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Mekhanik [1.2K]
3 years ago
11

Determine the acid dissociation constant for a 0.0250 M weak acid solution that has a pH of 2.37 . The equilibrium equation of i

nterest is​
Chemistry
1 answer:
Oksanka [162]3 years ago
3 0

Answer:

For part (a): pHsol=2.22

Explanation:

I will show you how to solve part (a), so that you can use this example to solve part (b) on your own.

So, you're dealing with formic acid, HCOOH, a weak acid that does not dissociate completely in aqueous solution. This means that an equilibrium will be established between the unionized and ionized forms of the acid.

You can use an ICE table and the initial concentration ofthe acid to determine the concentrations of the conjugate base and of the hydronium ions tha are produced when the acid ionizes

HCOOH(aq]+H2O(l]⇌ HCOO−(aq] + H3O+(aq]

I 0.20 0 0

C (−x) (+x) (+x)

E (0.20−x) x x

You need to use the acid's pKa to determine its acid dissociation constant, Ka, which is equal to

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Calculate the mass defect for the formation of phosphorus-31. The mass of a phosphorus-31 nucleus is 30.973765 amu. The masses o
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<u>Answer:</u> The mass defect for the formation of phosphorus-31 is 0.27399

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Mass defect is defined as the difference in the mass of an isotope and its mass number.

The equation used to calculate mass defect follows:

\Delta m=[(n_p\times m_p)+(n_n\times m_n)]-M

where,

n_p = number of protons

m_p = mass of one proton

n_n = number of neutrons

m_n = mass of one neutron

M = mass number of element

We are given:

An isotope of phosphorus which is _{15}^{31}\textrm{P}

Number of protons = atomic number = 15

Number of neutrons = Mass number - atomic number = 31 - 15 = 16

Mass of proton = 1.00728 amu

Mass of neutron = 1.00866 amu

Mass number of phosphorus = 30.973765 amu

Putting values in above equation, we get:

\Delta m=[(15\times 1.00728)+(16\times 1.00866)]-30.973765\\\\\Delta m=0.27399

Hence, the mass defect for the formation of phosphorus-31 is 0.27399

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the equation of the parabola is given as y = 1/4 (x-6)^2 +1. D

<h3>How to find the parabola</h3>

Given the focus(6,2) and the directrix, y = 0

Use the formula

\sqrt{(x-6)^{2} + (y-2)^{2} } } = (y - 0)

Find the square of both sides, square root is removed from the side with it and square added on the other side

(x-6)^2 + (y-2)^2 = (y- 0)^{2}

Expand the expression and bring all terms to one side

x^{2} - 6x - 6x + 36 +y^{2} -2y- 2y +4 = y^{2} - 0

Collect like terms

x^{2} - 12x -4y + 40 = 0

Make 'y' the subject of the formula

4y = x^{2} - 12x + 40

Divide through by 4 and substrate 4 from 40 to give a perfect quadratic equation

y = 1/4 x^{2} - 12x + 36 + 4

Divide factor by 4 to give 1

y = 1/4( x^{2} -6x -6x +36) + 1

Simplify the expanded quadratic equation

y = x (x -6) - 6 (x- 6), we have (x - 6) ^2

Then insert in place into previous equation

y = 1/4 (x-6)^2 + 1

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