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ehidna [41]
3 years ago
8

Determine the acid dissociation constant for a 0.020 m formic acid solution that has a ph of 2.74. formic acid is a weak monopro

tic acid and the equilibrium equation of interest is hcooh(aq) + h2o(l) ⇌ h3o+(aq) + hco2-(aq). enter your answer in exponential (e) format (sample 1.23e-4) with two decimal places and without units.
Chemistry
1 answer:
Setler [38]3 years ago
7 0

Answer:

Ka = 1.82x10⁻⁴

Explanation:

To solve this problem, we need a basis. In this case, we need the overall reaction and a ICE chart. So, let's write the overall reaction first:

HCOOH + H₂O <--------> H₃O⁺ + HCOO⁻    Ka = ?

Now that we have the overall reaction, we need to write the ICE chart. In this way we can determine what data do we have, and what do we have left to determine:

      HCOOH + H₂O <--------> H₃O⁺ + HCOO⁻    Ka = ?

i)        0.02                                0             0

e)       0.02-x                             x              x

The Ka expression is:

Ka = [H₃O⁺] [HCOO⁻] / [HCOOH]

Replacing the given data we have:

Ka = x² / 0.02 - x

Now, the value of x can be calculated because we already have the pH of the formic acid, and with it, we can calculate the [H₃O⁺] with the following expression:

[H₃O⁺] = 10^(-pH)

Replacing we have:

[H₃O⁺] = 10^(-2.74) = 1.82x10⁻³ M

This is the value of x, so replacing in the Ka expression, we can calculate then, the value of Ka:

Ka = (1.82x10⁻³)² / (0.02 - 1.82x10⁻³)

<h2>Ka = 1.82x10⁻⁴</h2>
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solniwko [45]

Answer:

The correct option is C

Explanation:

From the question we are told that

The reaction is

C_{16}H_{32}O_2(g) + 23O_2(g) \to 16 CO_2(g) + 16 H_2O(l)

Generally \Delta  H  =  \Delta  U + \Delta N*  RT

Here \Delta  H is the change in enthalpy

\Delta  U is the change in the internal energy

              \Delta N  is the difference between that number of moles of product and the number of moles of reactant

Looking at the reaction we can discover that the elements that was consumed and the element that was formed is O_2 and  CO_2 and this are both gases so the change would occur in the number of moles

So  

\Delta  H  =  \Delta  U + [16 -23]*  RT

\Delta  H  =  \Delta  U -7RT

The  negative sign in the equation tell us that the enthalpy\Delta_r H would be less than the Internal energy \Delta_r U

4 0
3 years ago
How many grams of Mg(NO3)2 would be produced? please show how you got the answer​
Len [333]

Answer:

148 grams of relative atomic mass

Explanation:

magnesium atomic mass : 24

nitrogen : 14

oxygen : 16

24 × 1

14 × 1 × 2

16 × 3 × 2

24 + 28 + 80 = 148 grams

7 0
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MrRissso [65]
The Krebs cycle uses acetyl CoA as a reactant.
4 0
3 years ago
Calculate the ph of a 0.40 m solution of sodium benzoate (nac6h5coo) given that the ka of benzoic acid (c6h5cooh) is 6.50 x 10-5
asambeis [7]
Hello!

The dissociation reaction for Benzoic Acid is the following:

C₆H₅COOH + H₂O ⇄ C₆H₅COO⁻ + H₃O⁺

The Ka expression is the following and we clear for the concentration of H₃O⁺(X) assuming that the dissociation is little so we can rule it out in the denominator of the equation:

Ka= \frac{[C_6H_5COO^{-}]*[H_3O^{+}] }{[C_6H_5COOH]}=\frac{X*X }{0,40 -X}  (assume: 0,40-X\approx0,40)\\  \\ X= \sqrt{0,40*6,50*10^{-5} }=0.00510M \\  \\ pH=-log([H_3O^{+}]=2,29

So, the pH of this Benzoic Acid solution is 2,29

Have a nice day!


3 0
3 years ago
Photoelectrons are removed with kinetic energy 1.864×10^-21 J, when photons of light with energy 4.23×10^-19 J fall on the metal
tensa zangetsu [6.8K]
The minimum energy required to remove an electron from a potassium metal can be obtained by subtracting the energy of the incident photons from the kinetic energy of the removed photoelectrons. Based from the given values, the following equation is obtained:

Minimum energy required = 4.23×10^-19 J - <span>1.864×10^-21 J
</span>
We then get 4.2114 x 10^-19 J as the minimum energy required to remove the electron. We then convert this into units of energy per mole. This is to be done by using Avogadro's number which result to the following equation:

Minimum energy required per mole = 4.2114 x 10^-19 J x 6.022 x 10^23 mol^-1

The final answer is then 253.608 kJ/mol
6 0
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