if 5ml of 3.0 x10^-5 M crystal violet solution is mixed with 2ml of 0.1M of NaOH, what is the concentration of crystal violet in this solution?
CV+ (aq ) + OH− (aq) = CVOH (aq )
We use the relation M1V1 = M2V2
M1 = Molarity of crystal violet = 3 x 10-5 M
V1 = Volume of crystal violet = 0.005 L = 0.005 L
M2 = Final molarity of solution = Unknown
V2 = Final volume = Volume of NaOH + Volume of crystal violet = 0.005 L + 0.002 L = 0.007 L
M2 = M1V1 / V2 = (3 x 10-5 x0.005) / 0.007
M2 = 2.14 x 10-5 M
if 5ml of 3.0 x10^-5 M crystal violet solution is mixed with 2ml of 0.1M of...
12. The reaction between 1.0 x 104 M crystal violet and 1.0 M NaOH was monitored with UV-Vis spectroscopy. The reaction can be considered as an elementary reaction. I C25H30N3+(aq) + OH-(aq)- C2sH310N3(aq) From the experiment after 1000 s the [CV+) decreased to 50% of the initial concentration. What is the time required to consume 90% of the crystal violet?! A) B) 3322 s 5336 s 1100 s 705 s 3001 s D) E)
React 5.00 mL of 2.00 x 10-5 M crystal violet (CV+) solution with 5.00 mL 2.00 x 10-2 M sodium hydroxide solution. Which reactant is in excess? By how much? What is the concentration of crystal violet immediately after mixing (i.e., before any reaction has taken place)?
The rate of the crystal violet/NaOH reaction is given by the following generalized rate law: Rate = k’[CV+]y where k’ is the pseudo rate constant defined by k’ = k [OH-]x , where k is the actual rate constant The actual rate constant can be calculated using the pseudo rate constant and the [OH-] present in the reaction mixture. If the reaction mixture is prepared by mixing 5.0 mL of 1.00 x 10-4 M crystal violet solution with 5.0 mL...
III ) Fe+3(aq) + SCN-(aq) ß→ Fe(SCN)+2(aq) (3) Prepare a stock solution by mixing 2mL each of 0.1M FeCl3 and 0.1M KSCN in 100mL graduated cylinder. Add enough water to make 100mL solution. Note the color. If you are the first one to use this station, make this solution and save it for the rest of the class. If you are using someone else's stock solution, make sure to make observations of each reactant. (4) *Add about 1mL of 0.1M...
7.00mL of 0.00750 M
Integrated Chemical Kinetics Determination of the Rate Law for the Crystal Violet/Hydroxide ion reaction Page 7 Pre-laboratory Assignment Name: Show all work in your lab notebook and box your final answers. Express to correct sig figs and include units 1) 7.00 mL of 0.00750 M CV solution was diluted to 10.0 mL. Calculate the Beer's law constant ko if the diluted solution's absorbance was measured to be 0.275 using a spectrophotometer. 2) 25.0 mL of a...
question 3 please
Integrated Chemical Kinetics Determination of the Rate Law for the Crystal Violet/Hydroxide ion reaction Page 7 Pre-laboratory Assignment Name: Show all work in your lab notebook and box your final answers, Express to correct sig figs and include units 1) 7.00 mL of 0.00750 M CV solution was diluted to 10,0 mL. Calculate the Beer's law constant Ks if the diluted solution's absorbance was measured to be 0.275 using a spectrophotometer. 2) 25.0 mL of a solution...
DATA I. Determining the Effect of Temperature lO 5 l.om Solution concentrations: [crystal violet]- [NaoH] = Ice Bath Temperature O.5℃ Reaction time 3:15 wn (trial 1) 3:27 trial2) Reaction time 4 sec-(trial 1) /s--(trial2) I. Determining the Rate Law for the reaction of NaOH and crystal violet. Hot Water Bath Temperature 55 C
a protein sample was diluted 1/5 and then 5ml of the diluted sample was mixed with 20 ml of reactants to give 25 ml total volume. from this 25 ml solution, 5ml were removed. this 5ml contain 3 mg of protein. what was the concentration of protein in the original sample
What volume (mL) of crystal violet stock solution 2.5 x 10-5M is required to prepare a concentration of 0.00000125 M to a total of 10 mL diluted solution?
I'm not sure if my answer to part d is correct. If it's not
then I need part d and part e answered. Thank you!
b. A solution is prepared by mixing 25.0 mls of 4.84 x 10% M crystal violet solution with 20.0 mLs of 0.10 M NaOH solution and diluted to a final volume of 100.0 mLs. Determine the limiting reagent. mol CV, 4.84 X10- M 5x 0.025 1.21 x 15 mol mol NaOH = 0.10 X 0.020...