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Aleonysh [2.5K]
3 years ago
6

A 5 mL sample of an unknown liquid is placed into a 250 mL flask that has had all of the air removed. The pressure measured in t

he flask is found to be 34.5kPa. In a seperate trial the student puts 214 mL of the same liquid into the same flask at the same conditions. What will the pressure measure (in kPa) in the second trial
Chemistry
1 answer:
Brrunno [24]3 years ago
8 0

Answer:

See explanation below

Explanation:

In this case, we can use the Boyle's law. Assuming that the temperature of both trial remains constant, then:

P₁V₁ = P₂V₂    (1)

You should note that this expression is usable when you are dealing with gases. However, we can treat this unknown liquid as a gas, because all the air on the flask is removed, and we can assume that the liquid can behave like an ideal gas.

So using the above expression, we can solve for P₂:

P₂ = P₁V₁ / V₂    (2)

In this case, we already have the values of presures and volume, so replacing in this expression:

P₂ = 34.5 * 5 / 214

<h2>P₂ = 0.806 kPa</h2>

This should be the pressure of the liquid.

Hope this helps

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Give the spectator ions for the reaction that occurs when aqueous solutions of H 2SO 4 and KOH are mixed.
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The spectator ions is: K^+   and  SO^{2-}_4

Explanation:

The equation of reaction between  H₂ SO₄ and KOH is:

H_2SO_{4(aq)} + 2KOH _{aq} \to K_2SO_{4(aq)} +2H_2O _{(l)}

Rewriting this equation as ionic;

[2H^{+} + SO^{2-}_4 + 2K^+ +2OH^- \to 2K^+  SO_4^{2-} + 2H_2O ]

Spectators ions are ions present on both sides of the ionic equation by the same quantity but do not take part in the net reaction.

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How many total atoms are in 0.290 g of P2O5?
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Step 1. Convert <em>grams of P₂O₅ to moles of P₂O₅</em>.

\text{Moles of P}_{2}\text{O}_{5} = \text{0.290 g } \text{P}_{2}\text{O}_{5} \times \frac{\text{1 mol }\text{P}_{2}\text{O}_{5}}{\text{141.94 g }\text{P}_{2}\text{O}_{5}} = \text{0.002 043 mol } \text{P}_{2}\text{O}_{5} \\

Step 2. Convert <em>moles of P₂O₅ to molecules of P₂O₅</em>.

\text{Molecules of } \text{P}_{2}\text{O}_{5} = \text{0.002 043 mol } \text{P}_{2}\text{O}_{5} \times \frac{6.022 \times 10^{23}\text{ molecules }\text{P}_{2}\text{O}_{5}}{\text{1 mol } \text{P}_{2}\text{O}_{5}}\\

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Step 3. Convert <em>molecules of P₂O₅ to atoms</em>.

There are seven atoms in 1 mol P₂O₅.

∴ \text{Total atoms} = 1.23\times 10^{21 }\text{ molecules }\text{P}_{2}\text{O}_{5} \times\frac{\text{7 atoms}}{\text{1 molecule }\text{P}_{2}\text{O}_{5}} = 8.61 \times 10^{21}\text{ atoms}\\

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