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2 years ago

Write the formula of the conjugate acid of HCO₂⁻.

Chemistry

1 answer:

Alex2 years ago

7 0

Taking into account the Brønsted-Lowry acid-base theory, the conjugate acid of HCO₂⁻ is H₂CO₂.

<h3>Brønsted-Lowry acid-base </h3>

The Brønsted-Lowry acid-base theory (or the Brønsted-Lowry theory) identifies acids and bases based on whether the species accepts or donates protons or H⁺.

According to this theory, acids are proton donors while bases are proton acceptors. That is, an acid is a species that donates an H⁺ proton while a base is a chemical species that accepts an H⁺ proton from the acid.

So, reactions between acids and bases are H⁺ proton transfer reactions.

<h3>Conjugate base and conjugate acid</h3>

Then, a conjugate base is an ion or molecule resulting from the acid that loses the proton, while a conjugate acid is an ion or molecule resulting from the base that gains the proton:

acid + base ⇄ conjugate base + conjugate acid

<h3>Conjugate acid of HCO₂⁻</h3>

Like a conjugate acid is an ion or molecule resulting from the base that gains the proton, the conjugate acid of HCO₂⁻ is H₂CO₂.

Learn more about the Brønsted-Lowry acid-base theory:

<u>brainly.com/question/12916250?referrer=searchResults</u>

<u>brainly.com/question/1191429?referrer=searchResults</u>

<u>brainly.com/question/4000152?referrer=searchResults</u>

<u>brainly.com/question/12808135?referrer=searchResults</u>

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Given the following information, what is the concentration of H2O(g) at equilibrium? [H2S](eq) = 0.671 M [O2](eq) = 0.587 M Kc =

MAVERICK [17]

<u>Answer:</u> The equilibrium concentration of water is 0.597 M

<u>Explanation:</u>

Equilibrium constant in terms of concentration is defined as the ratio of concentration of products to the concentration of reactants each raised to the power their stoichiometric ratios. It is expressed as $K_{c}$

For a general chemical reaction:

$aA+bB\rightleftharpoons cC+dD$

The expression for $K_{eq}$ is written as:

$K_{c}=\frac{[C]^c[D]^d}{[A]^a[B]^b}$

The concentration of pure solids and pure liquids are taken as 1 in the expression.

For the given chemical reaction:

$2H_2S(g)+O_2(g)\rightleftharpoons 2S(s)+2H_2O(g)$

The expression of $K_c$ for above equation is:

$K_c=\frac{[H_2O]^2}{[H_2S]^2\times [O_2]}$

We are given:

$[H_2S]_{eq}=0.671M$

$[O_2]_{eq}=0.587M$

$K_c=1.35$

Putting values in above expression, we get:

$1.35=\frac{[H_2O]^2}{(0.671)^2\times 0.587}$

$[H_2O]=\sqrt{(1.35\times 0.671\times 0.671\times 0.587)}=0.597M$

Hence, the equilibrium concentration of water is 0.597 M

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3 years ago

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almond37 [142]

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2 years ago

What is it called when moon is between setting and rising

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3 years ago

We can also perform a similar calculation for the mass defect and binding energy for nuclear reactions using the masses of the a

sukhopar [10]

Answer:

See Explanation

Explanation:

$\frac{235}{92} U + \frac{1}{0} n ---->\frac{137}{52} Te + \frac{97}{40} Zr +2\frac{1}{0} n$

Hence the mass defect is;

[235.04393 + 1.00867] - [ 136.92532 + 96.91095 + 2(1.00867)]

= 236.0526 - 235.85361

= 0.19899 amu

Since 1 amu = 1.66 * 10^-27 Kg

0.19899 amu = 0.19899 * 1.66 * 10^-27 = 3.3 * 10^-28 Kg

Binding energy = Δmc^2

Binding energy = 3.3 * 10^-28 Kg * (3 * 10^8)^2 = 2.97 * 10^-11 J

ii) $\frac{10}{5}B + \frac{1}{0}n-----> \frac{7}{3} Li + \frac{4}{2} He + Energy$

Hence the mass defect is;

[10.01294 + 1.00867] - [7.01600 + 4.00260]

= 11.02161 - 11.0186

= 0.00301 amu

Since 1 amu = 1.66 * 10^-27 Kg

0.00301 amu = 0.00301 * 1.66 * 10^-27 = 4.997 * 10^-30 Kg

Binding energy = Δmc^2

Binding energy = 4.997 * 10^-30 Kg * (3 * 10^8)^2 = 4.5 * 10^-13 J

7 0

2 years ago

A 200.0mL closed flask contains 2.000mol of carbon monoxide gas and 2.000mol of oxygen gas at the temperature of 300.0K. How man

max2010maxim [7]

Answer:

There will react 0.400 moles of oxygen.

Explanation:

<u>Step 1:</u> Data given

Volume of the closed flask = 200.00 mL = 0.2 L

Number of moles of CO = 2.000 mol

Number of moles of O2 = 2.000 mol

Temperature = 300.0 K

Pressure decreases with 10%

<u>Step 2</u>: The balanced equation

2CO(g)+O2(g)⟶2CO2(g)

<u>Step 3</u>: Calculate the initial pressure of the flask before the reaction

P = nRT/V

⇒ with n = the number of moles (2.000 moles CO + 2.000 moles O2 = 4.000 moles)

⇒ R is gas constant (0.08206 atm*L/mol*K)

⇒T = the temperature = 300.0K

⇒ V = the volume = 200.0 mL = 0.2 L

P = (4 * 0.08206*300)/0.2

P = 492.36 atm

<u>Step 4:</u> When the pressure is 10 % decreased:

The final pressure = 492.36 - 49.236 = 443.124 atm

<u>Step 5:</u> Calculate the number of moles

n = PV/RT

⇒ with n = the number of moles

⇒ with P = the pressure = 443.124 atm

⇒ V = the volume = 200.0 mL = 0.2 L

⇒ R is gas constant (0.08206 atm*L/mol*K)

⇒T = the temperature = 300.0K

n =(443.124*0.2)/(0.08206*300)

n = 3.6 moles = total number of moles

<u>Step 6:</u> Calculate number of moles

For the reaction :2CO(g) + O₂(g) ⟶ 2CO₂(g)

For each mole of O2 we have 2 moles of CO, to produce 2 moles of CO2

Moles CO = (2 -2X) moles

Moles O2 = (2-X) moles

Moles CO2 = 2X

The total number of moles (4 -X)= 3.6 moles

Where X are moles that react

X = 0.400 moles

There will react 0.400 moles of oxygen.

6 0

3 years ago