Given:
0.607 mol of the weak acid
0.609 naa
2.00 liters of solution
The solution for finding the ph of a buffer:
[HA] = 0.607 / 2.00 = 0.3035 M
[A-]= 0.609/ 2.00 = 0.3045 M
pKa = 6.25
pH = 6.25 + log 0.3045/ 0.3035 = 6.25 is the ph buffer prepared.
6-(2+4)
6-6
=0 Formal charge of O atoms a and b is 0
6-(1+6)
6-7
= -1 Formal charge for O atoms c is -1
Answer 0,0,-1
Answer:
D) 10.812
Explanation:
Given:
The two isotopes of boron (B).
Boron-10 has an abundance of 19.8% and Boron -11 has an abundance of 80.2%
To find:
The atomic mass of europium.
Solution:
Mass of Boron-10 = 10.013 amu
The percentage abundance of Boron-10 = 19.8%
The fractional abundance of Boron-10 =0.198
Mass of Boron-11 = 11.009 amu
The percentage abundance of Boron-11= 80.2%
The fractional abundance of Boron-11= 0.802
The average atomic mass of Boron = A.M
A.M = Mass of isotope x Fractional abundance of isotope
= 10.013amu x 0.198 +11.009amu x 0.802 = 10.81u
Therefore, D) 10.812 amu is the atomic mass of boron.
0.008 ÷ 51.3 = 0.0002
Sig Figs
1
0.0002
Decimals
4
0.0002
Scientific Notation
2 × 10-4
E-Notation
2e-4
Words
zero point zero zero zero two
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