Answer:
The concentration of species in 500 mL of a 2.104 M solution of sodium sulfate is 4.208 M sodium ion and 2.104 M sulfate ion. (option E)
Explanation:
Step 1: Data given
Volume = 500 mL = 0.500 L
The concentration sodium sulfate = 2.104 M
Step 2: The equation
Na2SO4 → 2Na+ + SO4^2-
For 1 mol Na2SO4 we have 2 moles sodium ion (Na+) and 1 mol sulfate ion (SO4^2-)
Step 3: Calculate the concentration of the ions
[Na+] = 2*2.104 M = 4.208 M
[SO4^2-] = 1*2.104 M = 2.104 M
The concentration of species in 500 mL of a 2.104 M solution of sodium sulfate is 4.208 M sodium ion and 2.104 M sulfate ion. (option E)
Pollution and rising temperatures
Answer:
5.22 moles of Na₂CO₃ have mass of 553.22 g
Explanation:
Given data:
Mass of Na₂CO₃ = ?
Number of moles of Na₂CO₃ = 5.22 mol
Solution:
Formula:
Mass = number of moles × molar mass
Molar mass of Na₂CO₃ = 105.98 g/mol
Mass = 5.22 mol × 105.98 g/mol
Mass = 553.22 g
Thus, 5.22 moles of Na₂CO₃ have mass of 553.22 g.
