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dlinn [17]
2 years ago
8

An oxybromate compound, kbrox, where x is unknown, is analyzed and found to contain 43.66 % br.

Chemistry
1 answer:
aniked [119]2 years ago
6 0

Answer is: an oxybromate compound is KBrO₄ (x = 4).

ω(Br) = 43.66% ÷ 100%.

ω(Br) = 0.4366; mass percentage of bromine.

If we take 100 grams of compound:

m(Br) = ω(Br) · 100 g.

m(Br) = 0.4366 · 100 g.

m(Br) = 43.66 g; mass of bromine.

n(Br) = m(Br) ÷ M(Br).

n(Br) = 43.66 g ÷ 79.9 g/mol,

n(Br) = 0.55 mol; amoun of bromine.

From chemical formula (KBrOₓ), amount of potassium is equal to amount of bromine: n(Br) = n(K).

m(K) = 0.55 mol · 39.1 g/mol.

m(K) = 21.365 g; mass of potassium in the compound.

m(O) = 100 g - 21.365 g - 43.66 g.

m(O) =34.97 g; mass of oxygen.

n(O) = 34.97 g ÷ 16 g/mol.

n(O) = 2.185 mol.

n(K) : n(Br) : n(O) = 0.55 mol : 0.55 mol : 2.185 mol /÷ 0.55 mol.

n(K) : n(Br) : n(O) = 1 : 1 : 4.

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The polymer formed from the monomer ch2=ch–cn is select one:
podryga [215]
Answer:
             None of the given options show polymer made up of H₂C=CH-CN (Acrylonitrile).

Explanation:
                   Acrylonitrile (H₂C=CH-CN) which is a monomer on self linkage results in a large chain polymer called as Polyacrylonitrile.
                   The structure of Polyacrylonitrile is as follow,

                                             --(H₂C-CHCN-)n--

Where n shows the number of Acrylonitrile units joined together in the formation of Polyacrylonitrile. This polymerization reaction can take place by different mechanisms including free radical mechanism, acid catalyzed addition or base catalyzed addition reaction.

The polymerization is shown below,

5 0
2 years ago
(2 pts) The solubility of InF3 is 4.0 x 10-2 g/100 mL. a) What is the Ksp? Include the chemical equation and Ksp expression. MW
Aleonysh [2.5K]

Answer:

a) Ksp = 7.9x10⁻¹⁰

b) Solubility is 6.31x10⁻⁶M

Explanation:

a) InF₃ in water produce:

InF₃ ⇄ In⁺³ + 3F⁻

And Ksp is defined as:

Ksp = [In⁺³] [F⁻]³

4.0x10⁻²g / 100mL of InF₃ are:

4.0x10⁻²g / 100mL ₓ (1mol / 172g) ₓ (100mL / 0.1L) = <em>2.3x10⁻³M  InF₃. </em>Thus:

[In⁺³] = 2.3x10⁻³M  InF₃ × (1 mol In⁺³ / mol InF₃) = 2.3x10⁻³M  In⁺³

[F⁻] = 2.3x10⁻³M  InF₃ × (3 mol F⁻ / mol InF₃) = 7.0x10⁻³M F⁻

Replacing these values in Ksp formula:

Ksp = [2.3x10⁻³M  In⁺³] × [7.0x10⁻³M F⁻]³ = <em>7.9x10⁻¹⁰</em>

<em></em>

b) 0.05 moles of F⁻ produce solubility of InF₃ decrease to:

7.9x10⁻¹⁰ = [x] [0.05 + 3x]³

Where x are moles of In⁺³ produced from solid InF₃ and 3x are moles of F⁻ produced from the same source. That means x is solubility in mol / L

Solving from x:

x = -0.018 → False solution, there is no negative concentrations.

x = 6.31x10⁻⁶M → Right answer.

Thus, <em>solubility is 6.31x10⁻⁶M</em>

3 0
2 years ago
6.
RUDIKE [14]

Answer:

THE CURRENT REQUIRED TO PRODUCE 193000 C OF ELECTRICITY IS 35.74 A.

Explanation:

Equation:

Al3+ + 3e- -------> Al

3 F of electricity is required to produce 1 mole of Al

3 F of electricity = 27 g of Al

If 18 g of aluminium was used, the quantity of electricity to be used up will be:

27 g of AL = 3 * 96500 C

18 G of Al = x C

x C = ( 3 * 96500 * 18 / 27)

x C = 193 000 C

For 18 g of Al to be produced, 193000 C of electricity is required.

To calculate the current required to produce 193 000 C quantity of electricity, we use:

Q = I t

Quantity of electricity = Current * time

193 00 = I * 1.50 * 60 * 60 seconds

I = 193 000 / 1.50 * 60 *60

I = 193 000 / 5400

I = 35.74 A

The cuurent required to produce 193,000 C of electricity by 18 g of aluminium is 35.74 A

3 0
2 years ago
Which statement is true about molarity and percent by mass?
Triss [41]

Answer : The correct option is, They are different units of concentration.

Explanation :

Molarity : It is defined as the number of moles of solute present in one liter of solution.

Formula of molarity :

\text{Molarity}=\frac{\text{Moles of solute}}{\text{Volume of solution in liters}}

Mass percent : It is defined as the mass of solute present in the mass of solution.

Formula of mass percent :

\text{Mass percent}=\frac{\text{Mass of solute}}{\text{Mass of solution}}\times 100

Hence, both are the different unit of concentration.

5 0
2 years ago
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mote1985 [20]
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9 0
2 years ago
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