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2 years ago

What substance is acting as the Brønsted-Lowry base in the forward reaction below? H2O + HCI ----> H3O+ CI-

Chemistry

1 answer:

yarga [219]2 years ago

6 0

H2O is the Bronsted-Lowry base because it accepts the hydrogen ion to become H3O after the reaction is complete.

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miss Akunina [59]

Answer:

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2 years ago

Read 2 more answers

EDTA EDTA is a hexaprotic system with the p K a pKa values: p K a1 = 0.00 pKa1=0.00 , p K a2 = 1.50 pKa2=1.50 , p K a3 = 2.00 pK

mihalych1998 [28]

Answer:

Check the explanation

Explanation:

When,

pH = -log[H+] = 3.30

[H+] = $5.0 X 10^{-4} M$

$Ka1 = 1 ; Ka2 = 0.0316 ; Ka3 = 0.01 ; Ka4 = 0.002 ; Ka5 = 7.4 X 10^{-7} ; Ka6 = 4.3 X 10^{-11}$

$alpha[Y^-4] = [H+]^6 + Ka1[H+]^5 + Ka1Ka2[H+]^4 + Ka1Ka2Ka3[H+]^3 + Ka1Ka2Ka3Ka4[H+]^2 + Ka1Ka2Ka3Ka4Ka5[H+] + Ka1Ka2Ka3Ka4Ka5Ka6$

= $1.56 X 10^{-20} + 3.12 X 10^{-17} + 2 X 10^{-15} + 4 X 10^{-14} + 1.6 X 10^{-13} + 2.34 X 10^{-16} + 2 X 10^{-23}$

= $2.02 X 10^{-13}$

When,

pH = -log[H+] = 10.15

[H+] = $7.08 X 10^{-11} M$

Ka1 = 1 ; Ka2 = 0.0316 ; Ka3 = 0.01 ; Ka4 = 0.002 ; Ka5 = $7.4 X 10^{-7}$ ; Ka6 = $4.3 X 10^-11$

$alpha[Y^{-4}] = [H+]^6 + Ka1[H+]^5 + Ka1Ka2[H+]^4 + Ka1Ka2Ka3[H+]^3 + Ka1Ka2Ka3Ka4[H+]^2 + Ka1Ka2Ka3Ka4Ka5[H+] + Ka1Ka2Ka3Ka4Ka5Ka6$

= $1.26 X 10^{-61} + 1.8 X 10^{-51} + 8.1 X 10^{-43} + 1.12 X 10^{-34} + 3.17 X 10^{-27} + 3.3 X 10^{-23} + 1.83 X 10^{-23}$

= $5.12 X 10^{-23}$

4 0

2 years ago

Which of the following is an example of a chemical change? *

Norma-Jean [14]

Answer:

hydrogen and oxygen to form water

3 0

2 years ago

Identifica los dos pares conjugados acido-base en cada una de las siguientes reacciones

aliya0001 [1]

Answer:

PULLIN OUT THE COUPE AT THE LOT?

Explanation:

TOLD EM F*&*& 12 F*(&*(& SWAT

8 0

2 years ago

How many grams of water can I convert from a liquid to a gas with 6768 joules?

vredina [299]

Answer:

The amount of water converted from liquid to gas with 6,768 joules is approximately 3.035 g

Explanation:

The amount of heat required to convert a given amount of liquid to gas at its boiling point is known as the latent heat of evaporation of the liquid

The latent heat of evaporation of water, Δ $H_v$ ≈ 2,230 J/g

The relationship between the heat supplied, 'Q', and the amount of water in grams, 'm', evaporated is given as follows

Q = m × Δ $H_v$

Therefore, the amount of water, 'm', converted from liquid to gas at the boiling point temperature (100°C), when Q = 6,768 Joules, is given as follows;

6,768 J = m × 2,230 J/g

∴ m = 6,768 J /(2,230 J/g) ≈ 3.035 g

The amount of water converted from liquid to gas with 6,768 joules = m ≈ 3.035 g.

8 0

2 years ago