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kolbaska11 [484]
3 years ago
6

A) For the following reaction, K > 1. Classify each of the reactants and products based on their strength as Bronsted-Lowry a

cids or bases.
C9H7N + HNO2Doublearrow.GIFC9H7NH+ + NO2-

a) HNO2 1) stronger acid

b) NO2- 2) weaker acid

c) C9H7NH+ 3) stronger base

d) C9H7N 4) weaker base

B)
For the following reaction, K < 1. Classify each of the reactants and products based on their strength as Bronsted-Lowry acids or bases.

C5H11NH+ + C6H5COO-Doublearrow.GIFC5H11N + C6H5COOH

a) C5H11NH+ 1) strongest acid

b) C6H5COO- 2) stongest base

c) C5H11N 3) weakest acid

d) C6H5COOH 4) weakest base
Chemistry
1 answer:
Aleks04 [339]3 years ago
4 0

Answer:

Explanation:

According to Bronsted-Lowry acids or base theory ,  the reagent capable of giving hydrogen ion or proton  will be acid and that which accepts hydrogen ion or proton  will be base .

C₉H₇N + HNO₂   ⇄    C₉H₇NH⁺ + NO₂⁻

If K > 1 , reaction is proceeding from left to right .

Hence HNO₂ is giving H⁺ or proton and C₉H₇N is accepting proton to form

C₉H₇NH⁺ .  

Hence HNO₂ is bronsted acid and C₉H₇N is bronsted base .

B )

when K < 1 , reaction above proceeds from right to left . That means

C₉H₇NH⁺ is giving H⁺ so it is a bronsted acid and NO₂⁻ is accepting H⁺ so it is a bronsted base .

Hence ,  NO₂⁻ is a bronsted base and C₉H₇NH⁺ is a bronsted acid .

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Find the amount of heat energy needed to convert 400 grams of ice at -38°C to steam at 160°C.
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The amount of heat energy needed to convert 400 g of ice at -38 °C to steam at 160 °C is 1.28×10⁶ J (Option D)

<h3>How to determine the heat required change the temperature from –38 °C to 0 °C </h3>
  • Mass (M) = 400 g = 400 / 1000 = 0.4 Kg
  • Initial temperature (T₁) = –25 °C
  • Final temperature (T₂) = 0 °
  • Change in temperature (ΔT) = 0 – (–38) = 38 °C
  • Specific heat capacity (C) = 2050 J/(kg·°C)
  • Heat (Q₁) =?

Q = MCΔT

Q₁ = 0.4 × 2050 × 38

Q₁ = 31160 J

<h3>How to determine the heat required to melt the ice at 0 °C</h3>
  • Mass (m) = 0.4 Kg
  • Latent heat of fusion (L) = 334 KJ/Kg = 334 × 1000 = 334000 J/Kg
  • Heat (Q₂) =?

Q = mL

Q₂ = 0.4 × 334000

Q₂ = 133600 J

<h3>How to determine the heat required to change the temperature from 0 °C to 100 °C </h3>
  • Mass (M) = 0.4 Kg
  • Initial temperature (T₁) = 0 °C
  • Final temperature (T₂) = 100 °C
  • Change in temperature (ΔT) = 100 – 0 = 100 °C
  • Specific heat capacity (C) = 4180 J/(kg·°C)
  • Heat (Q₃) =?

Q = MCΔT

Q₃ = 0.4 × 4180 × 100

Q₃ = 167200 J

<h3>How to determine the heat required to vaporize the water at 100 °C</h3>
  • Mass (m) = 0.4 Kg
  • Latent heat of vaporisation (Hv) = 2260 KJ/Kg = 2260 × 1000 = 2260000 J/Kg
  • Heat (Q₄) =?

Q = mHv

Q₄ = 0.4 × 2260000

Q₄ = 904000 J

<h3>How to determine the heat required to change the temperature from 100 °C to 160 °C </h3>
  • Mass (M) = 0.4 Kg
  • Initial temperature (T₁) = 100 °C
  • Final temperature (T₂) = 160 °C
  • Change in temperature (ΔT) = 160 – 100 = 60 °C
  • Specific heat capacity (C) = 1996 J/(kg·°C)
  • Heat (Q₅) =?

Q = MCΔT

Q₅ = 0.4 × 1996 × 60

Q₅ = 47904 J

<h3>How to determine the heat required to change the temperature from –38 °C to 160 °C</h3>
  • Heat for –38 °C to 0°C (Q₁) = 31160 J
  • Heat for melting (Q₂) = 133600 J
  • Heat for 0 °C to 100 °C (Q₃) = 167200 J
  • Heat for vaporization (Q₄) = 904000 J
  • Heat for 100 °C to 160 °C (Q₅) = 47904 J
  • Heat for –38 °C to 160 °C (Qₜ) =?

Qₜ = Q₁ + Q₂ + Q₃ + Q₄ + Q₅

Qₜ = 31160 + 133600 + 167200 + 904000 + 47904

Qₜ = 1.28×10⁶ J

Learn more about heat transfer:

brainly.com/question/10286596

#SPJ1

7 0
2 years ago
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