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hichkok12 [17]
3 years ago
14

Which member of each pair is the stronger base? part a ethylamine or aniline ethylamine or aniline ethylamine aniline request an

swer part b ethylamine or ethoxide ion ethylamine or ethoxide ion ethylamine ethoxide ion?

Chemistry
1 answer:
Vladimir79 [104]3 years ago
5 0
Lowery-Bronsted theory is applied here. Acc. to this theory Base accepts protons and Acids donate proton.

Part 1: 
Aniline is less basic than ethylamine because the lone pair on nitrogen (which accepts proton) is not localized. It resonates throughout the conjugated system of phenyl ring. Hence due to unavailability of electrons for accepting proton it is less basic compare to ethylamine. In ethyl amine the lone pair of electron is localized and available to abstract proton.

Part 2:
In this case the alkyl groups attached to -NH₂ (in ethylamine) and -O⁻ (in ethoxide are same (i.e. CH₃-CH₂-). Ethoxide is more basic than ethylamine because ethoxide is a conjugate base of ethanol (pKa value of ethanol = 15.9 very weak acid) and the conjugate base of weak acid is always a strong base. Secondly, the oxygen atom more Electronegative than Nitrogen atom can attract more electron cloud from alkyl group as compared to Nitrogen in ethylamine. Hence, oxygen in ethoxide attains greater electron cloud than the nitrogen in ethylamine. Therefore, it is more basic than ethylamine.

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Where does regional metamorphism usually occur ?
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Answer:

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Hope this Helps!

3 0
3 years ago
50cm3 of 1 mol/dm3 HCl at 30°C was mixed with 50cm3 of 1mol/dm3 NaOH at 30°C in a styrofoam calorimeter. The temperature of the
trapecia [35]

Answer:

-21 kJ·mol⁻¹  

Explanation:

Data:

                    H₃O⁺ +  OH⁻ ⟶ 2H₂O

       V/mL:    50         50  

c/mol·dm⁻³:   1.0         1.0

     

ΔT = 4.5 °C  

       C = 4.184 J·°C⁻¹g⁻¹

C_cal = 50 J·°C⁻¹

Calculations:

(a) Moles of acid

\text{Moles of acid} = \text{0.050 dm}^{3} \times \dfrac{\text{1.0 mol}}{\text{1 dm}^{3}} = \text{0.050 mol}\\\\\text{Moles of base} = \text{0.050 dm}^{3} \times \dfrac{\text{1.0 mol}}{\text{1 dm}^{3}} = \text{0.050 mol}

So, we have 0.050 mol of reaction

(b) Volume of solution

V = 50 dm³ + 50 dm³ = 100 dm³

(c) Mass of solution

\text{Mass of solution} = \text{100 dm}^{3} \times \dfrac{\text{1.00 g}}{\text{1 dm}^{3}} = \text{100 g}

(d) Calorimetry

There are three energy flows in this reaction.

q₁ = heat from reaction

q₂ = heat to warm the water

q₃ = heat to warm the calorimeter

q₁ + q₂ + q₃ = 0

     nΔH   +         mCΔT       + C_calΔT = 0

0.050ΔH + 100×4.184×4.5 +   50×4.5  = 0

0.050ΔH +          1883        +      225    = 0

                                  0.050ΔH + 2108 = 0

                                              0.050ΔH = -2108

                                                        ΔH = -2108/0.0500

                                                              = -42 000 J/mol

                                                              = -42 kJ/mol

This is the heat of reaction for the formation of 2 mol of water

The heat of reaction for the formation of mol of water is -21 kJ·mol⁻¹.

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3 years ago
1. Give the similarity (or similarities) between Cl and CI^(-1).
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Answer:

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2) both are the same element.

Explanation:

8 0
3 years ago
Is the formation of sulfur tetrafluoride monoxide endothermic or exothermic?
sleet_krkn [62]

endothermic means it absorbs energy while exothermic means it releases energy. remember this.

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2 years ago
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