Answer:
2.408x10^23 molecules
Explanation:
Avogadro's hypothesis gives us a background understanding that 1mole of any substance contains 6.02x10^23 molecules.
From the above conclusion,
1mole of N205 contains 6.02x10^23 molecules.
Therefore, 0.4mole of N2O5 will contain = 0.4 x 6.02x10^23 = 2.408x10^23 molecules
Answer:
Conformational isomers
Explanation:
Conformational isomers are temporarily different shapes of the same molecule and for this reason are not classified as isomers in some textbooks.
Hello!
To find the mass of 7.602 x 10^24 atoms of lead, we need to find the number of moles first, then we will be able to find the mass. To find the moles, we need to divide the total number of atoms by Avogadro's number, which is 6.02 x 10^23. Also, one mole of lead is equal to it's atomic mass, which is 207.2 grams.
1). Find the number of moles
(7.602 x 10^24) / (6.02 x 10^23) ≈ 12.62790
Using significant figures, there are 12.8 moles of lead.
2). Find the mass of lead
12.8 moles x 207.2 grams = 2652.16 grams
Using significant figures, the mass of the given amount of lead is 2650 grams. (Exact value is 2652.16 grams).
Answer:
Maximum number of moles of AlCl3 produced is 5 moles
Explanation:
First thing's first, let's bring out the balanced chemical equation.
2Al + 3Cl2 -> 2AlCl3
Before proceeding to calculating the maximum number of moles of AlCl3 that can be formed, we have to identify the limiting reagent.
Every 2 mole of AlCl3 requires 2 moles of Al and 3 moles of Cl2.
If all of the 5 moles of Al were to be used up, there would need to be 5 × (2 / 3) or 3.333 moles of Cl2. 6 moles of Cl2 is available, this means Al is our limiting reagent.
5 mol of Al * (2 mol of Cl2 / 3 mol of Al) = 3.33 mol of Cl2
From the equation, 2 mol of Al produces 2 mol of AlCl3. This means 5 mol of Al would produce x?
2 = 2
5 = x
x = (5 * 2 ) / 2
x = 10 /2 = 5
Maximum number of moles of AlCl3 produced is 5 moles
False, it would be considered 9.46 * 1,000,000,000,000,000