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uranmaximum [27]
3 years ago
15

If 498 mol of octane combusts, what volume of carbon dioxide is produced at 23.0 °c and 0.995 atm?

Chemistry
1 answer:
MAXImum [283]3 years ago
5 0
Octane on combustion yields CO₂ and H₂O;

                         2C₈H₁₈<span>  +  2 5O</span>₂    →    16 CO₂<span>  +  18 H</span>₂<span>O
</span>
According to equation,

                      2 moles Octane produces  =  16 moles of CO₂ 
So,
              498 moles Octane will produce  =  X moles of CO₂

Solving for X,
                                     X  =  (498 mol × 16 mol) ÷ 2 mol

                                     X  =  3984 moles of CO₂

Now,
Calculating for Volume,
As,
                                    P V  =  n R T
Solving for V,
                                        V  =  n R T / P       ------- (1)

Pressure  =  P  =  0.995 atm 
<span>Volume  =  V  =  ? </span>
<span>Moles  =  n  =  3984 </span>
<span>Gas Constant  =  R  =  0.0821 L.atm.K</span>⁻¹.<span>mol</span>⁻¹<span> </span>
<span>Temperature  =  T  =  23 + 273 = 296 K
</span>
Putting values in eq. 1

                V  =  (3984 mol × 0.0821 L.atm.mol⁻¹.K⁻¹ × 296 K) ÷ 0.995 atm

                                      Volume  =  973040.95 L
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The solubility of acetanilide is 12.8 g in 100 mL of ethanol at 0 ∘C, and 46.4 g in 100 mL of ethanol at 60 ∘C. What is the maxi
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Answer: 72.41% and 26.90% respectively.

Explanation:

At 60°C, you can dissolve 46.4g of acetanilide in 100mL of ethanol. If you lower the temperature, at 0°C, you can dissolve just 12.8g, which means (46.4g-12.8g)=33.6g of acetanilide must have precipitated from the solution.

We can calculate recovery as:

\%R=\frac{crystalized\ mass}{initial\ mass}*100 =\frac{33.6\ g}{46.4\ g}*100=72.41\%

So the answer to the first question is 72.41%.

For the second part just use the same formula, the mass of the precipitate is the final mass minus the initial mass, (171mg-125mg)=46mg.

\%R=\frac{crystalized\ mass}{initial\ mass}*100 =\frac{46\ mg}{171\ mg}*100=26.90\%

So the answer to the second question is 26.90%.

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2 years ago
A sample of krypton gas in a container of volume 1.90 L exerts a pressure of 0.553 atm at 21 Celsius. How many moles of gas are
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Explanation:

pv = nRT

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