How many milliliters of a 0.300M KOH solution contain 1.75×10−2 mol of KOH?
Molarity = moles solute / Litres solution Therefore Litres = moles / Molarity moles KOH = 1.75x10^-2 mol / 0.300 M = 0.0583 L = 58.3 ml (3 sig figs)
How many milliliters of 1.50 M KOH solution are needed to provide 0.110 mol of KOH?
Molarity = moles / volume in liters. By rearranging the equation we get that volume = moles/molarity. From here the numbers just need to be plugged in and solved. Volume (in liters) = .110 mol KOH / 1.5 M KOH = .073 L. To convert to mililiters just mulitply .073 L x 1000 mL/L - 73 ml
How many milliliters of a 0.330 M KOH solution contain 1.67×10−2 mol of KOH? need answer in ML?
How many milliliters of a 0.330 M KOH solution contain 1.67×10−2 mol of KOH? need answer in ML Please give me the answers and if possible try explaining it for me. Thanks in advance
How many milliliters of 1.20M KOH solution are needed to provide 0.100mol of KOH?
Okay, so this question is a simple matter of rearranging the equation for molarity (M) :Molarity=moles of solute/liters of solutionM=mol/L Solve for liters: L=mol/MNow it's a simple plug and play problem L=(0.100molKOH)/(1.20MKOH)L=0.08333Now convert L to mL. . . it takes 83.33mL of a 1.20M KOH solution to give .100moles of KOH
How many mL of a 0.200 M KOH solution contains 10.0 g of KOH?
10.0 g KOH / 56.11 g/mole = 0.178 mole of KOH liters x 0.200 moles/liter =0.178 moles liters = 0.891 mL = 891
How many milliliters of 8.00 M KOH must be added to neutralize the following solutions?
a) Both acids are strong and fully dissociate, Determine the number of moles of each acid first. Let C1, V1 = concentration and volume of HBr; C2, V2 = concentration and volume of H2SO4 moles HBr = n = (0.24 x 0.025) = 0.006 moles moles H2SO4 = n = (0.2 x 0.075) = 0.015 moles But since sulfuric is diprotic, it will create 0.015 x 2 = 0.03 moles of H+ ions Total moles of H+ ions from both acids = 0.006 + 0.03 = 0.036 moles moles of NaOH = n = CV = 8V One mole of NaOH makes one mole of OH- ions. To neutralize acids we need 0.036 moles of OH- (and hence NaOH). n = 8V = 0.036 V = n / 8 V = 0.036 / 8 = 0.0045 litres = 4.5 mls b) moles acid = CV = 0.3 x 0.045 = 0.0135 moles moles base = n = CV = 0.25 x 0.010 = 0.0025 moles there is excess acid (0.0135 - 0.0025 = 0.011 moles Hence the amount of base needed = V = n / C = 0.011 / 8 = 0.0014 L = 1.4 mls.
How many milliliters of 10.0 M KOH must be added to neutralize the following solutions?
When you neutralize and acid and base, H reacts with OH- to create water. Concentration = moles / volume Calculate how many moles of HBr you have. HBr dissociates into H and Br- so you know how many moles of H you have. Using the same equation, calculate the moles of LiOH. LiOH dissociates into Li and OH- so you know how many moles of OH- you have. The KOH must add the necessary moles of OH- that when added with the OH- moles from the LiOH to equal the moles of H Using the same equation concentration = moles / volume., you know the moles and the concentration, calculate the volume needed. Part B is done the same way, just different numbers but the same steps.
How many milliliters of water must be added to 50ml of 7.0 M of HCL so that the solution concentration is reduced to 0.35 M?
Molarity= (number of moles of that substance) / (Volume of solution in litres)7= (number of moles of HCl) / 0.05Thus,Number of moles of HCl = 7 x 0.05 = 0.35The number of moles of HCl will remain unchanged.Now, for new solution,0.35= 0.35 / (New volume of solution in litres)Thus,New volume = 1 litreThus, amount of water added = 1- 0.05 litre = 0.95 litre = 950 mlI am pretty sure that's how you do it!