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Helga [31]
2 years ago
12

Calculate the pOH of an aqueous solution of .0.073 M LiOH

Chemistry
1 answer:
Novosadov [1.4K]2 years ago
7 0

Considering the definition of pOH and strong base, the pOH of the aqueous solution is 1.14

The pOH (or potential OH) is a measure of the basicity or alkalinity of a solution and indicates the concentration of ion hydroxide (OH-).

pOH is expressed as the logarithm of the concentration of OH⁻ ions, with the sign changed:

pOH= - log [OH⁻]

On the other hand, a strong base is that base that in an aqueous solution completely dissociates between the cation and OH-.

LiOH is a strong base, so the concentration of the hydroxide will be equal to the concentration of OH-. This is:

[LiOH]= [OH-]= 0.073 M

Replacing in the definition of pOH:

pOH= -log (0.073 M)

<u><em>pOH= 1.14 </em></u>

In summary, the pOH of the aqueous solution is 1.14

Learn more:

  • <u>brainly.com/question/16032912?referrer=searchResults </u>
  • <u>brainly.com/question/13557815?referrer=searchResults</u>
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Explain what Happens at the saturation point when adding salt to water at room temperature.
lorasvet [3.4K]

Answer:

See below  

Step-by-step explanation:

You won't see much happening. The solution is saturated, so the salt will fall to the bottom of the container and sit there. It will not dissolve.

However, at the atomic level, Na⁺ and Cl⁻ ions are being pulled from the surface of the crystals and going into solution as hydrated ions. At other places, Na⁺ and Cl⁻ ions are returning to the surface of the crystals.

The process is

NaCl(s) ⇌ Na⁺(aq) + Cl⁻(aq)

The rates of the forward and reverse processes are equal, so you see no net change.

6 0
3 years ago
Which of the following frequencies corresponds to light with the longest wavelength? A) 3.00 times 10^13 s^-1 B) 4.12 times 10^5
Vsevolod [243]

Answer:

1) B. 4.12 times 10^5 s^-1

2) B. frequency-v

3) C. 1.18 times 10^15 s^-1

4) B. 4.39 times 10^-19 J

5) B. Energy is absorbed

6) C. 3

7) 1s^2  2s^2  2p^6  3s^2  3p^6  4s^2  3d^10  4p^6  5s^2  4d^10  5p^6     6s^2  

Explanation:

1)

The wavelength is inversely proportional to the frequency. Thus, the smallest frequency shall correspond to the longest wavelength. Thus, the correct answer is <u>B. 4.12 times 10^5 s^-1</u>

2)

<u>B. frequency-v</u> are wrongly paired because frequency is represented by f. Thus, correct pair will be frequency-f.

3)

The relationship between wavelength (λ) and frequency (f) is:

c = fλ

f = c/λ

where, c = speed of light = 3 x 10^8 m/s

f = (3x10^8 m/s) / (254 x 10^-9 m)

f = 1.18 x 10^15 s^-1

Thus, the correct option is <u>C. 1.18 times 10^15 s^-1.</u>

4)

The energy of photon is given as:

E = hc/λ

where, c = speed of light = 3 x 10^8 m/s

            h = Plank's Constant = 6.625 x 10^-34 J.s

E = (6.625 x 10^-34 J.s)(3 x 10^8 m/s) / (453 x 10^-9 m)

E = 4.39 x 10^-19 J

Thus, the correct option is <u>B. 4.39 times 10^-19 J</u>.

5)

Since, the higher energy levels away from nucleus have higher energies. So, in order to move an electron from lower to higher or distant energy level, it must absorb energy from external source. So, the correct option is <u>B. Energy is absorbed</u>.

6)

The P-Sub-level has three orbitals, each having two electrons. Thus, P-Sub-Level accommodates total of 6 electrons. The correct option is <u>C. 3</u>

7)

The electronic configuration of barium atom is:

<u>1s^2  2s^2  2p^6  3s^2  3p^6  4s^2  3d^10  4p^6  5s^2  4d^10  5p^6  6s^2  </u>

8 0
3 years ago
Instead of using ratios for back titrations we can also use molarities, if our solutions are standardized. A 0.196 g sample of a
kvv77 [185]

Answer:

The mass percent of Al(OH)₃ is 15.3%

Explanation:

The reaction is:

Al(OH)₃ + 3HCl = AlCl₃ + 3H₂O

The excess acid is neutralized with a solution of sodium hidroxide, in the reaction:

NaOH + HCl = NaCl + H₂O

The total moles of HCl is:

n_{HCl,total} =M_{HCl} *V_{HCl} =0.111*0.025=2.78x10^{-3} moles

From the second titration, the moles of excess of HCl is:

n_{HCl,excess} =n_{NaOH} =M_{NaOH} *V_{NaOH} =0.132*0.01105=1.46x10^{-3} moles

The difference between the total and excess of HCl, it can be know the moles that reacts with the aluminum hydroxide, is:

n_{HCl,reacts} =n_{HCl,total}-n_{HCl,excess} =2.78x10^{-3} moles-1.46x10^{-3} moles=1.32x10^{-3} moles

The ratio between HCl and Al(OH)₃ is 3:1. The MW for aluminum hydroxide is 78 g/mol, thus:

m_{Al(OH)3} =1.32x10^{-3} molesHCl*\frac{1molAl(OH)3}{3molesHCl} *\frac{78gAl(OH)3}{1molAl(OH)3} =0.03g

The percentage of Al(OH)₃ is:

Percentage-Al(OH)3=\frac{m_{Al(OH)3} }{m_{antiacid} } *100=\frac{0.03}{0.196} =15.3%

3 0
3 years ago
Cree que podríamos vivir sin leyes
astraxan [27]

Answer:

nop

Explanation:

porqUE todo sería un desastre

3 0
3 years ago
g The combustion of 1.877 1.877 g of glucose, C 6 H 12 O 6 ( s ) C6H12O6(s), in a bomb calorimeter with a heat capacity of 4.30
Mekhanik [1.2K]

Answer:

-2.80 × 10³ kJ/mol

Explanation:

According to the law of conservation of energy, the sum of the heat absorbed by the bomb calorimeter (Qcal) and the heat released by the combustion of the glucose (Qcomb) is zero.

Qcal + Qcomb = 0

Qcomb = - Qcal [1]

We can calculate the heat absorbed by the bomb calorimeter using the following expression.

Qcal = C × ΔT = 4.30 kJ/°C × (29.51°C - 22.71°C) = 29.2 kJ

where,

C: heat capacity of the calorimeter

ΔT: change in the temperature

From [1],

Qcomb = - Qcal = -29.2 kJ

The internal energy change (ΔU), for the combustion of 1.877 g of glucose (MW 180.16 g/mol) is:

ΔU = -29.2 kJ/1.877 g × 180.16 g/mol = -2.80 × 10³ kJ/mol

3 0
3 years ago
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