PH is calculated using <span>Handerson- Hasselbalch equation,
pH = pKa + log [conjugate base] / [acid]
Conjugate Base = Acetate (CH</span>₃COO⁻)
Acid = Acetic acid (CH₃COOH)
So,
pH = pKa + log [acetate] / [acetic acid]
We are having conc. of acid and acetate but missing with pKa,
pKa is calculated as,
pKa = -log Ka
Putting value of Ka,
pKa = -log 1.76 × 10⁻⁵
pKa = 4.75
Now,
Putting all values in eq. 1,
pH = 4.75 + log [0.172] / [0.818]
pH = 4.072
Answer:
Training officers in how to properly collect evidence
Explanation:
Forensic science is an interesting branch of science that involves the use of scientific procedures to solve a crime case. It encompasses collection of physical evidence from the crime scene and analyzing it in a laboratory using scientific means.
A forensic scientist is the individual in charge of performing these scientific procedures. His/her major role is to run the scientific analysis of the physical evidence brought in by the officers, however, he/she can also perform the task of training officers in how to properly collect evidence, in order not to damage the evidence or render it invalid for use.
Answer:
The answer to your question is: 1, 2, 1, 2
Explanation:
1 Fe(s) + 2 Na⁺(aq) → 1 Fe²⁺(aq) + 2 Na(s)
Fe⁰ - 2e⁻ ⇒ Fe⁺² Oxidases
Na⁺ + 1 e⁻ ⇒ Na⁰ Reduces
1 x ( 1 Fe⁰ ⇒ 1 Fe⁺²) Interchange number of
2 x ( 2Na⁺ ⇒ 2 Na⁰ ) electrons
Answer:
Equilibrium constant for 
is 0.5
Equilibrium constant for decomposition of 
is 
Explanation:
dissociates as follows:
initial 0.72 mol 0 0
at eq. 0.72 - 0.40 0.40 0.40
Expression for the equilibrium constant is as follows:
![k=\frac{[PCl_3][Cl_2]}{[PCl_5]}](https://tex.z-dn.net/?f=k%3D%5Cfrac%7B%5BPCl_3%5D%5BCl_2%5D%7D%7B%5BPCl_5%5D%7D)
Substitute the values in the above formula to calculate equilibrium constant as follows:
![k=\frac{[0.40/1][0.40/1]}{0.32/1} \\=\frac{0.40 \times 0.40}{0.32} \\=0.5](https://tex.z-dn.net/?f=k%3D%5Cfrac%7B%5B0.40%2F1%5D%5B0.40%2F1%5D%7D%7B0.32%2F1%7D%20%5C%5C%3D%5Cfrac%7B0.40%20%5Ctimes%200.40%7D%7B0.32%7D%20%5C%5C%3D0.5)
Therefore, equilibrium constant for is 0.5
Now calculate the equilibrium constant for decomposition of
It is given that 
is decomposed.
decomposes as follows:
initial 1.0 M 0 0
at eq. concentration of is:
![[NO_2]_{eq}=1-(0.000066) = 0.999934\ M](https://tex.z-dn.net/?f=%5BNO_2%5D_%7Beq%7D%3D1-%280.000066%29%20%3D%200.999934%5C%20M)
![[NO]_{eq}=6.6 \times 10^{-5}\ M](https://tex.z-dn.net/?f=%5BNO%5D_%7Beq%7D%3D6.6%20%5Ctimes%2010%5E%7B-5%7D%5C%20M)
![[O_2]_{eq}=3.3\times 10^{-5} = 3.3\times 10^{-5}\ M](https://tex.z-dn.net/?f=%5BO_2%5D_%7Beq%7D%3D3.3%5Ctimes%2010%5E%7B-5%7D%20%3D%203.3%5Ctimes%2010%5E%7B-5%7D%5C%20M)
Expression for equilibrium constant is as follows:
![K=\frac{[NO]^2[O_2]}{[NO_2]^2}](https://tex.z-dn.net/?f=K%3D%5Cfrac%7B%5BNO%5D%5E2%5BO_2%5D%7D%7B%5BNO_2%5D%5E2%7D)
Substitute the values in the above expression
![K=\frac{[6.6\times 10^{-5}]^2[3.3 \times 10^{-5}]}{[0.999934]^2} \\=1.79\times 10^{-14}](https://tex.z-dn.net/?f=K%3D%5Cfrac%7B%5B6.6%5Ctimes%2010%5E%7B-5%7D%5D%5E2%5B3.3%20%5Ctimes%2010%5E%7B-5%7D%5D%7D%7B%5B0.999934%5D%5E2%7D%20%5C%5C%3D1.79%5Ctimes%2010%5E%7B-14%7D)
Equilibrium constant for decomposition of is
Answer:
Groups of atoms that are added to carbon backbones and give them unique properties are known as <u>Functional Groups</u>.
Explanation:
In organic chemistry they are called as Functional Group because they are the active part of a molecule. These groups give a unique characteristic to molecule both chemically and physically. Also, each functional group represent a different class of compounds.
Examples:
S No. Functional Group Name
1 R--X Alkyl Halides
2 R--OH Alcohols
3 R--NH₂ Amines
4 R--O--R Ethers
5 R--CO--R Ketones
6 R--CO--H Aldehydes
7 R--CO--OH Carboxylic acids
8 R--CO--X Acid Halides
10 R--CO--NR₂ Acid Amides
11 R--CO-OR' Esters
