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2 years ago

When fluorine gas combines with water vapor, the following reactions occurs.

1 answer:

6 0

Hello!

The Chemical reaction is the following:

2F₂(g) + 2 H₂O(g) → O₂(g) + 4HF(g)

If we assume that all the gases are ideal gases, we can describe the coefficients of this reaction as Volume rather than moles, so we can apply the following equations:

$2 L F_2* \frac{2 L H_2O}{2 L F_2}=2 L H_2O \\ \\ 2 L F_2* \frac{1 mol O_2}{2 L F_2}=1 mol O_2 \\ \\ 2 L F_2* \frac{4 mol HF}{2 mol F_2}= 4 mol HF$

So, 2 L of Water Vapor reacts with the fluorine; and 1 L of Oxygen and 4 L of Hydrogen Fluoride are produced.

Have a nice day!

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<h3><u>moles of H2SO4</u></h3>

Avogadro's number (6.022 × 1023) is defined as the number of atoms, molecules, or "units of anything" that are in a mole of that thing. So to find the number of moles in 3.4 x 1023 molecules of H2SO4, divide by 6.022 × 1023 molecules/mole and you get 0.5646 moles but there are only 2 sig figs in the given so we need to round to 2 sig figs. There are 0.56 moles in 3.4 x 1023 molecules of H2SO4

Note the way this works is to make sure the units are going to give us moles. To check, we do division of the units just like we were dividing two fractions:

(molecules of H2SO4) = (molecules of H2SO4)/1 and so we have 3.4 x 1023/6.022 × 1023 [(molecules of H2SO4)/1]/[(molecules of H2SO4)/(moles of H2SO4)]. Now, invert the denominator and multiply:

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1 year ago

1. NaOH mass of a solution of 200g in which its percentage is 25%. What mass of sulfuric acid solution is needed to completely n

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Answer:

$m_{H2SO4$ = 61.25 g

$m_{Na2SO4}$ = 88.75 g

Explanation:

$m_{NaOH}$ = $\frac{200 . 25 }{100}$ = 50 g

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2 : 1 : 1 : 2

1.25 (moles)

⇒ $n_{H2SO4}$ = 1.25 × 1 ÷ 2 = 0.625 (moles) ⇒ $m_{H2SO4}$ = 0.625 × 98 = 61.25 g

$n_{Na2SO4}$ = 1.25 × 1 ÷ 2 = 0.625 (moles) ⇒ $m_{Na2SO4}$ = 0.625 × 142 = 88.75 g

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Explanation:

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