Answer: There are 0.5 grams of barium sulfate are present in 250 of 2.0 M solution.
Explanation:
Given: Molarity of solution = 2.0 M
Volume of solution = 250 mL
Convert mL int L as follows.
Molarity is the number of moles of solute present in liter of solution. Hence, molarity of the given solution is as follows.
Thus, we can conclude that there are 0.5 grams of barium sulfate are present in 250 of 2.0 M solution.
Answer : The concentration after 17.0 minutes will be,
Explanation :
The expression for first order reaction is:
where,
= concentration at time 't' (final) = ?
= concentration at time '0' (initial) = 0.100 M
k = rate constant =
t = time = 17.0 min = 1020 s (1 min = 60 s)
Now put all the given values in the above expression, we get:
Thus, the concentration after 17.0 minutes will be,
Answer:
0.0714 M for the given variables
Explanation:
The question is missing some data, but one of the original questions regarding this problem provides the following data:
Mass of copper(II) acetate:
Volume of the sodium chromate solution:
Molarity of the sodium chromate solution:
Now, when copper(II) acetate reacts with sodium chromate, an insoluble copper(II) chromate is formed:
Find moles of each reactant. or copper(II) acetate, divide its mass by the molar mass:
Moles of the sodium chromate solution would be found by multiplying its volume by molarity:
Find the limiting reactant. Notice that stoichiometry of this reaction is 1 : 1, so we can compare moles directly. Moles of copper(II) acetate are lower than moles of sodium chromate, so copper(II) acetate is our limiting reactant.
Write the net ionic equation for this reaction:
Notice that acetate is the ion spectator. This means it doesn't react, its moles throughout reaction stay the same. We started with:
According to stoichiometry, 1 unit of copper(II) acetate has 2 units of acetate, so moles of acetate are equal to:
The total volume of this solution doesn't change, so dividing moles of acetate by this volume will yield the molarity of acetate: