Answer: dispersion forces, dipole-dipole and hydrogen bonds
Explanation:
Answer:
It has been drawn and uploaded as an attachment. Please download it to see the structure.
Explanation:
The product formed as a result of the reaction of cyclohexene with H2 in presence of Pt (platinum) can be described as catalytic hydrogenation. Catalytic hydrogenation is defined as the process of hydrogen addition in the presence of a catalyst, which in this case is platinum.
Note that Cyclohexene (alkene) is a hydrocarbon molecule represented by the chemical formula, C6H10 .
It consists of a double bond. During the hydrogenation reaction, the alkene undergoes an addition reaction to give alkane which is a saturated hydrocarbon as the product.
The first step in order to derive the product is to draw the chemical structure of cyclohexene and identify the double bond present in it.
The final product can be derived by replacing the double bond with the single bond and satisfying all the valences of the carbon atom. The final product structure has been drawn and uploaded as an attachment. Please download it to see the structure.
Ans:
The structure of the cyclohexane thus, formed has been shown as follows with all the hydrogen atoms:
Answer:
- Alanine = 5.61 mmoles
- Leucine = 3.81 mmoles
- Tryptophan = 2.45 mmoles
- Cysteine = 4.13 mmoles
- Glutamic acid = 3.40 mmoles
Explanation:
Mass / Molar mass = Moles
Milimoles = Mol . 1000
500 mg / 1000 = 0.5 g
- Alanine = 0.5 g / 89 g/m → 5.61x10⁻³ moles . 1000 = 5.61mmoles
- Leucine = 0.5 g / 131 g/m → 3.81 x10⁻³ moles . 1000 = 3.81 mmoles
- Tryptophan = 0.5 g / 204 g/m → 2.45x10⁻³ moles . 1000 = 2.45 mmoles
- Cysteine = 0.5 g / 121 g/m → 4.13x10⁻³ moles . 1000 = 4.13 mmoles
- Glutamic acid = 0.5 g 147 g/m → 3.40x10⁻³ moles . 1000 = 3.4 mmoles
The standard formation equation for glucose C6H12O6(s) that corresponds to the standard enthalpy of formation or enthalpy change ΔH°f = -1273.3 kJ/mol is
C(s) + H2(g) + O2(g) → C6H12O6(s)
and the balanced chemical equation is
6C(s) + 6H2(g) + 3O2(g) → C6H12O6(s)
Using the equation for the standard enthalpy change of formation
ΔHoreaction = ∑ΔHof(products)−∑ΔHof(Reactants)
ΔHoreaction = ΔHfo[C6H12O6(s)] - {ΔHfo[C(s, graphite) + ΔHfo[H2(g)] + ΔHfo[O2(g)]}
C(s), H2(g), and O2(g) each have a standard enthalpy of formation equal to 0 since they are in their most stable forms:
ΔHoreaction = [1*-1273.3] - [(6*0) + (6*0) + (3*0)]
= -1273.3 - (0 + 0 + 0)
= -1273.3
Answer is: reducin agent in this reaction is thiosulfate (S₂O₃²⁻).
Balanced chemical reaction: 2S₂O₃²⁻(aq) + I₂(aq) → S₄O₆²⁻(aq) + 2I⁻<span>(aq).
Reducing agent is element or compound who loose electrons in chemical reaction. Sulfur in </span>thiosulfate change oxidation number from +2 to +5 tetrathionate anion (two<span> sulfur </span>atoms in the ion have oxidation state<span> 0 and two atoms have oxidation state +5).</span>
