Atomic mass (K)= 39.1 amu
therefore:
1 mol (k)---------------------39.1 g
x------------------------------- 2.25 g
x=(1 mol * 2.25 g) / 39.1 g=0.05754....≈0.06 moles
Answer: 0.06 moles.
Answer:
they are transfer from the towers
Explanation:
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:I think it could be A or B but I would choose A.
Explanation:
Answer: Volume of the 1M EtOH and water should be 0.75 ml and 9.25 ml respectively to obtain the working concentration.
Explanation:
According to the dilution law,
where,
= molarity of stock solution = 1M
= volume of stock solution = ?
= molarity of diluted solution = 0.075 M (1mM=0.001M)
= volume of diluted solution = 10 ml
Putting in the values we get:
Thus 0.75 ml of 1M EtOH is taken and (10-0.75)ml = 9.25 ml of water is added to make the volume 10ml.
Therefore, volume of the 1M EtOH and water should be 0.75 ml and 9.25 ml respectively to obtain the working concentration
