Answer:
Reaction follows frieldel-craft alkylation mechanism
Explanation:
- Reaction between benzene and chloroform in the presence of
follows friedel-craft alkylation of benzene - Firstly, one equivalent of benzene adds onto chloroform by replacing one equivalent of Cl from chloroform.
- Remaining two equivalent of Cl in chloroform are similarly replaced by two equivalent of benzene to produce triphenylmethane
- Reaction mechanism has been shown below
Answer:
Weaker
Explanation:
Since the charges on nucleus and electron are opposite in nature, an attractive force exists between them. It is true in generally, that when objects are made to move closer together in the direction of an attractive force, potential energy decreases (and increases whenever attracting objects are force to move apart).
Answer:
Electron transport produces 3 ATP molecule(s) per NADH molecule and 2 ATP molecules(s) perFADH 2 molecule.
Explanation:
The mechanism by which ATP is produced is explained by the theory of chemosmotic coupling.
This theory establishes that the synthesis of ATP in cellular respiration comes from an electrochemical gradient existing between the internal membrane and the space of the intermembrane of the mitochondria, through the use of the energy of NADH and FADH2 that have been formed by the rupture of molecules rich in energy, such as glucose.
Answer:
1. V₁ = 2.0 mL
2. V₁ = 2.5 mL
Explanation:
<em>You are provided with a stock solution with a concentration of 1.0 × 10⁻⁵ M. You will be using this to make two standard solutions via serial dilution.</em>
To calculate the volume required (V₁) in each dilution we will use the dilution rule.
C₁ . V₁ = C₂ . V₂
where,
C are the concentrations
V are the volumes
1 refers to the initial state
2 refers to the final state
<em>1. Perform calculations to determine the volume of the 1.0 × 10⁻⁵ M stock solution needed to prepare 10.0 mL of a 2.0 × 10⁻⁶ M solution.</em>
C₁ . V₁ = C₂ . V₂
(1.0 × 10⁻⁵ M) . V₁ = (2.0 × 10⁻⁶ M) . 10.0 mL
V₁ = 2.0 mL
<em>2. Perform calculations to determine the volume of the 2.0 × 10⁻⁶ M solution needed to prepare 10.0 mL of a 5.0 × 10⁻⁷ M solution.</em>
C₁ . V₁ = C₂ . V₂
(2.0 × 10⁻⁶ M) . V₁ = (5.0 × 10⁻⁷ M) . 10.0 mL
V₁ = 2.5 mL
