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3 years ago

How is the oxidation state of a transition metal determined from the chemical formula ?

Chemistry

2 answers:

asambeis [7]3 years ago

8 0

Answer: The subscript of the second ion identifies the oxidation state of the transition metal

Explanation: Apex

Vera_Pavlovna [14]3 years ago

7 0

Answer:

Explanation:

In a chemical formula, the oxidation state of transition metals can be determined by establishing the relationships between the electrons gained and that which is lost by an atom.

We know that for compounds to be formed, atoms would either lose, gain or share electrons between one another.

The oxidation state is usually expressed using the oxidation number and it is a formal charge assigned to an atom which is present in a molecule or ion.

To ascertain the oxidation state, we have to comply with some rules:

The algebraic sum of all oxidation numbers of an atom in a neutral compound is zero.
The algebraic sum of all the oxidation numbers of all atoms in an ion containing more than one kind of atom is equal to the charge on the ion.

For example, let us find the oxidation state of Cr in Cr₂O₇²⁻

This would be: 2x + 7(-2) = -2

x = +6

We see that the oxidation number of Cr, a transition metal in the given ion is +6.

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Mole used in a chemistry sentence

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3 years ago

A propane stove burned 470 grams propane and produced 625 grams of water (this is the actual yield) C3H8 +5O2=3CO2+4H20. What wa

Liula [17]

Answer:

81.3%

Explanation:

Step 1:

The balanced equation for the reaction:

This is shown below:

C3H8 + 5O2 —> 3CO2 + 4H2O

Step 2:

Data obtained from the question. This includes:

Mass of propane (C3H8) = 470 g

Actual yield of water (H2O) = 625 g

Percentage yield of water (H2O) =?

Step 3:

Determination of the mass of propane (C3H8) burned and the mass of water (H2O) produce from the balanced equation. This is illustrated below:

C3H8 + 5O2 —> 3CO2 + 4H2O

Molar Mass of C3H8 = (3x12) + (8x1) = 36 + 8 = 44g/mol

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

Mass of H2O from the balanced equation = 4 x 18 = 72g

From the balanced equation above,

44g of C3H8 was burned and 72g of H2O was produced.

Step 4:

Determination of the theoretical yield of H2O. This is illustrated below:

From the balanced equation above,

44g of C3H8 produced 72g of H2O.

Therefore, 470g of C3H8 will produce = (470x72)/44 = 769.09g of H2O.

Therefore, the theoretical yield of H2O is 769.09g

Step 5:

Determination of the percentage yield of water (H2O). This is illustrated below:

Actual yield of water (H2O) = 625g

theoretical yield of H2O = 769.09g

Percentage yield of water (H2O) =?

Percentage yield = Actual yield/Theoretical yield x100

Percentage yield = 625/769.09 x100

Percentage yield = 81.3%

Therefore, the percentage yield of water (H2O) is 81.3%

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3 years ago

I WILL GIVE BRAINLIEST!!!!!!!!

docker41 [41]

You can calculate the excess reactant by subtracting the mass of excess reagent consumed from the total mass of reagent given therefore,
The answer: Theoretical yield is 121.60 g of NH₃
Excess reactant is H₂
Rate limiting reactant is N₂
explanation: 100 g of Nitrogen
100 g of hydrogen
We are required to identify the theoretical yield of the reaction, the excess reactant and the rate limiting reagent.
We first write the equation for the reaction between nitrogen and hydrogen;
N₂ + 3H₂ → 2NH₃
From the reaction 1 mole of nitrogen reacts with 3 moles of Hydrogen gas.
Secondly we determine the moles of nitrogen gas given and hydrogen gas given;
Moles of Nitrogen gas
Moles = Mass ÷ Molar mass
Molar mass of nitrogen gas = 28.0 g/mol
Moles of Nitrogen gas = 100 g ÷ 28 g/mol 3.57 moles
Moles of Hydrogen gas
Molar mass of Hydrogen gas = 2.02 g/mol
Moles = 100 g ÷ 2.02 g/mol
= 49.50 moles
From the mole ratio given by the equation, 1 mole of nitrogen requires 3 moles of Hydrogen gas.
Thus, 3.57 moles of Nitrogen gas requires (3.57 × 3) 10.71 moles of Hydrogen gas.
This means, Nitrogen gas is the rate limiting reagent and hydrogen gas is the excess reactant.
Third calculate the theoretical yield of the reaction.
1 mole of nitrogen reacts to from 2 moles of ammonia gas
Therefore;
Moles of ammonia gas produced = Moles of nitrogen × 2
= 3.57 moles × 2
= 7.14 moles
But; molar mass of Ammonia gas is = 17.03 g/mol
Therefore;
Mass of ammonia gas produced = 7.14 moles × 17.03 g/mol
= 121.59 g
= 121.60 g
Thus, the theoretical amount of ammonia gas produced is 121.60 g

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