Answer:
a. add a little distilled water to see which layer the water adds to
Explanation:
The problem tells us to keep in mind two major aspects of the test: It has to be <em>simple</em>, as well as <em>non-destructive</em>:
- Adding distilled water can be made in under a minute, without requiring specialized laboratory equipment, unlike IR.
- It is also non-destructive, because the contents on either layer won't change due to the added distilled water.
Answer:
b. oxygen side being slightly negative and the hydrogen side being slightly positive.
Explanation:
The water molecule is a polar molecule, that is to say that its distribution of electronic density is different throughout the molecule.
In this way, in the water molecule there is a negative partial charge towards the oxygen atom and a positive partial charge towards the hydrogen atom.
This polar characteristic of the water molecule allows ions and other molecules to exhibit water solubility and is widely used in chemical reactions.
Answer:
isopropyl benzene (cumene)
Explanation:
The reaction of isopropyl chloride and AlCl3 and benzene belongs to the class of organic reactions known as the Friedel Kraft alkylation.
The mechanism of the reaction involves the formation of a tetrahedral complex [AlCl4]^-.
The electrophile now is the isopropyl group which attacks the benzene to yield the product.
The balanced equation between NaOH and H₂SO₄ is as follows
2NaOH + H₂SO₄ ---> Na₂SO₄ + 2H₂O
stoichiometry of NaOH to H₂SO₄ is 2:1
number of moles of NaOH moles reacted = molarity of NaOH x volume
number of NaOH moles = 0.08964 mol/L x 27.86 x 10⁻³ L = 2.497 x 10⁻³ mol
according to molar ratio of 2:1
2 mol of NaOH reacts with 1 mol of H₂SO₄
therefore 2.497 x 10⁻³ mol of NaOH reacts with - 1/2 x 2.497 x 10⁻³ mol of H₂SO₄
number of moles of H₂SO₄ reacted - 1.249 x 10⁻³ mol
Number of H₂SO₄ moles in 34.53 mL - 1.249 x 10⁻³ mol
number of H₂SO₄ moles in 1000 mL - 1.249 x 10⁻³ mol / 34.53 x 10⁻³ L = 0.03617 mol
molarity of H₂SO₄ is 0.03617 M
Properties of a solution that depend only on the ratio of the number of particles of solute and solvent in the solution are known as colligative properties. For this problem, we use boiling point elevation concept.
ΔT(boiling point) = (Kb)mi
ΔT(boiling point) = (0.51 C-kg / mol )(4.0 mol / 2.05 kg ) (2)
ΔT(boiling point) = 1.99 C
T(boiling point) = 101.99 C
