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Neporo4naja [7]

3 years ago

13

A gas sample contained in cylinder equipped with a moveable piston occupied 300.0 mL at a pressure of 20.0 atm. what would be th

e final pressure if the volume were increased to 500.0 mL at constant temperature?

1 answer:

7nadin3 [17]3 years ago

6 0

P1 × V1 = P2 × V2
20 × 300 = P2 × 500
6000 = P2 × 500
12 = P2

The final pressure is 12.0 atm.

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For the following reaction, 4.34 grams of sulfur dioxide are mixed with excess oxygen gas . The reaction yields 3.89 grams of su

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Answer:

5.42g, 71.77%

Explanation:

First, we have to write out the balanced chemical equation. The unbalanced equation can be written as “SO2+O2 -> SO3” and to balance it, we can see that having two mols of SO2 and two mols of SO3 will make each side have the same amount of mols per element on each side. So the balanced chemical equation is “2SO2 + O2 -> 2SO3”

Now, we want to solve for the theoretical yield in grams of SO3. To do this, we have to use dimensional analysis. We convert g SO2 into mols SO2 using the molar mass of the elements. Then we convert mols of SO2 into mols of SO3 using the balanced equation. Once we’ve done that, we can convert mols of SO3 into grams of SO3.

You should know how to look up the molar mass of elements on the periodic table by now. Find the masses and set up the terms so they cancel like so:

$4.34g \times \frac{1mol \: so2}{64.07g \: so2} \times \frac{2 \: mol \: so3}{2 \: mol \: so2} \times \frac{80.07gso3}{1 \: mol \: so3}$

Doing the math, we get 5.42g so3 as the theoretical yield. This is the most amount that you could ever get if the world was a perfect place. But alas, it isn’t and mistakes are gonna happen, so the number is going to be less than that. So the best we can do, is to figure out the percent yield that we got.

In a lab scenario, this was calculated to be 3.89 g as stated by the problem. The percent composition formula is

$| \frac{result}{expected \: result} | \times 100$

and plugging the numbers into it, we get:

$| \frac{3.89}{5.42} | \times 100 = 71.77\%$

make sure to follow the decimal/significant figure rules of your instructor, but only round at the end. My professor didn't care too much thankfully, but some professors do

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3 years ago

The balanced combustion reaction for C6H6 is 2C6H6(l)+15O2(g)⟶12CO2(g)+6H2O(l)+6542 kJ If 8.600 g C6H6 is burned and the heat pr

jenyasd209 [6]

Answer : The final temperature of the water is, $22.5166^oC$

Explanation : Given,

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First we have to calculate the moles of benzene.

$\text{Moles of benzene}=\frac{\text{Mass of benzene}}{\text{Molar mass of benzene}}=\frac{8.600g}{78g/mol}=0.1103mole$

Now we have to calculate the energy of combustion.

The given balanced chemical reaction is:

$2C_6H_6(l)+15O_2(g)\rightarrow 12CO_2(g)+6H_2O(l)+6542 kJ$

According to reaction,

As, 2 moles of benzene gives 6542 kJ of energy on combustion.

So, 0.1103 mole of benzene gives $\frac{6542 kJ}{2}\times 0.1103=360.7913kJ$ of energy on combustion.

Now we have to calculate the final temperature of the water.

Formula used : $q_w=m_w\times c_w\times \Delta T=m_w\times c_w\times (T_{final}-T_{initial})$

where,

$q_w$ = heat released = 360.7913 kJ = 36079.13 J

$m_w$ = mass of water = 5691 g

$c_w$ = specific heat of water= $4.18J/g^oC$

$T_{final}$ = final temperature = ?

$T_{initial}$ = initial temperature = $21^oC$

Now put all the given values in the above formula, we get:

$36079.13J=5691g\times 4.18J/g^oC\times (T_{final}-21^oC)$

$T_{final}=22.5166^oC$

Therefore, the final temperature of the water is, $22.5166^oC$

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