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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😂😂😂
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?
The final volume of the solution will be 68.18 mL. This can be calculated using the formula for dilution, which is M1V1 = M2V2. Solving for V2 (final volume) gives V2 = (M1V1) / M2 = (8.25 M * 25 mL) / 2.40 M = 68.18 mL.
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
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
What is the molarity of the NaOH solution if 25.0 mL of the solution is required to neutralize 35.0 mL of a 0.20 M HCl solution?
To find the molarity of the NaOH solution, use the formula M1V1 M2V2. Given that V1 25.0 mL, V2 35.0 mL, and M2 0.20 M, you can calculate M1 to be 0.14 M.
A 40Kg football player leaps through the air to collide with and tackle a 60Kg player heading toward him also in the air. If the 40Kg player is heading to the right at 7M per S and the 60Kg player...?
The collision would involve momentum conservation, where the total momentum before the collision is equal to the total momentum after the collision. By using the equation (m1v1) + (m2v2) = (m1v1') + (m2v2'), you can solve for the final velocities. The direction and speed of the players after the collision would depend on the angle and intensity of the impact and the individual masses and velocities.
What is the molarity of a hydrochloric acid solution prepare by diluting 200.0mL of 0.500 M HCl to a total volume of 1.00 L?
The molarity of the final solution can be calculated using the formula: M1V1 = M2V2. By plugging in the values, we get (0.500 M)(0.200 L) = M2(1.00 L). Solving for M2 gives us 0.100 M for the final molarity of the solution.
What does 4 equals M M stand for?
4 Minute Mile
What volume of 8.0 M H2SO4 is required to prepare 200.0 ml of a 0.50 M?
To prepare a 200ml solution of 0.5M H2SO4, you need to add 12.5ml of 8M H2SO4 and then dilute it up to 200ml. Well, this is how to calculate it, I'll go straight to the equation. use the M1V1 = M2V2 equation, M1 = 8.0 M V1 = Volume needed M2 = 0.5 M V2 = 200 ml 8.0 M x V1 = 0.5 M x 200 ml V1 = (0.5 M x 200 ml)/ 8.0 M = 12.5 ml Cheers :)
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
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 Ml of 14.5 M NH3 are needed to prepare 2.00 L of a 1.00 M solution?
To prepare a 1.00 M solution from a 14.5 M NH3 solution, you'll need to dilute it. By using the formula: M1V1 = M2V2, you can calculate the volume of 14.5 M NH3 needed. Plugging in the values, you get 138 mL of 14.5 M NH3 needed.
