Use the Henderson-Hasselbach equation:
pH = pKa + log[base]/[acid]
pH = -log(1.7 x 10^-5) + log(0.590/0.130) = 5.43
Answer:
There is no picture, how can we help?
Explanation:
Answer:
D
Explanation:
We must study the reaction pictured in the question closely before we begin to attempt to answer the question.
Now, the reaction is a free radical reaction. This implies that only one electron is transferred. The transfer of one electron is shown using a half arrow rather than a full arrow. The both species are radicals (odd electron species) and contribute one electron each.
Hence we must show electron movements in both species using a half arrow.
Answer:
Succinic acid
Explanation:
The most common possibility is succinic acid
As it has decimals after whole no till hundredth it contains OH and C in most of the cases .
Let's check for succinic acid
- C4H_6O_6
- 4(12)+4(16)+6
- 64+48+6
- 118u
Yes approximately equal
Molecular formula is.
(CH_2)_2(CO_2H)_2
Answer:
320 g
Step-by-step explanation:
The half-life of Co-63 (5.3 yr) is the time it takes for half of it to decay.
After one half-life, half (50 %) of the original amount will remain.
After a second half-life, half of that amount (25 %) will remain, and so on.
We can construct a table as follows:
No. of Fraction Mass
half-lives t/yr Remaining Remaining/g
0 0 1
1 5.3 ½
2 10.6 ¼
3 15.9 ⅛ 40.0
4 21.2 ¹/₁₆
We see that 40.0 g remain after three half-lives.
This is one-eighth of the original mass.
The mass of the original sample was 8 × 40 g = 320 g
