Answer:
grams
Explanation:
Complete question is
You are asked to pre pare a pH = 4.00 buffer starting from 1.50 L of 0.0200 M solution of benzoic acid
(C6H5COOH)
and any amount you need of sodium benzoate
(C6H5COONa)
How many grams of sodium benzoate should be added to prepare the buffer? Neglect the small volume change that occurs when the sodium benzoate is added.
Solution
Given
pH of the buffer solution ![= 4](https://tex.z-dn.net/?f=%3D%204)
Concentration of C6H5COOH
M
Volume of the buffer solution
L
value for benzoic acid is ![6.3 * 10^ {-5}](https://tex.z-dn.net/?f=6.3%20%2A%2010%5E%20%7B-5%7D)
Concentration of sodium benzoate
![pH = - log Ka + log \frac{C6H5COONa}{C6H5COOH}](https://tex.z-dn.net/?f=pH%20%3D%20-%20log%20Ka%20%2B%20log%20%5Cfrac%7BC6H5COONa%7D%7BC6H5COOH%7D)
Substituting the given values we get
![log \frac{C6H5COONa}{C6H5COOH} = 4 + log (6.3 * 10^ {-5})\\log \frac{C6H5COONa}{C6H5COOH} = -0.20\\\frac{C6H5COONa}{C6H5COOH} = 10^{-0.2}\\{C6H5COONa} = 10^{-0.2} * 0.02\\{C6H5COONa} = 0.63 * 0.02\\{C6H5COONa} = 0.0126 M](https://tex.z-dn.net/?f=log%20%5Cfrac%7BC6H5COONa%7D%7BC6H5COOH%7D%20%20%3D%204%20%2B%20log%20%286.3%20%2A%2010%5E%20%7B-5%7D%29%5C%5Clog%20%5Cfrac%7BC6H5COONa%7D%7BC6H5COOH%7D%20%20%3D%20-0.20%5C%5C%5Cfrac%7BC6H5COONa%7D%7BC6H5COOH%7D%20%3D%2010%5E%7B-0.2%7D%5C%5C%7BC6H5COONa%7D%20%3D%2010%5E%7B-0.2%7D%20%2A%200.02%5C%5C%7BC6H5COONa%7D%20%3D%20%200.63%20%2A%200.02%5C%5C%7BC6H5COONa%7D%20%3D%20%200.0126%20M)
Number of moles in sodium benzoate
![= 0.0126 * 1.5 \\= 0.0189 Mol](https://tex.z-dn.net/?f=%3D%200.0126%20%2A%201.5%20%5C%5C%3D%200.0189%20Mol)
Mass of sodium benzoate
![0.0189 mol * 144.147 \frac{g}{mol} \\= 2.72 g](https://tex.z-dn.net/?f=0.0189%20mol%20%2A%20144.147%20%5Cfrac%7Bg%7D%7Bmol%7D%20%5C%5C%3D%202.72%20g)
The answer is A. An sp3 hybridized atom will have a total of 4 single bonds and lone pairs. Since carbon here is attached to four hydrogens, it must be sp3 hybridized. The C-H bonds are also the least polar bonds here, even though there is a slight amount of electronegativity difference between the two.
The atomic number is the number of protons meaning that there are 14 protons in an atom of silicon. The mass number is the sum of the protons and neutrons. 28 - 14 = 14. There are 14 protons and 14 neutrons in silicon.
Answer:
EMPIRICAL FORMULA:
C₂H₃O₂
Explanation:
Mass of compound = 4.647 g
Mass of CO₂ = 8.635 g
Mass of H₂O = 1.767 g
Empirical formula = ?
Solution:
Percentage of C = 8.635/ 4.647 × 12/44 ×100
Percentage of C = 1.86× 12/44 ×100 = 50.7
Percentage of H = 1.767/4.647 × 2/ 18 × 100
Percentage of H = 0.38 × 2/ 18 × 100
Percentage of H = 4.18
Percentage of O = 100 - (50.7+4.18)
Percentage of O = 100 - 54.88
Percentage of O = 45.12
Number of grams atom:
Number of grams atom of C = 50.7/ 12 = 4.23
Number of grams atom of H = 4.18 /1 = 4.18
Number of grams atom of O = 45.12/ 16 = 2.82
Atomic ratio:
C : H : O
4.23/2.82 : 4.18/2.82 : 2.82/2.82
1 : 1.5 : 1
C : H : O (1 : 1.5 : 1)
C : H : O 2(1 : 1.5 : 1)
C : H : O (2 : 3 : 2)
EMPIRICAL FORMULA:
C₂H₃O₂