<u>Answer:</u> The mass of sodium chloride solution present is 0.256 grams.
<u>Explanation:</u>
We are given:
39.0 % of sodium in sodium chloride solution
This means that 39.0 grams of sodium is present in 100 grams of sodium chloride solution
Mass of sodium given = 100 mg = 0.1 g (Conversion factor: 1 g = 1000 mg)
Applying unitary method:
If 39 grams of sodium metal is present in 100 grams of sodium chloride solution
So, if 0.1 grams of sodium metal will be present in =
of sodium chloride solution.
Hence, the mass of sodium chloride solution present is 0.256 grams.
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Answer: 44g
Explanation: The formular for finding Moles is ;
Moles = Mass / Molar Mass or Formular Mass.
Base on this question; Moles = 10, Mass = 440g, and Formular Mass = ?
Making 'Formular Mass', subject of the formular; we thus have;
Formular mass = Mass / Moles = 440/ 10 = 44g
Answer:
a.option is the correct answer
Answer:
13.94moles of Na₂O
Explanation:
The balanced reaction expression is given as:
4Na + O₂ → 2Na₂O
Given parameters:
Number of moles of O₂ = 6.97moles
Unknown:
Number of moles of Na₂O
Solution:
To solve this problem;
1 mole of O₂ will produce 2 moles of Na₂O ;
6.97 moles of O₂ will produce 6.97 x 2 = 13.94moles of Na₂O