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emmainna [20.7K]
3 years ago
15

1) According to the balanced chemical equation below; how many atoms of silver (Ag) will be produced from 100 g of copper (Cu) w

ith excess silver nitrate (AgNO3)? Cu + 2 AgNO3 ⇒ Cu(NO3) 2 + 2 Ag
Chemistry
1 answer:
raketka [301]3 years ago
4 0

Answer:

There will be produced 1.89x10²⁴ atoms of Ag

Explanation:

This is the reaction:

Cu + 2 AgNO₃ ⇒ Cu(NO₃)₂ + 2 Ag

1 mol of copper reacts with 2 moles of silver nitrate to produce solid silver and 1 mol of copper nitrate.

Ratio is 1:2 so from the moles I have of Cu, I'll produce the double of moles of silver.

100 g / 63.55 g / m = 1.57 moles of Cu

1.57 moles of Cu will produce 3.14 moles of Ag.

Let's calculate the number of atoms, by NA

1 mol of Ag has NA atoms (6.02x10²³)

3.14 moles of Ag will have (3.14  . NA) = 1.89x10²⁴ atoms of Ag

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1)Identify the atoms that are participating in a covalent bond.
2)Draw each atom by using its element symbol. The number of valence electrons is shown by placing up to two dots on each side of the element symbol, with each dot representing a single valence electron.
3)Predict the number of covalent bonds each atom will make using the octet rule.
4)Draw the bonding atoms next to each other, showing a single covalent bond as either a pair of dots or a line representing a shared valence electron pair. If the molecule forms a double or triple bond, use two or three lines to represent the shared electron pairs, respectively.

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3 years ago
Um Test
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Suspensions

Explanation:

Suspensions are heterogeneous mixtures that contains large particles that can settle out or be filtered.

  • Suspensions are mixtures of small insoluble particles of a solid in a liquid or gas.
  • Examples are:
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  2. muddy water
  3. harmattan

The particles in suspension can settle on standing

Learn more:

Suspension brainly.com/question/1557970

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8 0
3 years ago
Consider the balanced equation below. 4NH3 + 3O2 --> 2N2 + 6H2O What is the mole ratio of NH3 to N2?
AURORKA [14]
The balanced equation given is:
4NH3 + 3O2 .....> 2N2 + 6H2O

From this equation, we can note that 4 moles of NH3 are required to produce 2 moles of N2.

Therefore, the mole ratio of NH3 to N2 is 4:2 which can be simplified into 2:1
4 0
3 years ago
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Is metal rusting a chemical reaction? Yes or no
mars1129 [50]

Answer: Yes it is. Rust would be considered a chemical reaction because it changes the chemical makeup of the metal.

Explanation:

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3 years ago
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Ammonia and oxygen react to form nitrogen and water.
Nata [24]

Answer:

A. 19.2 g of O2.

B. 3.79 g of N2.

C. 54 g of H2O.

Explanation:

The balanced equation for the reaction is given below:

4NH3(g) + 3O2(g) → 2N2+ 6H2O(g)

Next, we shall determine the masses of NH3 and O2 that reacted and the masses of N2 and H2O produced from the balanced equation.

This is illustrated below:

Molar mass of NH3 = 14 + (3x1) = 17 g/mol

Mass of NH3 from the balanced equation = 4 x 17 = 68 g

Molar mass of O2 = 16x2 = 32 g/mol

Mass of O2 from the balanced equation = 3 x 32 = 96 g

Molar mass of N2 = 2x14 = 28 g/mol

Mass of N2 from the balanced equation = 2 x 28 = 56 g

Molar mass of H2O = (2x1) + 16 = 18 g/mol

Mass of H2O from the balanced equation = 6 x 18 = 108 g

Summary:

From the balanced equation above,

68 g of NH3 reacted with 96 g of O2 to produce 56 g of N2 and 108 g of H2O.

A. Determination of the mass of O2 needed to react with 13.6 g of NH3.

This is illustrated below:

From the balanced equation above,

68 g of NH3 reacted with 96 g of O2.

Therefore, 13.6 g of NH3 will react with = (13.6 x 96)/68 = 19.2 g of O2.

Therefore, 19.2 g of O2 are needed for the reaction.

B. Determination of the mass of N2 produced when 6.50 g of O2 react.

This is illustrated below:

From the balanced equation above,

96 g of O2 reacted to produce 56 g of N2.

Therefore, 6.5 g of O2 will react to produce = (6.5 x 56)/96 = 3.79 g of N2.

Therefore, 3.79 g of N2 were produced from the reaction.

C. Determination of the mass of H2O formed from the reaction of 34 g of NH3.

This is illustrated below:

From the balanced equation above,

68 g of NH3 reacted to 108 g of H2O.

Therefore, 34 g of NH3 will react to produce = (34 x 108)/68 = 54 g of H2O.

Therefore, 54 g of H2O were obtained from the reaction.

4 0
3 years ago
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