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malfutka [58]
2 years ago
11

Calculate the solubility at 25°C of CuBr in pure water and in a 0.0120M CoBr2 solution. You'll find Ksp data in the ALEKS Data t

ab. Round both of your answers to 2 significant digits. solubility in pure water: ×7.210−4gL solubility in 0.0120 M CoBr2 solution: ×5.210−7gL
Chemistry
1 answer:
iragen [17]2 years ago
8 0

Answer:

S = 7.9 × 10⁻⁵ M

S' = 2.6 × 10⁻⁷ M

Explanation:

To calculate the solubility of CuBr in pure water (S) we will use an ICE Chart. We identify 3 stages (Initial-Change-Equilibrium) and complete each row with the concentration or change in concentration. Let's consider the solution of CuBr.

    CuBr(s) ⇄ Cu⁺(aq) + Br⁻(aq)

I                       0             0

C                     +S           +S

E                       S             S

The solubility product (Ksp) is:

Ksp = 6.27 × 10⁻⁹ = [Cu⁺].[Br⁻] = S²

S = 7.9 × 10⁻⁵ M

<u>Solubility in 0.0120 M CoBr₂ (S')</u>

First, we will consider the ionization of CoBr₂, a strong electrolyte.

CoBr₂(aq) → Co²⁺(aq) + 2 Br⁻(aq)

1 mole of CoBr₂ produces 2 moles of Br⁻. Then, the concentration of Br⁻ will be 2 × 0.0120 M = 0.0240 M.

Then,

    CuBr(s) ⇄ Cu⁺(aq) + Br⁻(aq)

I                       0           0.0240

C                     +S'           +S'

E                       S'            0.0240 + S'

Ksp = 6.27 × 10⁻⁹ = [Cu⁺].[Br⁻] = S' . (0.0240 + S')

In the term (0.0240 + S'), S' is very small so we can neglect it to simplify the calculations.

S' = 2.6 × 10⁻⁷ M

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Wet steam at 1100 kPa expands at constant enthalpy (as in a throttling process) to 101.33 kPa, where its temperature is 105°C. W
nikklg [1K]

Answer:

There are 3 steps of this problem.

Explanation:

Step 1.

Wet steam at 1100 kPa expands at constant enthalpy to 101.33 kPa, where its temperature is 105°C.

Step 2.

Enthalpy of saturated liquid Haq = 781.124 J/g

Enthalpy of saturated vapour Hvap = 2779.7 J/g

Enthalpy of steam at 101.33 kPa and 105°C is H2= 2686.1 J/g

Step 3.

In constant enthalpy process, H1=H2 which means inlet enthalpy is equal to outlet enthalpy

So, H1=H2

     H2= (1-x)Haq+XHvap.........1

    Putting the values in 1

    2686.1(J/g) = {(1-x)x 781.124(J/g)} + {X x 2779.7 (J/g)}

                        = 781.124 (J/g) - x781.124 (J/g) = x2779.7 (J/g)

1904.976 (J/g) = x1998.576 (J/g)

                     x = 1904.976 (J/g)/1998.576 (J/g)

                     x = 0.953

So, the quality of the wet steam is 0.953

                   

7 0
3 years ago
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Radioactive radium has a half-life of approximately 1,599 years. The initial quantity is 13 grams. How much (in grams) remains a
Luda [366]

The quantity of substance remains after 850 years is 8.98g if the half life of radioactive radium is 1,599 years.

<h3>What is half life period? </h3>

The time taken by substance to reduce to its half of its initial concentration is called half life period.

We will use the half- life equation N(t)

N e^{(-0.693t) /t½}

Where,

N is the initial sample

t½ is the half life time period of the substance

t2 is the time in years.

N(t) is the reminder quantity after t years .

Given

N = 13g

t = 350 years

t½ = 1599 years

By substituting all the value, we get

N(t) = 13e^(0.693 × 50) / (1599)

= 13e^(- 0.368386)

= 13 × 0.691

= 8.98

Thus, we calculated that the quantity of substance remains after 850 years is 8.98g if the half life of radioactive radium is 1,599 years.

learn more about half life period:

brainly.com/question/20309144

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4 0
1 year ago
An aqueous solution in a 55 gallon (208 l drum), characterized by minimal buffering capacity, received 4kg of phenol and 1.5 kg
lianna [129]

An aqueous solution in a 55 gallon (208 l drum), characterized by minimal buffering capacity, received 4kg of phenol and 1.5 kg of sodium phenate. What is the ph of the solution. The pka of phenol = 9.98. Mw of phenol and sodium phenate are 94 g/mol and 116 g/mol, respectively.


Volume of solution = 55 gallons = 208.2 L [ 1 gallon = 3.78 L]

moles of phenol = mass / molar mass = 4000 g / 94 = 42.55 moles

moles of sodium phenate = mass / molar mass = 1500 / 116 = 12.93 moles

pKa of phenol = 9.98

We know that the pH of buffer is calculated using Hendersen Hassalbalch's equation

pH = pKa + log [salt] / [acid]

volume is same for both the sodium phenate and phenol has we can directly take the moles of each in the formula

pH = 9.98 + log [12.93 / 42.55] = 9.46



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3 years ago
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vivado [14]

Answer:3.2

Mass:32

Molecular weight: 10

Moles: 3.2

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Bas_tet [7]
<h2>Carbon-dioxide</h2>

Explanation:

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