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3 years ago

When a sample of ca(s) loses 1 mole of electrons in a reaction with a sample of o2(g) the oxygen?

2 answers:

6 0

The reaction involved in present case is:

Net Reaction: Ca + 1/2 O2 → CaO. ..................(1)

In terms of oxidation and reduction, the reaction can be shown at

Oxidation: Ca → Ca2+ + 2e- .................(2)
Reduction: 1/2O2 + 2e- → O2-...................(3)

From, reaction 1 it can be seen that 1 mol of Ca reacts with 1/2 mol of O2 to form 1 mol of CaO.

From, reaction 2 it can be seen that 1 mol of Ca, generates 2 mol of e-.

Thus, when 1/2 mol of Ca is used in reaction, it will lose 1 mol of electrons.

Mandarinka [93]3 years ago

3 0

Gains 1 mole of electrons

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In which of the following does nitrogen have the lowest oxidation state?

Leokris [45]

Nh4cl (the oxidation number of N=-3) so it would be the right answer

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1 year ago

Which of these compounds would you expect to have the lowest melting point?

Leya [2.2K]

Answer:

Hmmm. #3

Explanation:

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3 years ago

If 100.0g of nitrogen is reacted with 100.0g of hydrogen, what is the theoretical yield of the reaction? What is the excess reac

Shkiper50 [21]

theoretical yield of ammonia (NH₃) = 121.38 g

The limiting reactant is nitrogen (N₂) and the excess reactant is hydrogen (H₂).

Explanation:

We have the following chemical reaction in which nitrogen react with hydrogen to produce ammonia:

N₂ + 3 H₂ → 2 NH₃

Now we need to calculate the number of moles of each reactant:

number of moles = mass / molar weight

number of moles of N₂ = 100 / 28 = 3.57 moles

number of moles of H₂ = 100 / 2 = 50 moles

We see from the chemical reaction that 3 moles of H₂ react with 1 mole of N₂ so 50 moles of H₂ react with 16.67 moles of N₂ which is way more than the available N₂ quantity of 3.57 moles, so the limiting reactant is nitrogen (N₂) and the excess reactant is hydrogen (H₂).

Knowing this we devise the following reasoning:

if 1 mole of N₂ produces 2 moles of NH₃

then 3.57 moles of N₂ produces X moles of NH₃

X = (3.57 × 2) / 1 = 7.14 moles of NH₃

mass = number of moles × molar weight

mass of NH₃ = 7.14 × 17 = 121.38 g (theoretical yield)

Learn more about:

limiting reactant

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7 0

3 years ago

Which energy profile best shows that the enthalpy of formation of CS2 is 89.4 KJ/mol?

Anna11 [10]

Answer:

Option C. Energy Profile D

Explanation:

Data obtained from the question include:

Enthalpy change ΔH = 89.4 KJ/mol.

Enthalpy change (ΔH) is simply defined as the difference between the heat of product (Hp) and the heat of reactant (Hr). Mathematically, it is expressed as:

Enthalpy change (ΔH) = Heat of product (Hp) – Heat of reactant (Hr)

ΔH = Hp – Hr

Note: If the enthalpy change (ΔH) is positive, it means that the product has a higher heat content than the reactant.

If the enthalpy change (ΔH) is negative, it means that the reactant has a higher heat content than the product.

Now, considering the question given, the enthalpy change (ΔH) is 89.4 KJ/mol and it is a positive number indicating that the heat content of the product is higher than the heat content of the reactant.

Therefore, Energy Profile D satisfy the enthalpy change (ΔH) for the formation of CS2 as it indicates that the heat content of product is higher than the heat content of the reactant.

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3 years ago

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Explanation:

<h2>The number of energy levels (n) increases, so there is a greater distance between the nucleus and the outermost orbital.</h2>

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1 year ago