Compared to a 26-gram sample of NaCl(s) at STP, a 52-gram sample of NaCl(s) at STP has(1) a different density(2) a different gra
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Answer: The net ionic equation is
Explanation:
A double displacement reaction involves the exchange of ions. Chemicals that dissolve in water are marked with the symbol (aq), while those that do not dissolve and remain solid are shown with (s) after their formulas.
The ion-based representation of the equation is:
"Spectator ions" are the ions that do not participate in the chemical reaction, appearing on both sides of the equation in ionic form.
Ammonium and chlorate ions are present on both sides; thus, they do not factor into the net ionic equation.
Therefore, the net ionic equation is:
Answer:
1) This dilution plan will yield a 200μM solution.
2) This dilution plan will not yield a 200μM solution.
3) This dilution plan will not yield a 200μM solution.
4) This dilution plan will yield a 200μM solution.
5) This dilution plan will yield a 200μM solution.
Explanation:
Convert the initial molarity into molar form as shown.
Let's examine the following serial dilution processes.
1)
Dilute 5.00 mL of the stock solution to 500 mL. Then take 10.00 mL of this new solution and dilute it further to 250 mL.
Concentration of 500 mL solution:
10 mL of this solution is further diluted to 250 mL
Convert μM:
Thus, this dilution scheme will yield a 200μM solution.
2)
Dilute 5.00 mL of the stock solution to 100 mL. Then take 10.00 mL of this new solution and dilute to 1000 mL.
Concentration of 100 mL solution:
10 mL of this solution is further diluted to 1000 mL
Convert μM:
Thus, this dilution scheme will not yield a 200μM solution.
3)
Dilute 10.00 mL of the stock solution to 100 mL, followed by diluting 5 mL of that new solution to 100 mL.
Concentration of 100 mL solution:
5 mL of this solution is diluted to 1000 mL
Convert μM:
Thus, this dilution scheme will not yield a 200μM solution.
4)
Dilute 5 mL of the stock solution to 250 mL. Then take 10 mL of this new solution and further dilute it to 500 mL.
Concentration of 250 mL solution:
10 mL of this solution is further diluted to 500 mL
Convert μM:
Thus, this dilution scheme will yield a 200μM solution.
5)
Dilute 10 mL of the stock solution to 250 mL. Then take another 10 mL of this new solution and dilute it to 1000 mL.
Concentration of 250 mL solution:
10 mL of this solution is further diluted to 1000 mL
Convert μM:
Thus, this dilution scheme will yield a 200μM solution.