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2 years ago

How many total atoms are in 0.240 g of P2O5?

1 answer:

3 0

The answer is 1.02 × 10²⁵ atoms.

1. Calculate molar mass (Mr) of P2O5 which is the sum of atomic masses (Ar) of its elements:
Ar(P) = 31 g/mol
Ar(O) = 16
Mr(P2O5) = 2 * Ar(P) + 5 * Ar(O) = 2 * 31 + 5 * 16 = 62 + 80 = 142 g/mol

2. Calculate number of moles (n) which is the quotient of a sample mass (m) and the molar mass (Mr):
n = m/Mr
m = 0.240 g
Mr = 142 g/mol
n = 0.240 / 142 = 0.0017 mol

3. Avogadro's number is the number of units atoms in 1 mole of substance:
<span>6.023 × 10²³ atoms per 1 mol
x atoms are per 0.0017 mol

</span>6.023 × 10²³ atoms : 1 mol = x atoms : 0.0017 mol
x = 6.023 × 10²³ atoms * 0.0017 mol : 1 mol
x = 0.0102 × 10²³ = 1.02 × 10² × 10²³ = 1.02 × 10²⁵ atoms

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A certain half-reaction has a standard reduction potential +0.80 V . An engineer proposes using this half-reaction at the anode

Strike441 [17]

Answer:

a. Minimum 1.70 V

b. There is no maximum.

Explanation:

We can solve this question by remembering that the cell potential is given by the formula

ε⁰ cell = ε⁰ reduction - ε⁰ oxidation

Now the problem states the cell must provide at least 0.9 V and that the reduction potential of the oxidized species 0.80 V, thus

ε⁰ reduction - ε⁰ oxidation ≥ ε⁰ cell

Since ε⁰ oxidation is by definition the negative of ε⁰ reduction , we have

ε⁰ reduction - ( 0.80 V ) ≥ 0.90 V

⇒ ε⁰ reduction ≥ 1.70 V

Therefore,

(a) The minimum standard reduction potential is 1.70 V

(b) There is no maximum standard reduction potential since it is stated in the question that we want to have a cell that provides at leat 0.9 V

8 0

3 years ago

Man's scientific theories are often in correct because his knowledge is

Alik [6]

Answer:

Incomplete

Explanation:

I wouldn't say that a scientific theory is wrong.

It is the best we have at the time, based on all available knowledge .

As each new discovery is made, scientists modify the theory to fit the new knowledge.

The scientific method is a never-ending process.

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3 years ago

An organic compound was extracted into dichloromethane and then the aqueous layer is shaken with saturated sodium chloride solut

Ilya [14]

Answer: option A. to decrease the solubility of the organic product in water

Explanation: sodium chloride solution act as a drying agent to remove water from an organic compound that is in solution. The salt water works to pull the water from the organic layer to the water layer,therby decreasing

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3 years ago

Given the balanced equation representing a reaction:

dimaraw [331]

The answer is (4) synthesis. Synthesis reaction means that two or more reactants combine directly to one production. Substitution or single replacement means that one element of a compound is replaced by another element. Double replacement means that two ionic reactants exchange ions to form two new productions.

3 0

3 years ago

g a solution is made by mixing 500.0 mL of 0.037980.03798 M Na2sO4 Na2sO4 with 500.0 mL of 0.034280.03428 M NaOH NaOH . Complete

Mandarinka [93]

Answer:

The concentration of the sodium and arsenate ions at the end of the reaction in the final solution

[Na⁺] = 0.05512 M

[HAsO₄²⁻] = 0.00185 M

[AsO₄³⁻] = 0.01714 M

Explanation:

Complete Question

A solution is made by 500.0 mL of 0.03798 M Na₂HAsO₄ with 500.0 mL of 0.03428 M NaOH. Complete the mass balance expressions for the sodium and arsenate species in the final solution.

Na₂HAsO₄ + NaOH → Na₃AsO₄ + H₂O

From the information provided in this question, we can calculate the number of moles of each reactant at the start of the reaction and we then determine which reagent is in excess and which one is the limiting reagent (in short supply and determines the amount of products to be formed)

Concentration in mol/L = (Number of moles) ÷ (Volume in L)

Number of moles = (Concentration in mol/L) × (Number of moles)

For Na₂HAsO₄

Concentration in mol/L = 0.03798 M

Volume in L = (500/1000) = 0.50 L

Number of moles = 0.03798 × 0.5 = 0.01899 mole

For NaOH

Concentration in mol/L = 0.03428 M

Volume in L = (500/1000) = 0.50 L

Number of moles = 0.03428 × 0.5 = 0.01714 mole

Since the NaOH is in short supply, it is evident that it is the limiting reagent and Na₂HAsO₄ is in excess.

Na₂HAsO₄ + NaOH → Na₃AsO₄ + H₂O

0.01899 0.01714 0 0 (At time t=0)

(0.01899 - 0.1714) | 0 → 0.01714 0.01714 (end)

0.00185 | 0 → 0.01714 0.01714 (end)

Hence, at the end of the reaction, the following compounds have the following number of moles

Na₂HAsO₄ = 0.00185 mole

This means Na⁺ has (2×0.00185) = 0.0037 mole at the end of the reaction and (HAsO₄)²⁻ has 0.00185 mole at the end of the reaction

NaOH = 0 mole

Na₃AsO₄ = 0.01714 moles

This means Na⁺ has (3×0.01714) = 0.05142 mole at the end of the reaction and (AsO₄)³⁻ has 0.01714 mole at the end of the reaction

H₂O = 0.01714 moles

So, at the end of the reaction

Na⁺ has 0.0037 + 0.05142 = 0.05512 mole

(HAsO₄)²⁻ has 0.00185 mole

(AsO₄)³⁻ has 0.01714 mole.

And since the Total volume of the reaction setup is now 500 mL + 500 mL = 1000 mL = 1 L

Hence, the concentration of the sodium and arsenate ions at the end of the reaction is

[Na⁺] = 0.05512 M

[HAsO₄²⁻] = 0.00185 M

[AsO₄³⁻] = 0.01714 M

Hope this Helps!!!

8 0

3 years ago

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