A mixture is made by adding 50.0 mL of 0.20 M NaOH(aq) to 50.0 mL of water. At 25.0 °C, what is its pH?

13

Explanation:

From the dilution formula;

C1V1 = C2V2

C1= concentration of the stock solution = 0.2 M

V1 = volume of the stock solution = 50.0 ml

C2= concentration of the diluted solution= the unknown

V2= volume of diluted solution= 100 ml

C2= C1V1/V2

C2= 0.2 × 50/100

C2= 0.1 M

But

pOH= -log[OH^-]

But [OH^-] = 0.1 M

pOH= -log [0.1]

pOH= 1

Since;

pH +pOH = 14

pH= 14 - pOH

pH= 14- 1

pH= 13

i think its 100.

hope this !

percent abundance for cu-63 can be shown as:

63 x + 65 (1-x) = 62.9296

65 - 62.9296 = 2 x

x = 1.035

thus, percent abundance for cu-63 is 1.035

percent abundance for cu-65 can be shown as:

63 x + 65 (1-x) = 64.9278

65 - 64.9278 = 2 x

x = 0.0361

thus, percent abundance for cu-63 is 0.0361

the mass spectrum of cu-63 showing the intense peak as, this isotope is found in nature more abundantly around 92.38 % in comparison to cu-65 which is found 62.58 % naturally