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yulyashka [42]
3 years ago
10

Solid calcium hydroxide is dissolved in water until the pH of the solution is 10.44. Determine the concentration of the calcium

hydroxide in the solution.
Chemistry
1 answer:
Orlov [11]3 years ago
6 0

Answer:

1.38x10⁻⁴ M

Explanation:

First, we <u>calculate the pOH of the solution</u>, using the <em>given pH</em>:

  • pOH = 14 - pH
  • pOH = 14 - 10.44
  • pOH = 3.56

Now we <u>calculate the concentration of OH⁻</u>, using<em> the pOH</em>:

  • pOH = -log[OH⁻]
  • [OH⁻] = 10^{-pOH} = 2.75x10⁻⁴ M

Following the reaction Ca(OH)₂ → Ca⁺² + 2OH⁻, we divide [OH⁻] by 2 to calculate [Ca(OH)₂]:

  • [Ca(OH)₂] = 2.75x10⁻⁴ M / 2
  • [Ca(OH)₂] = 1.38x10⁻⁴ M
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Tanya [424]

Answer: This is a typical acid/base equilibrium problem, that involves the use of logarithms.

Explanation:We assume that both nitric acid and hydrochloric acid dissociate to give stoichiometric

H

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Moles of nitric acid:

26.0

×

10

−

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⋅

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×

8.00

⋅

m

o

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⋅

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a

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.

And, moles of hydrochloric acid:

88.0

×

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−

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⋅

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×

5.00

⋅

m

o

l

⋅

L

−

1

=

0.440

⋅

m

o

l

H

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(

a

q

)

.

This molar quantity is diluted to

1.00

L

. Concentration in moles/Litre =

(

0.208

+

0.440

)

⋅

m

o

l

1

L

=

0.648

⋅

m

o

l

⋅

L

−

1

.

Now we know that water undergoes autoprotolysis:

H

2

O

(

l

)

⇌

H

+

+

O

H

−

. This is another equilibrium reaction, and the ion product

[

H

+

]

[

O

H

−

]

=

K

w

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K

w

=

10

−

14

at

298

K

.

So

[

H

+

]

=

0.648

⋅

m

o

l

⋅

L

−

1

;

[

O

H

−

]

=

K

w

[

H

+

]

=

10

−

14

0.648

=

?

?

p

H

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−

log

10

[

H

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]

=

−

log

10

(

0.648

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=

?

?

Alternatively, we know further that

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H

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H

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Answer link  

4 0
3 years ago
Write the balanced chemical equations for the decomposition of solid calcium hydroxide into solid calcium (ii) oxide (lime) and
Lesechka [4]

Calcium oxide (CaO) or lime in solid form can be prepared from the decomposition of calcium hydroxide {Ca(OH)_{2} in to lime (CaO) and water (H_{2}O at high temperature. The reaction is an endothermic reaction. That is heat is absorbed in this reaction process. One mole of calcium hydroxide decomposed into one mole of calcium oxide and one mole of water. The balanced reaction can be shown as-CaCO_{3} (solid) → CaO (solid) + H_{2}O (liquid). The heat of the reaction is (+) 63.7 kJ/mole of CaO.

3 0
3 years ago
What protects the Earth from meteorites and the sun's rays? (4 points)
inna [77]
I believe the answer is D. The atmosphere, different layers of the atmosphere can help in different ways! The stratosphere helps protect us from the sun's rays; The Mesophere is what helps protect us from meteorites! It's so cold that they burn up in that area.
5 0
3 years ago
If 36 000 kg of full cream milk containing 4% fat is to be separated in a 6 hour period into skim milk with 0.45% fat and cream
Yuri [45]

Answer:

B=5522.33kg/h

C=478.11kg/h

Explanation:

Hi! It's a mass balance. First we have to determine the inflow.

mass flow rate = 36000kg / 6h = 6000kg / h

We define the input variable

- input flow (A) = 6000kg / h

-XgA = percentage of fat in A = 0.04

We define output variables.

- skim milk (B)

-creme (C)

-XgB = fat percentage at B = 0.0045

-XgC = percentage of fat in C = 0.45

Then we can start with the balance.

As a general rule, the mass balance is:

Input = Output

Balance sheet

1) A = B + C

Fat balance

2) A * XgA = B * XgB + C * XgC

Now we can solve.

We replace and clear B in equation 2

6000kg / h * 0.04 = B * 0.0045 + C * 0.45

B = (240kg / h) /0.045-C*0.45/0.0045

3) B = 53333.33kg / h-C * 100

We replace equation 3 in 1 and clear C

A = B + C

6000kg / h = 53333.33kg / h-C * 100 + C

C=(6000kg/h-53333.33kg/h)/(-99)

C=478.11kg/h

We replace C in equation 3 and calculate B

B = 53333.33kg / h-478.11kg/h * 100

B=5522.33kg/h

Then we have the values ​​of the outflows.

C=478.11kg/h

B=5522.33kg/h

7 0
3 years ago
A common radio wavelength observed coming from astronomical objects is 21 cm. What temperature is associated with this radiation
qaws [65]

Answer:

The temperature associated with this radiation is 0.014K.

Explanation:

If we assume that the astronomical object behaves as a black body, the relation between its <em>wavelength</em> and <em>temperature</em> is given by Wien's displacement law.

\lambda_{max}=\frac{b}{T}

where,

λmax is the wavelength at the peak of emission

b is Wien's displacement constant (2.89×10⁻³ m⋅K)

T is the absolute temperature

For a wavelength of 21 cm,

T=\frac{b}{\lambda _{max} } = \frac{2.89 \times 10^{-3} m.K  }{0.21m} =0.014K

8 0
4 years ago
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