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3 years ago

How many O atoms are present in 2.0 moles of Na3PO4?

Chemistry

1 answer:

andrezito [222]3 years ago

4 0

Find out how many molecules there are in 2moles

To do this multiply avagadro's constant by 2 (6.02x10^23 * 2)

This gives 1.204x10^24 molecules

For every molecule you have 4 oxygen atoms so do 1.204x10^24 * 4

Answer = 4.816x10^24

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4 years ago

Complete and balance the following equation: NO2−(aq)+Cr2O72−(aq)→Cr3+(aq)+NO3−(aq)(acidic solution) Express your answer as a ne

Georgia [21]

Answer:

Part A: Cr₂O₇²⁻(aq) + 8H+ + 3NO₂⁻(aq) → 2Cr³⁺(aq) + 3NO₃⁻(aq) + 4H₂O.

Part B: 2HNO₃(aq) + 2S(s) + H₂O(l) → N₂O(g) + 2H₂SO₃(aq).

Part C: 2Cr₂O₇²⁻(aq) + 3CH₃OH(aq) + 16H⁺ → 4Cr³⁺(aq) + 3HCO₂H(aq) + 11H₂O.

Part D: 4BrO₃⁻(aq) + 5N₂H₄(aq) + 4H⁺(aq) → 2Br₂(I) + 5N₂(g) + 12H₂O(aq).

Part E: NO₂⁻(aq) + 2Al(s) + OH⁻(aq) + H₂O(aq) → NH₃(aq) + 2AlO₂⁻(aq).

Part F: H₂O₂(aq) + ClO₂(aq) → ClO₂⁻(aq) + O₂(g) + 2H⁺(aq).

Explanation:

Part A: Complete and balance the following equation: NO₂⁻(aq) + Cr₂O₇²⁻(aq) → Cr³⁺(aq) + NO₃⁻(aq) (acidic solution). Express your answer as a net chemical equation including phases.

To balance and write the net chemical equation, we should write the two-half reactions:

The two half reactions are:

The reduction reaction: Cr₂O₇²⁻(aq) + 14H⁺ + 6e → 2Cr³⁺(aq) + 7H₂O.

The oxidation reaction: H₂O(aq) + NO₂⁻(aq) → NO₃⁻(aq) + 2H⁺(aq) + 2e.

Before adding the reactions number of electron gained must be equal to number of electrons lost. we multiply the reduction reaction by 1 (be the same) and the oxidation reaction by 3 (3H₂O(aq) + 3NO₂⁻(aq) → 3NO₃⁻(aq) + 6H⁺(aq) + 6e) to equalize the no. of electrons in the two-half reactions.

Add up both reactions:

Cr₂O₇²⁻(aq) + 14H⁺ + 6e → 2Cr³⁺(aq) + 7H₂O.

3H₂O(aq) + 3NO₂⁻(aq) → 3NO₃⁻(aq) + 6H⁺(aq) + 6e.

So, the net redox reaction will be:

Cr₂O₇²⁻(aq) + 8H⁺ + 3NO₂⁻(aq) → 2Cr³⁺(aq) + 3NO₃⁻(aq) + 4H₂O.

Part B: Complete and balance the following equation: S(s) + HNO₃(aq) → H₂SO₃(aq) + N₂O(g) (acidic solution) Express your answer as a net chemical equation including phases.

To balance and write the net chemical equation, we should write the two-half reactions:

The two half reactions are:

The oxidation reaction: S(s) + 3H₂O(l) → H₂SO₃(aq) + 4H⁺(aq) + 4e.

The reduction reaction: 2HNO₃(aq) + 8H⁺(aq) + 8e → N₂O(g) + 5H₂O(l).

Before adding the reactions number of electron gained must be equal to number of electrons lost. we multiply the oxidation reaction by 2 (2S(s) + 6H₂O(l) → 2H₂SO₃(aq) + 8H⁺(aq) + 8e) and the reduction reaction by 1 (be the same) to equalize the no. of electrons in the two-half reactions.

