The number of neon moles that occupy a volume of 14.3 l at STP is calculated as follows
At STP 1 mole = 22.4 liters
what about 14.3 liters
by cross multiplication
= (1 mole x 14.3 l)/22.4 l =0.638 moles of neon
Answer:
For part (a): pHsol=2.22
Explanation:
I will show you how to solve part (a), so that you can use this example to solve part (b) on your own.
So, you're dealing with formic acid, HCOOH, a weak acid that does not dissociate completely in aqueous solution. This means that an equilibrium will be established between the unionized and ionized forms of the acid.
You can use an ICE table and the initial concentration ofthe acid to determine the concentrations of the conjugate base and of the hydronium ions tha are produced when the acid ionizes
HCOOH(aq]+H2O(l]⇌ HCOO−(aq] + H3O+(aq]
I 0.20 0 0
C (−x) (+x) (+x)
E (0.20−x) x x
You need to use the acid's pKa to determine its acid dissociation constant, Ka, which is equal to
Answer:
The pH does not decrease drastically because the NaOH reacts with the <u>D) Benzoic acid</u> present in the buffer solution.
Explanation:
The hydroxide ions will react with acidic part of the solutions, it means the benzoic acid, so it will form the conjugate base, the benzoate ion.
Answer:
See Explanation
Explanation:
The equation of the reaction;
KHSO4(aq) + KOH(aq) -------> K2SO4(aq) + H2O(l)
Number of moles of KHSO4 = 49.6 g/136.169 g/mol = 0.36 moles
Since the reaction is in a mole ratio of 1:1, 0.36 moles of K2SO4 is produced.
Number of moles of KOH = 25.3 g/56.1056 g/mol = 0.45 moles
Since the reaction is 1:1, 0.45 moles of K2SO4 is produced
Hence K2SO4 is the limiting reactant.
Mass of K2SO4 formed = 0.36 moles of K2SO4 * 174.26 g/mol = 62.7 g
So;
1 mole of KHSO4 reacts with 1 mole of KOH
0.36 moles of KHSO4 reacts with 0.36 * 1/1 = 0.36 moles of KOH
Amount of excess KOH = 0.45 moles - 0.36 moles = 0.09 moles
Mass of excess KOH = 0.09 moles * 56.1056 g/mol = 5 g of excess KOH
