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How many ml of 250 M NaOH are required to neutralize 30.4 ml of 152 M HCl?
M1V1=M2V2... By plugging in, you get 18.48 mL of NaOH
How can you prepare 250cm3 of 0.5m mgso4 from stock solution of 2.5m mgso4?
Just put both of them together use your common sense duh😂😂😂
If 50 ml of 50 M Hydrogen Cloride is used to completely neutralize 25 ml of KOH solution what is the molarity of the base?
m1v1 = m2v2 50 mL * 50 M = 25 mL * X M X = 100
How do you find magnitude of initial momentum?
INITIAL MOMENTUM = FINAL MOMENTUM ∑M1V1 + M2V2 +… + MnVn = ∑ M1V1 + M1V1 +… + MnVn + or momentum=mass x acceleration unit for momentum=kg x m/sec its confusing...
If 25.00ml of 50m naoh is used to titrate 26.25ml of hcl to the equivalence point what is the concentration of the hcl?
First balance the equation of the reaction: NaOH + HCl ---> NaOH + H2O The ratio of moles is 1:1 M1V1 = M2V2 (molarity1 x volume1) = (molarity2 x volume2) M2 = M1V1 / V2 molarity2 = (molarity1 x volume1) / (volume2) M of HCl = 50M x 0.025 litres / 0.02625 litres = 47.6 M
How many millilieters of 5.5 m naoh are needed to prepare 300 ml of 1.2 m naoh?
Use M1V1=M2V2 Where M is the concentration (5.5 m for M1 and 1.2 m for M2) and V is volume V1 is 300 ml and V2 is your unknown. Using this calculation for other questions be sure that units are all the same. So all molarities and all mL in this example.
How do you solve a dilution problem in Chem?
The key formula for solving a dilution problem is M1V1=M2V2 (alternately, MAVA=MBVB) where concentration is M (measured in Molars, a unite of concentration-->Molars=moles solute/Liters solution) and the volume of solution is V. M1V1 represents the inital conditions (pre-dilution), and M2V2 denotes the final conditions (post-dilution). Plug in the three values you know to find the fourth value, which you are solving for. Ex. 750mLs of a 0.5 M NaC2H3O2 solution is diluted with 250mL of H2O. What is the new concentration of NaC2H3O2? Answer: (0.5M)x(.750L)=(MB)x(.750L+.250L) 0.375ML=(MB)x(1L) MB=0.375M
How many ltiers of 30 M HCl could be made with 3.0 L of 50 MHCl?
To do this use the formula Where M1 = concentration of what you have and M2 is the concentration you are trying to achieve. And V1 is the volume you have and V2 is the volume you want. So in this case we have 3.0L(50 M HCl) = X L(30 M HCl) and solve for X. (X = 5 L) M1V1 = M2V2
What will be the final volume of a solution prepared by diluting 25 mL of 8.25 M sodium hydroxide to a concentration of 2.40?
For all dilution/ concentration problems you use the simple equation: M1V1 = M2V2 2.40*V2 = 8.25*25 V2 = (8.25*25)/2.40 V2 = 85.9mL Final volume will be 86mL.
How do you dilute a 10 M H2SO4 to 1M H2SO4?
Remember M1V1=M2V2, where M is molarity and V is volume. M1/M2=V2/V1, 10/1=v2/v1, For diluting the acid, we can add acid to water. So, assuming that 10M H2SO4 is having 1ml of water, we should add 1M of H2So4 to 10ml of water.
What does 4 equals M M stand for?
4 Minute Mile
How much water must be added to 588 mL of 0.222 M HCl to produce a 0.126 M solution?
To determine how much water must be added, we need to calculate the volume of the 0.222 M HCl that needs to be diluted to 0.126 M. The equation to use is M1V1 = M2V2. Rearranging the equation, we have V2 = (M1V1)/M2. Plugging in the values, we get V2 = (0.222 M)(588 mL)/(0.126 M) = 1037 mL. Therefore, 1037 mL of water must be added to the 588 mL of 0.222 M HCl to produce a 0.126 M solution.
