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AnnyKZ [126]
3 years ago
9

A 2 mole sample of F2(g) reacts with excess NaOH(aq) according to the equation above. If the reaction is repeated with excess Na

OH(aq) but with 1 mole of F2(g), which of the following is correct?
Group of answer choices

The amount of OF2(g) produced is doubled.

The amount of OF2(g) produced is halved.

The amount of NaF(aq) produced remains the same.

The amount of NaF(aq) produced is doubled.
Chemistry
1 answer:
77julia77 [94]3 years ago
6 0

Answer:

The amount of NaF produced is doubled.

(d) is correct option.

Explanation:

Given that,

A 2 mole sample of F₂ reacts with excess NaOH according to the equation.

The balance equation is

2F_{2}+2NaOH\Rightarrow 2NaF +H_{2}O+OF_{2}

If the reaction is repeated with excess NaOH but with 1 mole of F₂

The balance equation is

F_{2}+2NaOH\Rightarrow 2NaF +2OH

Hence, The amount of NaF produced is doubled.

(d) is correct option.

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erma4kov [3.2K]

Answer:

Explanation:

Fe⁺²(aq) + ClO₂(aq) → Fe⁺³(aq) + ClO₂⁻(aq)

Here oxidation number of Fe is increased from +2 to +3 , so Fe is oxidised .

The oxidation number of Cl is reduced from + 4 to +3  so Cl is reduced .

So ClO₂(aq) is oxidising agent and Fe⁺²(aq) is reducing agent .

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3 years ago
You need to make an aqueous solution of 0.207 M calcium acetate for an experiment in lab, using a 300 mL volumetric flask. How m
jeka94

Answer:

We have to add 9.82 grams of calcium acetate

Explanation:

Step 1: Data given

Molarity of the calcium acetate solution = 0.207 M

Volume = 300 mL = 0.300 L

Molar mass calcium acetate = 158.17 g/mol

Step 2: Calculate moles calcium acetate

Moles calcium acetate = molarity * volume

Moles calcium acetate = 0.207 M * 0.300 L

Moles calcium acetate = 0.0621 moles

Step 3: Calculate mass calcium acetate

Mass calcium acetate = moles * molar mass

Mass calcium acetate = 0.0621 moles * 158.17 g/mol

Mass calcium acetate = 9.82 grams

We have to add 9.82 grams of calcium acetate

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Answer:

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Explanation:

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