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zlopas [31]
3 years ago
9

Calculate the Empirical Formula for the following compound: 0.300 mol of S and 0.900 mole of O.

Chemistry
1 answer:
ziro4ka [17]3 years ago
6 0

Answer:

\boxed {\boxed {\sf SO_3}}

Explanation:

An empirical formula shows the smallest whole-number ratio of the atoms in a compound.

So, we must calculate this ratio. Since we are given the amounts of the elements in moles, we can do this in just 2 steps.

<h3>1. Divide </h3>

The first step is division. We divide the amount of moles for both elements by the <u>smallest amount of moles</u>.

There are 0.300 moles of sulfur and 0.900 moles of oxygen. 0.300 is smaller, so we divide both amounts by 0.300

  • Sulfur: 0.300/0.300= 1
  • Oxygen: 0.900/0.300= 3

<h3>2. Write Empirical Formula</h3>

The next step is writing the formula. We use the numbers we just found as the subscripts. These numbers go after the element's symbol in the formula. Remember sulfur is S and there is 1 mole and oxygen is O and there are 3 moles.

  • S₁O₃

This formula is technically correct, but we typically remove subscripts of 1 because no subscript implies 1 representative unit.

  • SO₃

\bold {The \  empirical \ formula \ for \ the \  compound \ is  \ SO_3}}

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Igoryamba

Explanation:

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A Cu2+ solution is prepared by dissolving a 0.4749 g piece of copper wire in acid. The solution is then passed through a Walden
Luda [366]

Answer:

Concentration of Cr_2O_7^{2-} = 0.03101 M

Concentration of MnO_4^- = 0.03721 M

Explanation:

A)

The reduction for Cr_2O_7^{2-} is;

Cr_2O_7^{2-} + 14 H ^+ _{(aq)}  + 6 e^- -----> 2 Cr^{3+} _{(aq)}+7H_2O _{(l)}

Cu^+_{(aq)} -----> Cu^{2+} _{(aq)} + 1 e^-

6 moles of Cu ^+ = 1 mole of Cr_2O_7^{2-}

number of moles of Cu reacted = \frac{mass \ of \ Cu \ wire }{ molecular weigh tof \ Cu wire }

number of moles of Cu reacted = \frac{0.4749}{63.55}

number of moles of Cu reacted = 0.00747 mole

number of moles of Cr_2O_7^{2-}reacted = \frac{0.00747}{6}

number of moles of Cr_2O_7^{2-}reacted = 0.001245 mole

Concentration of Cr_2O_7^{2-} = \frac{number \ of moles }{Volume}

Given that the volume = 40.15 mL = 40.15 *10^{-3}; we have:

Concentration of Cr_2O_7^{2-} = \frac{0.001245}{40.15*10^{-3}}

Concentration of Cr_2O_7^{2-} = 0.03101 M

B)

The reduction for MnO_4^- is;

MnO_4^- + 8H^+ + 5 e^- -----> Mn^{2+} + 4H_2O

Cu^+_{(aq)} -----> Cu^{2+} _{(aq)} + 1 e^-

5 moles of Cu ^+ = 1 mole of Cr_2O_7^{2-}

number of moles of Cu reacted = \frac{mass \ of \ Cu \ wire }{ molecular weigh tof \ Cu wire }

number of moles of Cu reacted = \frac{0.4749}{63.55}

number of moles of Cu reacted = 0.00747 mole

number of moles of MnO_4^- reacted = \frac{0.00747}{5}

number of moles of MnO_4^- reacted = 0.001494 mole

Concentration of MnO_4^- = \frac{number \ of moles }{Volume}

Given that the volume = 40.15 mL = 40.15 *10^{-3}; we have:

Concentration of MnO_4^- = \frac{0.001494 }{40.15*10^{-3}}

Concentration of MnO_4^- = 0.03721 M

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According to the ideal gas equation

PV = nRT

where

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In the problems there are two identical cylinders means, V₁ = V₂

The pressure inside each of two identical cylinders is equal to atmospheric pressure means, P₁ = P₂

Both gases are at the same temperature, means T₁ = T₂

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\frac{P_1 V_1}{P_2 V_2} \:=\: \frac{n_1 T_1}{n_2 T_2}

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So the total amount of hydrogen gas is equal to total amount of ideal nitrogen gas.

Learn more about the ideal gas equation here: brainly.com/question/27870704

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