Add up both reactions:

2S(s) + 6H₂O(l) → 2H₂SO₃(aq) + 8H⁺(aq) + 8e.

2HNO₃(aq) + 8H⁺(aq) + 8e → N₂O(g) + 5H₂O(l).

So, the net redox reaction will be:

2HNO₃(aq) + 2S(s) + H₂O(l) → N₂O(g) + 2H₂SO₃(aq).

Part C: Complete and balance the following equation: Cr₂O₇²⁻(aq) + CH₃OH(aq) → HCO₂H(aq) + Cr³⁺(aq) (acidic solution), Express your answer as a net chemical equation including phases.

To balance and write the net chemical equation, we should write the two-half reactions:

The two half reactions are:

The reduction reaction: Cr₂O₇²⁻(aq) + 14H⁺ + 6e → 2Cr³⁺(aq) + 7H₂O.

The oxidation reaction: CH₃OH(aq) + H₂O(aq) → HCO₂H(aq) + 4H⁺(aq) + 4e.

Before adding the reactions number of electron gained must be equal to number of electrons lost. we multiply the reduction reaction by 2 (2Cr₂O₇²⁻(aq) + 28H⁺ + 12e → 4Cr³⁺(aq) + 14H₂O) and the reduction reaction by 3 (3CH₃OH(aq) + 3H₂O(aq) → 3HCO₂H(aq) + 12H⁺(aq) + 12e) to equalize the no. of electrons in the two-half reactions.

Add up both reactions:

2Cr₂O₇²⁻(aq) + 28H⁺ + 12e → 4Cr³⁺(aq) + 14H₂O.

3CH₃OH(aq) + 3H₂O(aq) → 3HCO₂H(aq) + 12H⁺(aq) + 12e.

So, the net redox reaction will be:

2Cr₂O₇²⁻(aq) + 3CH₃OH(aq) + 16H⁺ → 4Cr³⁺(aq) + 3HCO₂H(aq) + 11H₂O.

Part D: Complete and balance the following equation: BrO₃⁻(aq) + N₂H₄(aq) → Br₂(l) + N₂(g)(acidic solution), Express your answer as a net chemical equation including phases.

To balance and write the net chemical equation, we should write the two-half reactions:

The two half reactions are:

The oxidation reaction: N₂H₄(aq) → N₂(g) + 4e + 4H⁺(aq).

The reduction reaction: 2BrO₃⁻(aq) + 10e + 12H⁺(aq) → Br₂(I) + 6H₂O(aq).

Before adding the reactions number of electron gained must be equal to number of electrons lost. we multiply the oxidation reaction by 5 (5N₂H₄(aq) → 5N₂(g) + 20e + 20H⁺(aq)) and the reduction reaction by 2 (4BrO₃⁻(aq) + 20e + 24H⁺(aq) → 2Br₂(I) + 12H₂O(aq)) to equalize the no. of electrons in the two-half reactions.

Add up both reactions:

5N₂H₄(aq) → 5N₂(g) + 20e + 20H⁺(aq).

4BrO₃⁻(aq) + 20e + 24H⁺(aq) → 2Br₂(I) + 12H₂O(aq).

So, the net redox reaction will be:

4BrO₃⁻(aq) + 5N₂H₄(aq) + 4H⁺(aq) → 2Br₂(I) + 5N₂(g) + 12H₂O(aq).

Very Important Note:

Due to the answer exceeds 5000 character, kindly find the answer of part E and F are in the attached word file with also other prats.

Download docx

6 0

4 years ago

Q: A wind-up toy that is not wound up...

olga55 [171]

Does not have energy

4 0

3 years ago

The two pieces of chlothing have charges and they each other and Electrons move from the wool to

Fantom [35]

? Not quite sure what you mean here.

8 0

3 years ago

Compute the freezing point of this solution:

likoan [24]

Answer:

Freezing point = 1.25

Explanation:

If we increase the concentration of the solution, the concentration of H+ does not change.

Convert 2.5% in to decimal

2.5% = 2.5 ÷100

= 0.025

The freezing point = 0.025 × 50

= 1.25

8 0

3 years ago