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Phoenix [80]
2 years ago
15

Be sure to answer all parts. Which product(s) form(s) at each electrode in the aqueous electrolysis of the following salts? Sele

ct as many as apply. (a) FeI2 Cathode: Anode: I− (aq) I2 (s) Fe (s) Fe2+ (aq) I− (aq) I2 (s) Fe (s) Fe2+ (aq) (b) K3PO4 Cathode: Anode: O2 (g) H2 (g) H+ (aq) OH− (aq) O2 (g) H2 (g) H+ (aq) OH− (aq)
Chemistry
1 answer:
zlopas [31]2 years ago
4 0

Answer:

FeI2

Anode

Fe^2+

Cathode

I2

K3PO4

Anode

H^+ and K^+

Cathode

H2

Explanation:

In electrolysis, there are two electrodes present, the cathode and the anode. The anode is the positive electrode. This is where oxidation occurs and positive ions are formed. Species give up a electrons at the anode and the electrons travel towards the cathode.

At the cathode (negative electrode where reduction occurs), species accept electrons according to their position in the electrochemical series. In the case of K3PO4, both hydrogen and potassium ions arrive at the cathode but hydrogen in preferentially discharged due to the position of the two ions in the electrochemical series.

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If a pharmacist added 12 g of azelaic acid diluent should be used to prepare 8 fluidto 50 g of an ointment containing 15% ounces
natta225 [31]
Answer is: 31,45%.
mrs₁(C₉H₁₆O₄-<span>azelaic acid) = 12g.
mr</span>₂(C₉H₁₆O₄) = 50g.
ω₂(C₉H₁₆O₄) = 15% = 0,15.
mrs₂(C₉H₁₆O₄) = mr₂·ω₂ = 50g·0,15 = 7,5g.
mrs₃(C₉H₁₆O₄) = mrs₁ + mr₂ = 12g + 7,5g = 19,5g.
mr₃ = mr₂ + mr₂ = 50g + 12g = 62g.
ω₃ = mrs₃÷mr₃ = 19,5g ÷ 62g = 31,45% = 0,3145.



7 0
2 years ago
Scientists determine that the wooden beams of a sunken ship have 67 % of the concentration of Carbon-14 that is found in the lea
DerKrebs [107]

Answer:

The age of ship is 1900 years.

Explanation:

The half life period of carbon - 14 is 5730 years

Ship have 67 % of the concentration of Carbon-14

And the ship lost (100- 67 = 33%) 33% of carbon - 14.

Therefore,

The age of ship = lost percentage X half life period of carbon -14

= 33 \times 5730 yrs = 1890.9yrs\equiv 1900\,years

Therefore, The age of ship is 1900 years.

3 0
2 years ago
Combustion analysis of a 13.42-g sample of the unknown organic compound (which contains only carbon, hydrogen, and oxygen) produ
Kisachek [45]

<u>Answer:</u> The empirical and molecular formula for the given organic compound is C_9H_{12}O and C_{18}H_{24}O_2

<u>Explanation:</u>

The chemical equation for the combustion of hydrocarbon having carbon, hydrogen and oxygen follows:

C_xH_yO_z+O_2\rightarrow CO_2+H_2O

where, 'x', 'y' and 'z' are the subscripts of Carbon, hydrogen and oxygen respectively.

We are given:

Mass of CO_2=39.01g

Mass of H_2O=10.65g

We know that:

Molar mass of carbon dioxide = 44 g/mol

Molar mass of water = 18 g/mol

<u>For calculating the mass of carbon:</u>

In 44 g of carbon dioxide, 12 g of carbon is contained.

So, in 39.01 g of carbon dioxide, \frac{12}{44}\times 39.01=10.64g of carbon will be contained.

<u>For calculating the mass of hydrogen:</u>

In 18 g of water, 2 g of hydrogen is contained.

So, in 10.65 g of water, \frac{2}{18}\times 10.65=1.18g of hydrogen will be contained.

Mass of oxygen in the compound = (13.42) - (10.64 + 1.18) = 1.6 g

To formulate the empirical formula, we need to follow some steps:

  • <u>Step 1:</u> Converting the given masses into moles.

Moles of Carbon =\frac{\text{Given mass of Carbon}}{\text{Molar mass of Carbon}}=\frac{10.64g}{12g/mole}=0.886moles

Moles of Hydrogen = \frac{\text{Given mass of Hydrogen}}{\text{Molar mass of Hydrogen}}=\frac{1.18g}{1g/mole}=1.18moles

Moles of Oxygen = \frac{\text{Given mass of oxygen}}{\text{Molar mass of oxygen}}=\frac{1.6g}{16g/mole}=0.1moles

  • <u>Step 2:</u> Calculating the mole ratio of the given elements.

For the mole ratio, we divide each value of the moles by the smallest number of moles calculated which is 0.1 moles.

For Carbon = \frac{0.886}{0.1}=8.86\approx 9

For Hydrogen = \frac{1.18}{0.1}=11.8\approx 12

For Oxygen = \frac{0.1}{0.1}=1.99\approx 2

  • <u>Step 3:</u> Taking the mole ratio as their subscripts.

The ratio of C : H : O = 9 : 12 : 1

The empirical formula for the given compound is C_9H_{12}O

For determining the molecular formula, we need to determine the valency which is multiplied by each element to get the molecular formula.

The equation used to calculate the valency is :

n=\frac{\text{Molecular mass}}{\text{Empirical mass}}

We are given:

Mass of molecular formula = 272.38 g/mol

Mass of empirical formula = 136 g/mol

Putting values in above equation, we get:

n=\frac{272.38g/mol}{136g/mol}=2

Multiplying this valency by the subscript of every element of empirical formula, we get:

C_{(2\times 9)}H_{(2\times 12)}O_{(2\times 1)}=C_{18}H_{24}O_2

Hence, the empirical and molecular formula for the given organic compound is C_9H_{12}O and C_{18}H_{24}O_2

6 0
2 years ago
The population of rabbits in a forest is decreasing. Foxes in that region are now competing with each other for the limited numb
sveta [45]
Rabbits are eaten by predators such as foxes and wild dogs. If rabbit number decline predators ( foxes ) compete for the rest of the resources in forest. This is the competition between the members of same species or : interspecific competition. Foxes compete for the same resource ( rabbits ).
Answer: Interspecific competition.  
6 0
1 year ago
Read 2 more answers
Which of the following species contains manganese with the highest oxidation number?
ioda

In NaMnO₄, Mn has the highest oxidation number.

The question is incomplete, the complete question is;

Which of the following species contains manganese with the highest oxidation number?

A) Mn

B) MnF₂

C) Mn₃(PO₄)₂

D) MnCl₄

E) NaMnO₄

In order to ascertain the specie that contains manganese with the highest oxidation number, we must calculate the oxidation number of manganese in each of the species one after the other.

1) For Mn, the oxidation number of Mn is zero because the atom is uncombined.

2) For MnF₂;

Mn has an oxidation number of +2

3) For Mn₃(PO₄)₂

Mn has an oxidation number of +2

4) For MnCl₄

Mn has an oxidation number of +4

5) For NaMnO₄

Mn has an oxidation number of +7

Hence in NaMnO₄, Mn has the highest oxidation number.

Learn more: brainly.com/question/10079361

7 0
1 year ago
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