An 11 M KOH solution indicates there are 11 moles of KOH per liter of water. 1 mole of KOH has a volume of 27.4 mL, so to account for the added volume it is necessary to add 15.75 moles of KOH per liter of water, or 884 grams per liter.
Dissolve 5,6106 g KOH in 1 L demineralized water at 20 0C.
1 cm is .01m so 600cm*.01m = 6m =] -ray
When KOH reacts with HCl, these products are formed. This is a neutralization reaction. KOH is a base while HCl is an acid.
The answer is 12,831 g KOH.
500cm
because KOH is hygroscopic
1 cm is .01m so 600cm*.01m = 6m =] -ray
.01m = 1cm
dissolve KOH powder in ethanol
6
I assume you mean 7.25 M KOH, but the process is the same anyway.Molarity = moles of solute/Liters of solution ( 400 ml = 0.4 Liters )7.25 M KOH = X moles KOH/0.4 Liters= 2.9 moles KOH (56.108 grams/1 mole KOH)= 163 grams potassium hydroxide needed===============================
1% solution of KOH contains 1g of KOH in 100g of solution. This means that you need to mix 1g of KOH and 99g of water.
K+ and OH- is KOH -----------
When KOH reacts with HCl, these products are formed. This is a neutralization reaction. KOH is a base while HCl is an acid.
i have Audi 01N and use it
0.800 M KOH (1mol/2.5L)(56.1g KOH/1mol)It's set up stoichiometrically, but hard to show that here... 0.800 M KOH / 2.5 L x 56.1gYour answer should come out 17.952 g KOH. If you're following sig figs, then your answer should come out 18. g KOH
Molar mass of KOH = 39.1+16.0+1.0 = 56.1Amount of KOH = mass of sample / molar mass = 32/56.1 = 0.570mol
Kokiriko is pronounced as "koh-ki-ree-koh".