Answer:
Nitrogen (ii) oxide
Explanation:
To know the IUPAC name for NO, we shall determine the oxidation number of N in NO.
NOTE: The oxidation number of oxygen (O) is always – 2.
Thus the oxidation number of N in NO can be obtained as follow:
N + O = 0 (ground state)
N + (– 2) = 0
N – 2 = 0
Collect like terms
N = 0 + 2
N = +2
Thus, the oxidation number of Nitrogen (N) in NO is +2.
Therefore, the IUPAC name for NO is Nitrogen (ii) oxide
<span>BaCl2+Na2SO4---->BaSO4+2NaCl
There is 1.0g of BaCl2 and 1.0g of Na2SO4, which is the limiting reagent?
"First convert grams into moles"
1.0g BaCl2 * (1 mol BaCl2 / 208.2g BaCl2) = 4.8 x 10^-3 mol BaCl2
1.0g Na2SO4 * (1 mol Na2SO4 / 142.04g Na2SO4) = 7.0 x 10^-3 mol Na2SO4
(7.0 x 10^-3 mol Na2SO4 / 4.8 x 10^-3 mol BaCl2 ) = 1.5 mol Na2SO4 / mol BaCl2
"From this ratio compare it to the equation, BaCl2+Na2SO4---->BaSO4+2NaCl"
The equation shows that for every mol of BaCl2 requires 1 mol of Na2SO4. But we found that there is 1.5 mol of Na2SO4 per mol of BaCl2. Therefore, BaCl2 is the limiting reagent.</span>
Answer:
The answer to your question is Volume = 2.94 L
Explanation:
Data
Concentration = 6.2 M
Number of moles = 18.25
Volume = ?
Process
1.- Use the Molarity formula to solve this problem
Molarity = Number of moles / Volume
- Solve for volume
Volume = Number of moles / Molarity
2.- Substitution
Volume = 18.25 / 6.2
3.- Simplification
Volume = 2.94 L
Explanation:
Number of moles is defined as mass divided by molar mass.
Mathematically, No. of moles = 
It is known that molar mass of water (
) is 18 g/mol and it is given that mass is 97.3 grams.
Hence, calculate the number of moles of water as follows.
No. of moles =
= 
= 5.40 mol
Thus, we can conclude that there are 5.40 moles of H2O are equivalent to 97.3 grams of H2O.
