gas is generated.
The only justification for assuming hydrochloric acid is used is that it's a fairly common acid that's easy to obtain. But as shown above with 2 alternative acids, the amount of hydrogen gas generated is very dependent upon the exact chemical reaction occurring and asking "How many grams of hydrogen are produced if 30.0 g of zinc reacts?" is a rather silly question unless you specify EXACTLY what the reaction is.</span>
The mass of gas A is 11.56. i got this answer by 0.68 multiplied by 17. because in this problem the key word is times.
Answer:
643g of methane will there be in the room
Explanation:
To solve this question we must, as first, find the volume of methane after 1h = 3600s. With the volume we can find the moles of methane using PV = nRT -<em>Assuming STP-</em>. With the moles and the molar mass of methane (16g/mol) we can find the mass of methane gas after 1 hour as follows:
<em>Volume Methane:</em>
3600s * (0.25L / s) = 900L Methane
<em>Moles methane:</em>
PV = nRT; PV / RT = n
<em>Where P = 1atm at STP, V is volume = 900L; R is gas constant = 0.082atmL/molK; T is absolute temperature = 273.15K at sTP</em>
Replacing:
PV / RT = n
1atm*900L / 0.082atmL/molK*273.15 = n
n = 40.18mol methane
<em>Mass methane:</em>
40.18 moles * (16g/mol) =
<h3>643g of methane will there be in the room</h3>
