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)
Hey there!:
Volume in mL :
1.71 L * 1000 => 1710 mL
Density = 0.921 g/mL
Therefore:
Mass = Density * Volume
Mass = 0.921 * 1710
Mass = 1574.91 g
Answer:
The elements in Group 2 (beryllium, magnesium, calcium, strontium, barium, and radium) are called the alkaline earth metals (see Figure below). These elements have two valence electrons, both of which reside in the outermost s sublevel. The general electron configuration of all alkaline earth metals is ns
The answer is -11.2 fahrenheit.
Answer:
Unless you mistyped the question, the answer should be 25+
