This problem could be solved easily using the Henderson-Hasselbach equation used for preparing buffer solutions. The equation is written below:
pH = pKa + log[(salt/acid]
Where salt represents the molarity of salt (sodium lactate), while acid is the molarity of acid (lactic acid).
Moles of salt = 1 mol/L * 25 mL * 1 L/1000 mL = 0.025 moles salt
Moles of acid = 1 mol/L* 60 mL * 1 L/1000 mL = 0.06 moles acid
Total Volume = (25 mL + 60 mL)*(1 L/1000 mL) = 0.085 L
Molarity of salt = 0.025 mol/0.085 L = 0.29412 M
Molarity of acid = 0.06 mol/0.085 L = 0.70588 M
Thus,
pH = 3.86 + log(0.29412/0.70588)
pH = 3.48
Answer:
6.022 × 10²² atoms
Explanation:
Generally 1 mol of any element contains 6.02×10^23 atoms. The number 6.022 × 10²³ is known as Avogadro's number.
Mass of Aluminium = 2.70g
Molar mass = 27g/mol
Number of moles = Mass / Molar mass = 2.70 / 27 = 0.1 mol
1 mol = 6.022 × 10²³
0.1 mol = x
x = 6.022 × 10²³ * 0.1 = 6.022 × 10²² atoms
The hydrocarbon is used in excess.
<h3><u>Explanation</u>:</h3>
The bromination of an arene is not simple as bromination of an alkane. This is because the carbocation or free radicle formation in benzene is a very energy consuming process. This is why a lewis base like aluminium bromide or ferric bromide is used. The ferric bromide takes in the bromine radicle and forms the brominium cation which helps in the formation of electrophile. Now this electrophile brominium cation attacks the benzene ring and forms a temporary sp3 hybrid carbon intermediate. Then the hydrogen is taken by the FeBr4- forming HBr and regenerating the FeBr3 as well as Aromaticity of the arene species at the same time. Here hydrocarbon is used in excess just to prevent the chances of multiple substitution in the same arene molecule.
Answer:
it has made the study of chemistry systematic and easy. it acts as an aid to memory
Answer: 51.9961 g/mol, don't know if it helps :)
Explanation:
