According to the reaction equation:
CH3COO- + H+ → CH3COOH
initial 0.25 0.15
change - 0.025 + 0.025
Equ (0.25-0.025) (0.15 + 0.025)
first, we have to get moles acetate and moles acetic acid:
moles of acetate = 0.25 - 0.025 = 0.225 moles
∴ [CH3COO-] = 0.225 mol / 1 L = 0.225 M
moles of acetic acid = 0.15 + 0.025 = 0.175 moles
∴ [ CH3COOH] = 0.175 mol / 1L = 0.175 M
Pka = -㏒ Ka
= -㏒ 1.8 x 10^-5
= 4.74
from H-H equation we can get the PH value:
PH = Pka + ㏒ [acetate / acetic acid]
PH = 4.74 + ㏒[0.225/0.175]
∴ PH = 4.8
<span>1.15x10^24 molecules of hypothetical substance b
Making the assumption that each molecule in hypothetical substance a reacts to produce a single molecule of hypothetical substance b, then the number of molecules of substance b will be the number of moles of substance a multiplied by avogadro's number. So
Moles hypothetical substance a = 29.9 g / 15.7 g/mol = 1.904458599 moles
This means that we should also have 1.904458599 moles of hypothetical substance b. And to get the number of atoms, multiply by 6.0221409x10^23, so:
1.904458599 * 6.0221409x10^23 = 1.146892x10^24 molecules.
Rounding to 3 significant figures gives 1.15x10^24</span>
Answer:
It indicates the number 4
Explanation:
IV=4
Why don’t you have friends?
The half life of an isotope is the amount of time it will take for the given isotope to decay till the point where it reaches half of what amount you had started with.
If the half life of Ra-229 is 4 minutes then every 4 minutes the sample is reduced to half of what you had before
4 minutes.) 50/2 Ra-229 --> 25g Ra-229
8 minutes.) 25/2 Ra-229 --> 12.5 Ra-229
12 minutes.) 12.5/2 Ra-229 --> 6.25 Ra-229 remaining
<em>I beleive this is the correct answer</em>
