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Nookie1986 [14]
3 years ago
15

How many nitrogen atoms are in 10.2 g of copper(ii) nitrate?

Chemistry
1 answer:
timofeeve [1]3 years ago
5 0
To determine the number of nitrogen atoms in the given amount of a compound, we need to know the number of moles of nitrogen. In order to come to this, its is important tht we know the chemical formula of the compound which would be Cu(NO3)2 (molar mass = 187.56 g/mol). When we obtain the moles of nitrogen, we use Avogadro's number to convert the number of moles to atoms. 

Moles Cu(NO3)2 = 10.2 g Cu(NO3)2 ( 1 mol / 187.56 g ) = 0.0544 mol 
Moles N = 0.0544 mol Cu(NO3)2 ( 2 mol N / mol Cu(NO3)2 ) = 0.109 mol
Atoms N = 0.109 mol N ( 6.022 x20^23 atoms N / mol N ) = 6.55x10^22 atoms N 
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Indicate the direction of polarity of each of the following covalent bonds. Please just write a + or - not the bonds.
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Answer:

1. C+ ---- O-

2. O+ ---- Cl-

3. O+ ----- F-

4. C+ ----- N-

5. Cl- ----- C+

6. S- ----- H+

7. S+ ----- Cl -

Explanation:

Electronegativity determines the polarity . There may be two atoms in a bond with high electronegativity, in such cases the positive charge is given to atom with comparatively lower electronegativity. Electronegativity determines the easiness with which an atom attract electrons in a chemical bond. A polar bond is formed when the difference in the electronegativity of two combining atoms is between 0.4 and 1.7. The correct direction is

1. C+ ---- O-

2. O+ ---- Cl-

3. O+ ----- F-

4. C+ ----- N-

5. Cl- ----- C+

6. S- ----- H+

7. S+ ----- Cl -

5 0
3 years ago
Question 1(Multiple Choice Worth 4 points) (04.05 LC) What is the oxidation number of manganese in MnO41−? +3 +4 +7 +8 Question
Elan Coil [88]

1) Answer is: the oxidation number of manganese in MnO₄⁻ is +7.

Permanganate anion has negative charge 1-.

Oxygen (O) in permanganate anion has oxidation number -2.

x + 4 · (-2) = -1.

x - 8 = -1.

x = +7; oxidation nzmber of manganese.

Oxidation number shows the degree of oxidation of an atom in a chemical compound.

2) Answer is: Zn + Cu2+ → Zn2+ + Cu.

In this chemical reaction, there is transfer of electrons from zinc (Zn) to copper (Cu). Zinc change oxidation number from 0 to +2 (lost electrons) and copper change oxidation number from +2 to 0 (gain electrons).

Oxidation half reaction: Zn⁰ → Zn²⁺ + 2e⁻.

Reduction half reaction: Cu²⁺ + 2e⁻ → Cu⁰.

In other chemical reactions, there is no change of oxidation number of elements.

3) Answer is: Br2 is the oxidizing agent because its oxidation number decreases.

Balanced chemical reaction: 2Al + 3Br₂ → 2AlBr₃.

In this chemical reaction, aluminium change oxidation number from 0 to +3 (lose electrons) and bromine change oxidation number from 0 to -1 (gain electrons, reduced).

Oxidizing agent is a substance that has the ability to oxidize other substances, to cause them to lose electrons.

In oxido-reduction reaction, at least one element lose and one element gain electrons.

4) Answer is: Br2 + 2Cl− → Cl2 + 2Br⁻.

Oxidation is increase of oxidation number.

In this balanced chemical reaction, chlorine change oxidation number from -1 (Cl⁻) to oxidation number 0 (Cl₂).

Oxidation half reaction: 2Cl⁻ → Cl₂⁰ + 2e⁻.

Reduction half reaction: Br₂ + 2e⁻ → 2Br⁻.

In other chemical reactions, chlorine is reduced.

5) Answer is: Observation 1 is a result of copper ions moving into the solution.

The reactivity series is a series of metals from highest to lowest reactivity. Metal higher in the reactivity series will displace another.

Copper (Cu) is higher in activity series than silver (Ag), so copper lose electron and silver gain electrons.

Copper is oxidized (increase of oxidation number) and silver is reduced.

3 0
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