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3 years ago

What is the ph of a solution of 0.650 m k2hpo4, potassium hydrogen phosphate?

Chemistry

1 answer:

Pie3 years ago

3 0

phosphate?

At 25 ∘C phosphoric acid, H3PO4, has the following equilibrium constants:

Ka1= 7.5×10^-3

Ka2= 36.2×10^−8

Ka3 = 4.2×10^−13

This is a question about the dissociation of weak polyprotic acids.

When pKas of polyprotic intermediates have a difference of 2 or more you just average them using the equation: pH = (pKa2 + pKa3) / 2

pKa2 = -log(Ka2) ; pKa3 = -log(Ka3)

So, for this problem, REGARDLESS OF THE SUPPLIED CONCENTRATION , the answer is:

pH = (7.2076+12.3767) / 2

pH = 9.79

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Mercury’s atomic emission spectrum is shown below. Estimate the wavelength of the orange line. What is its frequency? What is th

lions [1.4K]

The wavelength of the orange line is 610 nm, the frequency of this emission is 4.92 x 10¹⁴ Hz and the energy of the emitted photon corresponding to this orange line is 3.26 x 10⁻¹⁹ J.

"Your question is not complete, it seems to be missing the diagram of the emission spectrum"

the diagram of the emission spectrum has been added.

From the given chart;

The wavelength of the atomic emission corresponding to the orange line is 610 nm = 610 x 10⁻⁹ m

The frequency of this emission is calculated as follows;

c = fλ

where;

c is the speed of light = 3 x 10⁸ m/s
f is the frequency of the wave
λ is the wavelength

$f = \frac{c}{\lambda } \\\\f = \frac{3\times 10^8}{610 \times 10^{-9}} \\\\f = 4.92 \times 10^{14} \ Hz$

The energy of the emitted photon corresponding to the orange line is calculated as follows;

E = hf

where;

h is Planck's constant = 6.626 x 10⁻³⁴ Js

E = (6.626 x 10⁻³⁴) x (4.92 x 10¹⁴)

E = 3.26 x 10⁻¹⁹ J.

Thus, the wavelength of the orange line is 610 nm, the frequency of this emission is 4.92 x 10¹⁴ Hz and the energy of the emitted photon corresponding to this orange line is 3.26 x 10⁻¹⁹ J.

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6 0

1 year ago

Why is Cellular Respiration so important?

Nesterboy [21]

I say it’s A. Hopefully

5 0

3 years ago

g If 50.0 mL of a 0.75 M acetic acid solution is titrated with 1.0 M sodium hydroxide, what is the pH after 10.0 mL of NaOH have

V125BC [204]

Answer:

pH = 2.66

Explanation:

Acetic Acid + NaOH → Sodium Acetate + H₂O

First we calculate the number of moles of each reactant, using the given volumes and concentrations:

0.75 M Acetic acid * 50.0 mL = 37.5 mmol acetic acid

1.0 M NaOH * 10.0 mL = 10 mmol NaOH

We calculate how many acetic acid moles remain after the reaction:

37.5 mmol - 10 mmol = 27.5 mmol acetic acid

We now calculate the molar concentration of acetic acid after the reaction:

27.5 mmol / (50.0 mL + 10.0 mL) = 0.458 M

Then we calculate [H⁺], using the following formula for weak acid solutions:

[H⁺] = $\sqrt{C*Ka}=\sqrt{0.458M*1.76x10^{-5}}$

[H⁺] = 0.0028

Finally we calculate the pH:

pH = -log[H⁺]

pH = 2.66

8 0

2 years ago

If 5.100 g of c6h6 is burned and the heat produced from the burning is added to 5691 g of water at 21 °c, what is the final temp

emmasim [6.3K]

C6H12 = 6x12 + 6x1 = 78.
The equation indicates that 2x78 = 156g benzene will produce 6542kJ.
Using proportions you can then calculate that
x/6542kJ = 7.9g / 156g
x = 331.3kJ = 331300J.

heat = mass x ΔT x 4.18J/g°
ΔT = 331300J / (5691g x 4.18J/g°) = 13.9°

final temp = 21 + 14° = 35°C

4 0

3 years ago

Please help me with this

Zina [86]

Answer:

Explanation:

it breaks down a simple sugar into a type of energy their cells can use

6 0

3 years ago

Read 2 more answers