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alisha [4.7K]
3 years ago
6

A galvanic cell consists of a Cu(s)|Cu2+(aq) half-cell and a Cd(s)|Cd2+(aq) half-cell connected by a salt bridge. Oxidation occu

rs in the cadmium half-cell. The cell can be represented in standard notation as
Chemistry
1 answer:
Kryger [21]3 years ago
6 0

Answer:

Cd(s)|Cd^{2+}(aq) || Cu^{2+}(aq)|Cu(s)

Explanation:

A galvanic cell is composed of two electrodes immersed in a suitable electrolyte and connected via a salt bridge. One of the electrodes serves as a cathode where reduction or gain of electrons takes place. The other half cell functions as an anode where oxidation or loss of electrons occurs.

The representation is given by writing the anode on left hand side followed by its ion with its molar concentration. It is followed by a salt bridge. Then the cathodic ion with its molar concentration is written and then the cathode.

As it is given that cadmium acts as anode, it must be on the left hand side and copper must be on right hand side.

Cd(s)|Cd^{2+}(aq) || Cu^{2+}(aq)|Cu(s)

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Chlorine forms a number of oxides with the following oxidation numbers: 1, 3, 4, 6, and 7. These compounds are stable and neutra
SashulF [63]

Answer:

Cl_2O, Cl_2O_3, ClO_2, ClO_3, Cl_2O_7

Explanation:

Empirical formula of the compound is the simplest ratio of elements present in the compound.

Empirical formula of compounds of chlorine with oxygen is as follows:

Compounds in which oxidation state of Cl is +1

Cl_2O

Compounds in which oxidation state of Cl is +3

Cl_2O_3

Compounds in which oxidation state of Cl is +4

ClO_2

Compounds in which oxidation state of Cl is +6

ClO_3

Compounds in which oxidation state of Cl is +7

Cl_2O_7

8 0
3 years ago
If a substance changes from one phase to another, is it still same substance? why
shutvik [7]
When you boil water, you aren't changing the elements. You're just making water vapor. However, when you burn paper, it becomes carbon (mostly). So physical changes will not change the substance, only chemical changes will.
5 0
2 years ago
If 38.5 grams of potassium react with excess oxygen gas, how many grams of potassium oxide can be produced? 4K + O2 yields 2K2O
Lera25 [3.4K]

Answer:

46.40 g.

Explanation:

  • It is a stichiometric problem.
  • The balanced equation of the reaction: 4K + O₂ → 2K₂O.
  • It is clear that 4.0 moles of K reacts with 1.0 mole of oxygen produces 2.0 moles of K₂O.
  • We should convert the mass of K (38.5 g) into moles using the relation:

<em>n = mass / molar mass,</em>

n = (38.5 g) / (39.098 g/mol) = 0.985 mole.

<em>Using cross multiplication:</em>

4.0 moles of K produces → 2.0 moles of K₂O, from the stichiometry.

0.985 mole of K produces → ??? moles of K₂O.

∴ The number of moles of K₂O produced = (0.985 mole) (2.0 mole) / (4.0 mole) = 0.4925 mole ≅ 0.5 mole.

  • Now, we can get the mass of K₂O:

∴ mass = n x molar mass = (0.5 mole) (94.2 g/mol) = 46.40 g.

6 0
2 years ago
Radioactive decay can be described by the following equation where is the original amount of the substance, is the amount of the
soldi70 [24.7K]

Answer:

Iron remains = 17.49 mg

Explanation:

Half life of iron -55 = 2.737 years (Source)

t_{1/2}=\frac {ln\ 2}{k}

Where, k is rate constant

So,  

k=\frac {ln\ 2}{t_{1/2}}

k=\frac {ln\ 2}{2.737}\ year^{-1}

The rate constant, k = 0.2533 year⁻¹

Time = 2.41 years

[A_0] = 32.2 mg

Using integrated rate law for first order kinetics as:

[A_t]=[A_0]e^{-kt}

Where,  

[A_t] is the concentration at time t

[A_0] is the initial concentration

So,  

[A_t]=32.2\times e^{-0.2533\times 2.41}\ mg

[A_t]=32.2\times e^{-0.610453}\ mg

[A_t]=17.49\ mg

<u>Iron remains = 17.49 mg</u>

8 0
3 years ago
What are the two parts of the third Step?
spin [16.1K]

Answer:

A

Explanation:

7 0
2 years ago
Read 2 more answers
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