Answer:
Mass = 29.23 g
Explanation:
Given data:
Volume of solution = 814.2 mL 814.2/1000 = 0.8142 L)
Molarity of solution = 0.227 M
Mass of solute in gram = ?
Solution:
Molarity = number of moles / volume in L
By putting values,
0.227 M = number of moles / 0.8142 L
Number of moles = 0.227 M × 0.8142 L
Number of moles = 0.184 mol
Mass in gram:
Mass = number of moles × molar mass
Molar mass of calcium acetate = 158.17 g/mol
Mass = 0.184 mol × 158.17 g/mol
Mass = 29.23 g
Is the number of protons in an atom, or the total positive charge of an atom.
The empirical formula for a compound is KClO3
Explanation
find the moles of each element
moles = % composition/molar mass
molar mass of of potassium =39g/mol ,chlorine = 35.5 g/mol, oxygen =16 g/mol
moles of potassium = 31.9 / 39 = 0.818 moles
moles of chlorine = 28.9/35.5 = 0.814 moles
moles of oxygen = 39.2/ 16 = 2.45 moles
find the mole ratio by dividing with the smallest mole = 0.814 moles
potassium = 0.818/0.814 =1
chlorine = 0.814/0.814 = 1
oxygen = 2.45 /0.814 =3
the empirical formula is therefore = KClO3
6.2 grams of CO2 = 1.408786739226764 moles
Answer:
The mass of Na₂O that can be produced by the chemical reaction of 4.0 grams of sodium with excess oxygen in the reaction is 5.39 grams.
Explanation:
You know the balanced reaction:
4 NA + O₂ ⟶ 2 Na₂O
By stoichiometry of the reaction (that is, the relationship between the amount of reagents and products in a chemical reaction) react and are produced:
- Na: 4 moles
- O₂: 1 mole
- Na₂O: 2 moles
Being:
the molar mass of the compounds participating in the reaction is:
- Na: 23 g/mole
- O₂: 2*16 g/mole= 32 g/mole
- Na₂O: 2*23 g/mole +16 g/mole= 62 g/mole
Then by stoichiometry of the reaction they react and are produced:
- Na: 4 moles* 23 g/mole= 92 g
- O₂: 1 mole*32 g/mole= 32 g
- Na₂O: 2 moles* 62 g/mole= 124 g
Then you can apply the following rule of three: if 92 grams of Na produce 124 grams of Na₂O, 4 grams of Na, how much mass of Na₂O does it produce?

mass of Na₂O=5.39 g
<em><u>The mass of Na₂O that can be produced by the chemical reaction of 4.0 grams of sodium with excess oxygen in the reaction is 5.39 grams.</u></em>