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3 years ago

Which formula is an empirical formula?

Chemistry

2 answers:

marta [7]3 years ago

7 0

the answer is B......

poizon [28]3 years ago

6 0

<h2>Hello!</h2>

The answer is:

The empirical formula is the option B. $NH_{3}$

The empirical formula of a compound is the simplest formula that can be written. On the opposite, the molecular formula involves a variant of the same compound, but it can be also simplified to an empirical formula.

$MolecularFormula=n(EmpiricalFormula)$

We are looking for a formula that cannot be simplified by dividing the number of molecules/atoms that conforms the compound.

Let's discard option by option in order to find which formula is an empirical formula (cannot be simplified)

A. $N_{2}O_{4}$

It's not an empirical formula, it's a molecular formula since it can be obtained by multiplying the empirical formula of the same compound.

$N_{2}O_{4}=2(NO_{2})$

B. $NH_{3}$

It's an empirical formula since it cannot be obtained by the multiplication of a whole number and the simplest formula. It's the simplest formula that we can find of the compound.

C. $C_{3}H_{6}$

It's not an empirical formula, it's a molecular formula since it can be obtained by multiplying the empirical formula of the same compound.

$C_{3}H_{6}=3(CH_{2})$

D. $P_{4}O_{10}$

It's not an empirical formula, it's a molecular formula since it can be obtained by multiplying the empirical formula of the same compound.

$P_{4}O_{10}=2(P_{2}O_{5})$

Hence, the empirical formula is the option B. $NH_{3}$

Have a nice day!

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1.8 × 10² s

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Calculate the mass, in grams, of Ag2CrO4 that will precipitate when 50.0mL of 0.20M AgNO3 solution is mixed with 40.0mL of 0.10M

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Answer:

1.327 g Ag₂CrO₄

Explanation:

The reaction that takes place is:

2AgNO₃(aq) + K₂CrO₄(aq) → Ag₂CrO₄(s) + 2KNO₃(aq)

First we need to <em>identify the limiting reactant</em>:

We have:

0.20 M * 50.0 mL = 10 mmol of AgNO₃

0.10 M * 40.0 mL = 4 mmol of K₂CrO₄

If 4 mmol of K₂CrO₄ were to react completely, it would require (4*2) 8 mmol of AgNO₃. There's more than 8 mmol of AgNO₃ so AgNO₃ is the excess reactant. <em><u>That makes K₂CrO₄ the limiting reactant</u></em>.

Now we <u>calculate the mass of Ag₂CrO₄ formed</u>, using the <em>limiting reactant</em>:

4 mmol K₂CrO₄ * $\frac{1mmolAg_2CrO_4}{1mmolK_2CrO_4} *\frac{331.73mg}{1mmolAg_2CrO_4}$ = 1326.92 mg Ag₂CrO₄

1326.92 mg / 1000 = 1.327 g Ag₂CrO₄

7 0

3 years ago