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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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Answer:

(a) adding 0.050 mol of HCl

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A buffer is defined as the mixture of a weak acid and its conjugate base -or vice versa-.

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Thus:

<em>(a) adding 0.050 mol of HCl:</em> The addition of 0.050moles of HCl produce the reaction of 0.050 moles of NaF producing HF. That means after the reaction, all NaF is consumed and you will have in solution just the weak acid <em>destroying the buffer</em>.

(b) adding 0.050 mol of NaOH: The NaOH reacts with HF producing more NaF. Would be consumed just 0.050 moles of HF -remaining 0.050 moles of HF-. Thus, the buffer <em>wouldn't be destroyed</em>.

(c) adding 0.050 mol of NaF: The addition of conjugate base <em>doesn't destroy the buffer</em>

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How much heat is evolved in converting 1.00 mol of steam at 155.0 ∘c to ice at -50.0 ∘c? the heat capacity of steam is 2.01 j/(g
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C is the specific heat J/g.°C

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Q = Mw *[ ( Csteam * ΔTsteam)+(Cw*ΔTw) + (Cice  * ΔT ice)]

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