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iVinArrow [24]
3 years ago
14

what volume (L) of fluorine gas is required to react with 2.31 g of calcium bromide to form calcium fluoride and bromine gas at

8.19 atm and 35 degrees celsius? (Correct asnwer: 3.57x10^-3) but i got 3.57x10^-2
Chemistry
1 answer:
MrRa [10]3 years ago
5 0

Hey there!:

Balanced equation:

F2(g) + CaBr2(s) ---> CaF2 + Br2

1 mol F2 = 1 mol CaBr2

Calculation:

2.31 g CaBr2 * (1 mol CaBr2 / 199.886 g CaBr2)  * (1 mol F2 / 1 mol CaBr2)

= 0.0115565 moles of F2:

Assuming F2 is an ideal  

gas at these conditions:

P*V = n*R*T

Solving for V:

V = (n *R* T) / P

where :

n = 0.0115565 moles

R = 0.08206 atm·L/mol·K

Temperature in K = 35 + 273.15 => 308.15 K

P = 8.19 atm

Substituting numbers into V = (n x R x T) / P:

V = 0.0115565 * 0.08206 * 308.15 / 8.19

V = 0.29222 / 8.19

V = 0.0357 or 3.57*10⁻² L

You are correct!!!


Hope that helps!

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  4. Take another polident tablet and this time put it into a different cup, and crush it. Set it aside.
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A site in Pennsylvania receives a total annual deposition of 2.688 g/mof sulfate from fertilizer and acid rain. The ratio by mas
sertanlavr [38]

According to the statement

2.12 x 10^4 lbs pounds of CaCO₃ are needed to neutralize this acid

<h3>What is neutralization?</h3>

A chemical reaction in which an acid and a base react quantitatively with each other is known as neutralization or neutralization. In a water reaction, neutralization ensures that there is no excess of hydrogen or hydroxide ions in the solution.

<h3>According to the given information:</h3>

The equation of the neutralization reaction between H2SO4 and CaCO3.

CaCO3 + H2SO4 → CaSO4 + H2CO3

H2CO3 dissociate to water and carbon dioxide.

        CaCO3 + H2SO4 → CaSO4  + H2O + CO2

Now solving for the mass of CaCO3 needed to neutralize the acid.

mass of CaCO3 = 9460 Kg H2SO4  × \frac{1000 \mathrm{~g}}{1 \mathrm{~kg}} \times \frac{1 \mathrm{~mol} \mathrm{H}_2 \mathrm{SO} 4}{98.1 \mathrm{gH}_2 \mathrm{SO}_4} \times \frac{1 \mathrm{~mol} \mathrm{CaCO}\left(\mathrm{O}_3\right.}{1 \mathrm{~mol} \mathrm{H}_2 \mathrm{SO}_4}\times \frac{100.1 \mathrm{~g} \mathrm{CaCO}_3}{1 \mathrm{~mol} \mathrm{CaCO}_3} \times \frac{2.205 \mathrm{lb}}{1000 \mathrm{~g}}

= 21284.56606

mass of CaCO3 =  2.12 x 10^4 lbs

2.12 x 10^4 lbs pounds of CaCO₃ are needed to neutralize this acid.

To know more about neutralization visit:

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