v1= initial volume
c1= initial concentration
v2= final volume
c2= final concentration
For example, you have 10mL of an unknown substance with a concentration of 0,5mol/L. If you add 50mL, what will the final concentration be.
V1= 10mL
C1= 0,5mol/L
V2= 60mL
C2= x
10/0,5=60/x You must start by putting everything in the same mesure. We'll use mL here. So 0,5-->1000mL= 50-->10mL
50x60= 300
300/10= 30
30 is your C2
This is a mathematical equation for determining dilutions. c1 * v1 = c2 * v2 initial concentration times initial volume equals final concentration time final volume.
Applying the equation for a dilution (c1.V1 = c2.V2) gives4.0(mol/L)*V1(L) = 2.5(mol/L)*4.5(L) so V1 = 2.5*4.5/4.0 = 2.813 L = 2.8 litres
Use the formula C1(V1) = C1(V2) to find the unknown concentration. Here, V1 = 20ml and V2 = 33.86ml. Also, C2 is 0.1368M Now plug in these values to find V1 Therefore we have: 20ml(V1) = (0.1368)(33.86) Therefore, V1 = 0.2316M The a 0.2316M solution of H2SO4 was required for the neutralization reaction
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 :)
preparation of 500 ml 1x TAE buffer,50ml of 10x buffer add to 450ml DI.water. to calculate just use MV=MV so 500mL * 1x= 10x * V then solve for V. add the amount of DI water you need to get the volume you calculated above.
You can dilute by adding distilled water. When diluting, be sure to add the solution to water several times instead of adding water to the solution (especially if it is highly concentrated).
5 * 10**-12 mol 32 * 10**-9 mol Concentration (M) * Volume (L) = mols C1*V1=C2*V2 (5*10**-12)*V1=(32*10**-9)*V2 (5*10**-12)*V1/(32*10**-9)=V2 (5*10**-3)*V1/32=V2 The volume of the 5 picomolar solution that you wish take = V1 The volume of the 32 nanomolar solution that you need to make V1 at 5pM concentration = V2 Take V2, and place into graduated cylinder and fill to V1.
Use the formula C1V1 = C2 V2 where, C1 is the concentration of the original = 2.050 mol/L, V1 is the volume you are trying to look for = x, C2 is the new concentration you want 0.8543 mol/L and V2 si the new volume you want 0.750 L So substituting into the formula you would get 2.050 x V1 = 0.8543 x 0.750 2.0502 v1 = 0.640725 V1 = 0.640625 / 2.050 V1 = 0.3125 L
(v1/t1) = (v2/t2)
By using the formula V1 x N1 = V2 x N2 Taking V1= 250 ml; N1= 0.35M; N2= 5.7M. V2 = volume of 5.7M needed to dilute V2 = V1 x N1 N2 = 250 x 0.35 = 15.35ml 5.7
( | V1 - V2 | / ((V1 + V2)/2) ) * 100
v1 = initial velocity v2 = final velocity
the equation for an ideal gas is pv / t = nr n * r is a constant for a closed system p pressure v volume t temperature in kelvin p1 v1 /t1 = p2 v2 /t2 if p1 = p2 v1/t1 = v2/t2 t2= v2/v1 *t1 directly proportional to the change in volume if v1 = v2 the same can be done and you will find that t is directly proportional to change in pressure. generally t is directly proportional to the product of pressure and volume. pv = nr t
p1 is pressure 1 v1 is volume 1p2 is pressure 2v2 is volume 2they are in the boyles law thing
it is very easy to prepare working solution from a stock solution we use the formula for this purpose which is: C1V1 = C2V2 C1 is the concentration of the stock solution V1 required volume from the stock solution C2 concentration of the working solution V2 volume of the working solution
By using the formula V1 x C1 = V2 x C2Taking V1= 100 mL; C1= 1.00M; C2= 0.31MV2 = volume (in mL) of 0.31MV2 = V1 x N1 / N2 = 100x1.000/0.31= 322.6 mLSo you have to fill 100 mL UP TO 322.6 mL (total volume)with about (or somewhat more) than 322-100 = 222 mL (volume to add)
Vector
v0=v1+v2