Almost always, the first way in which biology researchers present the results of their latest research is to discuss the results of previous research that they want to build off of.
BOTH OF THE ABOVE ANSWERS ARE CORRECT
Two electron pairs is the answer
Answer:
392.97 litres
Explanation:
From the equation of reaction, we can see that 1 mole of methane yielded 1 mole of carbon iv oxide. Hence, 15.9 moles of methane will yield 15.9 moles of carbon iv oxide.
At s.t.p one mole of a gas occupies a volume of 22.4L ,hence 15.9 moles of a gas will occupy a volume of 22.4 × 15.9 which equals
356.16L.
Now, we can use the general gas equation to get the volume produced at the values given.
We have the following values;
V1 = 356.16L P1= 1 atm ( standard pressure) T1 = 273K ( standard temperature) V2 = ? T2 = 23.7 + 273 = 296.7K P2 = 0.985 atm
The general form of the general gas equation is given as :
(P1V1)T1 = (P2V2)/T2
After substituting the values , we get V2 to be 392.97Litres
Answer:
Mass = 65.4 g
Explanation:
Given data:
Mass of Al₂O₃ = 25 g
Mass of AlCl₃ = ?
Solution:
Chemical equation:
Al₂O₃ + 6HCl → 2AlCl₃ + 3H₂O
Number of moles of Al₂O₃:
Number of moles = mass/ molar mass
Number of moles = 25 g/ 101.96 g/mol
Number of moles = 0.2452 mol
Now we will compare the moles of Al₂O₃ and AlCl₃.
Al₂O₃ : AlCl₃
1 : 2
0.2452 : 2×0.2452 = 0.4904 mol
Mass of AlCl₃:
Mass = number of moles × molar mass
Mass = 0.4904 mol × 133.34 g/mol
Mass = 65.4 g
