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katovenus [111]

2 years ago

13

Assuming equal concentrations of conjugate base and acid, which one of the following mixtures is suitable for making a buffer so

lution with an optimum pH of 9.2−9.3? CH3COONa/CH3COOH (Ka = 1.8 x 10^−5) NH3/NH4Cl (Ka = 5.6 x 10^−10) NaOCl/HOCl (Ka = 3.2 x 10^−8) NaNO2/HNO2 (Ka = 4.5 x 10^−4) NaCl/HCl

1 answer:

BartSMP [9]2 years ago

4 0

Answer:

NH₃/NH₄Cl

Explanation:

We can calculate the pH of a buffer using the Henderson-Hasselbalch's equation.

$pH=pKa+log\frac{[base]}{[acid]}$

If the concentration of the acid is equal to that of the base, the pH will be equal to the pKa of the buffer. The optimum range of work of pH is pKa ± 1.

Let's consider the following buffers and their pKa.

CH₃COONa/CH3COOH (pKa = 4.74)

NH₃/NH₄Cl (pKa = 9.25)

NaOCl/HOCl (pKa = 7.49)

NaNO₂/HNO₂ (pKa = 3.35)

NaCl/HCl Not a buffer

The optimum buffer is NH₃/NH₄Cl.

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Explain the effects of nh3 and hcl on the cuso4 solution in terms of le chatelier's principle

Fittoniya [83]

The Principle of Le Chatelier states that if a system in equilibrium is subjected to a disturbance, the system will react in such a way that it will diminish the effect of that disturbance. Thus, when the concentration of one of the substances in an equilibrium system is changed, the equilibrium varies in such a way that it can compensate for this change.

For example, if the concentration of one of the reactants is increased, the equilibrium shifts to the right or to the side of the products. Also, if you add more reagents, the reaction will move even more to the right until the balance is re-established again, increasing the quantity of products.

In this way, adding HCl to a solution of CuSO4 will produce the following reaction:

CuSO4 (aq) + 2HCl (aq) ⇔ CuCl2 (aq) + H2SO4 (aq)

Initially the solution of CuSO4 in water will be blue, but when adding HCl the solution will change color to green, since the aqueous solutions of CuCl2 are green. By adding more HCl this color will intensify as the balance shifts to the right, producing more CuCl2 and H2SO4.

On the other hand, adding NH3 to a solution of CuSO4 will produce the following reaction

CuSO4 (aq) + 4NH3 (aq) ⇔ [Cu(NH3)4] SO4 (s)

Thus, by adding NH3 to the CuSO4 solution we will observe the formation of a precipitate corresponding to [Cu(NH3) 4] SO4. <u>When adding more NH3, the formation of more precipitate will be observed as the equilibrium moves to the right, producing a greater quantity of [Cu (NH3) 4] SO4.</u>

6 0

2 years ago

What is the kinetic energy acquired by the electron in hydrogen atom, if it absorbs a light radiation of energy 1.08x101 J. (A)

Delvig [45]

Explanation:

The given data is as follows.

Energy of radiation absorbed by the electron in hydrogen atom = $1.08 \times 10^{-17} J$

As energy is absorbed as a photon. Hence, frequency will be calculated will be as follows.

E = $h \nu$

$1.08 \times 10^{-17} J$ = $6.626 \times 10^{-34} Js \times \nu$

$\nu$ = $0.163 \times 10^{17} s^{-1}$

or, $\nu$ = $1.63 \times 10^{16} s^{-1}$

It is known that, $\nu = \frac{c}{\lambda}$

$1.63 \times 10^{16} s^{-1} = \frac{3 \times 10^{8} m/s}{\lambda}$

$\lambda$ = $1.84 \times 10^{-8} m$

And, according to De-Broglie equation $\lambda = \frac{h}{p}$

as, p = $m \times \nu$

So, $\lambda = \frac{h}{m \times \nu}$

$m \times \nu = \frac{6.626 \times 10^{-34} Js}{1.84 \times 10^{-8} m}$

= $3.6 \times 10^{-26} J/m$

Now, on squaring both the sides we get the following.

$(m \times \nu)^{2}$ = $(3.6 \times 10^{-26} J/m)^{2}$

= $12.96 \times 10^{-52}$

$m \times \nu^{2} = \frac{12.96 \times 10^{-52}}{m}$

where, m = mass of electron

So, $m \times \nu^{2} = \frac{12.96 \times 10^{-52}}{m}$

= $\frac{12.96 \times 10^{-52}}{9.1 \times 10^{-31}}$

= $1.42 \times 10^{-21}$ J

Since, K.E = $\frac{1}{2}m \nu^{2}$

= $\frac{1.42 \times 10^{-21} J}{2}$

= $0.71 \times 10^{-21} J$

Thus, we can conclude that kinetic energy acquired by the electron in hydrogen atom is $7.1 \times 10^{-22} J$ .

4 0

2 years ago

CuSO4(aq) + 2NaOH(aq) Cu(OH)2(s) + Na2SO4(aq)

REY [17]

Your answer is right.

Important elements to consider:

- to use the balanced equation (which you did)
- divide the masses of each compound by the correspondant molar masses (which you did)
- compare the theoretical proportions with the current proportions

Theoretical: 2 mol of Na OH : 1 mol of CuSO4
Then 4 mol of NaOH need 2 mol of CUSO4.

Given that you have more than 2 mol of of CUSO4 you have plenty of it and the NaOH will consume first, being this the limiting reagent.

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2 years ago

What is the temperature (in K) of 16.45 moles of methane gas in a 4.95 L container at 4.68 atm?

AlexFokin [52]

Using the ideal gas law: PV=nRT
P is pressure; V is volume; n is the amount in moles; R=0.082; T is temperature in K.

(4.68)*(4.95)=(16.45)*(0.0821)*T
Solve for T.
T=17.15

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How many atoms of H are in 2.8 mol N2H4?

yanalaym [24]

11.2 atoms H (also mind helping me with my most recent math question on my account?)

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2 years ago