True
After the process of Ore processing, Enrichment, Fuel production and being passed through the reactor core the last remaining step is spent fuel disposal.
Answer:
The answer is 98.07848. We assume you are converting between grams H2SO4 and mole. You can view more details on each measurement unit: This compound is also known as Sulfuric Acid. The SI base unit for amount of substance is the mole. 1 grams H2SO4 is equal to 0.010195916576195 mole.
<u>Quick conversion chart of moles H2SO3 to grams</u>
1 moles H2SO3 to grams = 82.07908 grams
2 moles H2SO3 to grams = 164.15816 grams
3 moles H2SO3 to grams = 246.23724 grams
4 moles H2SO3 to grams = 328.31632 grams
5 moles H2SO3 to grams = 410.3954 grams
6 moles H2SO3 to grams = 492.47448 grams
7 moles H2SO3 to grams = 574.55356 grams
8 moles H2SO3 to grams = 656.63264 grams
9 moles H2SO3 to grams = 738.71172 grams
10 moles H2SO3 to grams = 820.7908 grams
Answer:
Multiply the number of moles in the product by the molecular weight of the product to determine the theoretical yield.
Explanation:
For example:
If you created 0.5 moles of Aluminium Oxide the molecular weight of Aluminium Oxide is 101.96g/mole, so you would get 50.98g as the theoretical yield.
So multiply,..
101.96x0.5= 50.98
This is the correct way to calculate the theoretical yield
......
Answer:
ΔH°rxn = - 433.1 KJ/mol
Explanation:
- CH4(g) + 4Cl2(g) → CCl4(g) + 4HCl(g)
⇒ ΔH°rxn = 4ΔH°HCl(g) + ΔH°CCl4(g) - 4ΔH°Cl2(g) - ΔH°CH4(g)
∴ ΔH°Cl2(g) = 0 KJ/mol.....pure element in its reference state
∴ ΔH°CCl4(g) = - 138.7 KJ/mol
∴ ΔH°HCl(g) = - 92.3 KJ/mol
∴ ΔH°CH4(g) = - 74.8 KJ/mol
⇒ ΔH°rxn = 4(- 92.3 KJ/mol) + (- 138.7 KJ/mol) - 4(0 KJ/mol) - (- 74.8 KJ/mol)
⇒ ΔH°rxn = - 369.2 KJ/mol - 138.7 KJ/mol - 0 KJ/mol + 74.8 KJ/mol
⇒ ΔH°rxn = - 433.1 KJ/mol
