Answer:
1.58x10⁻⁵
2.51x10⁻⁸
0.0126
63.10
Explanation:
Phenolphthalein acts like a weak acid, so in aqueous solution, it has an acid form HIn, and the conjugate base In-, and the pH of it can be calculated by the Handerson-Halsebach equation:
pH = pKa + log[In-]/[HIn]
pKa = -logKa, and Ka is the equilibrium constant of the dissociation of the acid. [X] is the concentrantion of X. Thus,
i) pH = 4.9
4.9 = 9.7 + log[In-]/[HIn]
log[In-]/[HIn] = - 4.8
[In-]/[HIn] = 
[In-]/[HIn] = 1.58x10⁻⁵
ii) pH = 2.1
2.1 = 9.7 + log[In-]/[HIn]
log[In-]/[HIn] = -7.6
[In-]/[HIn] = 
[In-]/[HIn] = 2.51x10⁻⁸
iii) pH = 7.8
7.8 = 9.7 + log[In-]/[HIn]
log[In-]/[HIn] = -1.9
[In-]/[HIn] = 
[In-]/[HIn] = 0.0126
iv) pH = 11.5
11.5 = 9.7 + log[In-]/[HIn]
log[In-]/[HIn] = 1.8
[In-]/[HIn] = 
[In-]/[HIn] = 63.10
Answer:
Polar covalent bond.
Explanation:
When the bond is formed between the atoms by sharing the electrons the bond thus have covalent character. The atom with larger electronegativity attract the electron pair more towards it self and becomes partial negative while the other atom becomes partial positive. When the electronegativity difference is less than 0.4 the bond is non polar covalent.
When bonded atoms have greater electronegativity difference i.e 2 or greater than two the bond is ionic because electron is transfer from low electronegative atom to highest electronegative atom.
For example:
In water the electronegativity of oxygen is 3.44 and hydrogen is 2.2. That's why electron pair attracted more towards oxygen, thus oxygen becomes partial negative and hydrogen becomes partial positive.
In case of H₂, Cl₂, Br₂ the bond has very high covalent character because of zero electronegativity difference.
Ethanol is: flammable, liquid at room temperature, the boiling point is 78.37 ° C.
The temperature of a liquid can exceed its boiling point. An example is water. Although at ordinary pressure of 1 atm, the boiling point is 100 degrees, water can still exist in higher temperatures but this time in another state. Superheated steam is the term used for water whose temperature has higher than the boiling point
Answer:
Second step: 4-bromo-1-methyl-2-nitrobenzene.
Third step: 1.5-dibromo-2-methyl-3-nitrobenzene.
Explanation:
To solve this exercise I will use the concepts of electrophilic substitution. In these reactions, a functional group is displaced by an electrophile. In the attached image are the two main products.
