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Mila [183]
3 years ago
15

Use the drop-down menus to select the correct name for each of the organic compounds.

Chemistry
2 answers:
irakobra [83]3 years ago
6 0

Answer:

edge 2020

Explanation:

Molodets [167]3 years ago
5 0

Answer:

Here's what I get  

Explanation:

CH₃CH₂CH₂CH₂CH₂CH₃ —  hexane

CH₂=CHCH₂CH₂CH₂CH₃ —  hex-1-ene is the preferred IUPAC name (PIN). 1-Hexene is accepted

CH₃C≡CCH₃ —  but-2-yne (PIN); 2-butyne is accepted

CH₃CH(CH₃)CH₂CH₂CH₃ — 2-methylpentane

CH₃CH₂CHCICH₂CH₃ — 3-chloropentane

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2 years ago
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What is this equation?
DIA [1.3K]

2C_{7}H_{10} + 19O_{2} → 14CO_{2} + 10H_{2}O

8 0
2 years ago
Complete the dissociation reaction and the corresponding Ka equilibrium expression for each of the following acids in water. (Ty
anastassius [24]

Answer :

(A) The dissociation reaction of HC_2H_3O_2 will be:

HC_2H_3O_2(aq)\rightleftharpoons H^+(aq)+C_2H_3O_2^-(aq)

The equilibrium expression :

K_a=\frac{[H^+][C_2H_3O_2^-]}{[HC_2H_3O_2]}

(B) The dissociation reaction of Co(H_2O)_6^{3+} will be:

Co(H_2O)_6^{3+}(aq)\rightleftharpoons H^+(aq)+Co(H_2O)_5(OH)^{2+}(aq)

The equilibrium expression :

K_a=\frac{[H^+][Co(H_2O)_5(OH)^{2+}]}{[Co(H_2O)_6^{3+}]}

(C) The dissociation reaction of CH_3NH_3^+ will be:

CH_3NH_3^+(aq)\rightleftharpoons H^+(aq)+CH_3NH_2(aq)

The equilibrium expression :

K_a=\frac{[H^+][CH_3NH_2]}{[CH_3NH_3^+]}

Explanation :

Equilibrium constant : It is defined as the equilibrium constant. It is defined as the ratio of concentration of products to the concentration of reactants.

The equilibrium expression for the reaction is determined by multiplying the concentrations of products and divided by the concentrations of the reactants and each concentration is raised to the power that is equal to the coefficient in the balanced reaction.

As we know that the concentrations of pure solids and liquids are constant that is they do not change. Thus, they are not included in the equilibrium expression.

(A) The dissociation reaction of HC_2H_3O_2 will be:

HC_2H_3O_2(aq)\rightleftharpoons H^+(aq)+C_2H_3O_2^-(aq)

The equilibrium expression of HC_2H_3O_2 will be:

K_a=\frac{[H^+][C_2H_3O_2^-]}{[HC_2H_3O_2]}

(B) The dissociation reaction of Co(H_2O)_6^{3+} will be:

Co(H_2O)_6^{3+}(aq)\rightleftharpoons H^+(aq)+Co(H_2O)_5(OH)^{2+}(aq)

The equilibrium expression of Co(H_2O)_6^{3+} will be:

K_a=\frac{[H^+][Co(H_2O)_5(OH)^{2+}]}{[Co(H_2O)_6^{3+}]}

(C) The dissociation reaction of CH_3NH_3^+ will be:

CH_3NH_3^+(aq)\rightleftharpoons H^+(aq)+CH_3NH_2(aq)

The equilibrium expression of CH_3NH_3^+ will be:

K_a=\frac{[H^+][CH_3NH_2]}{[CH_3NH_3^+]}

3 0
3 years ago
Calculate the standard entropy of vaporization of ethanol at its boiling point, 352 K. The standard molar enthalpy of vaporizati
Korvikt [17]

Answer : The correct option is, (b) +115 J/mol.K

Explanation :

Formula used :

\Delta S=\frac{\Delta H_{vap}}{T_b}

where,

\Delta S = change in entropy

\Delta H_{vap} = change in enthalpy of vaporization = 40.5 kJ/mol

T_b = boiling point temperature = 352 K

Now put all the given values in the above formula, we get:

\Delta S=\frac{\Delta H_{vap}}{T_b}

\Delta S=\frac{40.5kJ/mol}{352K}

\Delta S=115J/mol.K

Therefore, the standard entropy of vaporization of ethanol at its boiling point is +115 J/mol.K

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