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kogti [31]
2 years ago
15

If the volume occupied by the air in a bicycle pump is 525 cm3, and the pressure changes from 73.2 kPa to 122.5 k.Pa as the pist

on is pushed down, what is the new volume occupied by the air?
Chemistry
1 answer:
Musya8 [376]2 years ago
6 0

Answer:

V_2=313.71\ cm^3

Explanation:

Given that,

Initial volume, V_1=525\ cm^3

The pressure changes from 73.2 kPa to 122.5 k.Pa.

We need to find the new volume occupied by the air. Let it is V₂. It can be calculated using Boyle's law such that,

P_1V_1=P_2V_2\\\\V_2=\dfrac{P_1V_1}{P_2}\\\\V_2=\dfrac{73.2\times 525}{122.5}\\\\V_2=313.71\ cm^3

So, the new volume is 313.71\ cm^3.

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An aqueous solution of 1.29 M ethanol, CH3CH2OH, has a density of 0.988 g/mL. The percent by mass of CH3CH2OH in the solution is
anzhelika [568]

Answer:

6%

Explanation:

Hello, for this case, we assume that the volume of the solution is 1L, thus, the mass is given by using the density as follows:

m_{solution}=1L*\frac{1000 mL}{1L}*\frac{0.988g}{1mL} =988g

Now, the mass of the ethanol:

m_{C_2H_5OH}=(1.29molC_2H_5OH/L*1L)*\frac{46gC_2H_5OH}{1molC_2H_5OH} \\m_{C_2H_5OH}=59.34g

Finally, the by mass percent is:

m/m=\frac{59.34g}{988g}*100\\%

%m=6%

Best regards.

4 0
2 years ago
Use the formula: PV=nRT<br><br> If P=12, V=22.4, R=.0821 and T=100, then what would n=?
Lina20 [59]
PV=nRT \\ (12)(22.4)=n(.0821)(100) \\ 268.8=8.21n \\ 268.8/8.21=8.21n/8.21 \\ n=32.74
n ≈ 32.74
Hope this helps!
7 0
2 years ago
What factors affect the dynamic state of equilibrium in a chemical reaction and how?
yanalaym [24]

Answer:

Only changes in temperature will influence the equilibrium constant K_c. The system will shift in response to certain external shocks. At the new equilibrium Q will still be equal to K_c, but the final concentrations will be different.

The question is asking for sources of the shocks that will influence the value of Q. For most reversible reactions:

  • External changes in the relative concentration of the products and reactants.

For some reversible reactions that involve gases:

  • Changes in pressure due to volume changes.

Catalysts do not influence the value of Q. See explanation.

Explanation:

\displaystyle K_c = {e}^{\Delta G/(R\cdot T)}.

Similar to the rate constant, the equilibrium constant K_c depends only on:

  • \Delta G the standard Gibbs energy change of the reaction, and
  • T the absolute temperature (in degrees Kelvins.)

The reversible reaction is in a dynamic equilibrium when the rate of the forward reaction is equal to the rate of the backward reaction. Reactants are constantly converted to products; products are constantly converted back to reactants. However, at equilibrium Q = K_c the two processes balance each other. The concentration of each species will stay the same.

Factors that alter the rate of one reaction more than the other will disrupt the equilibrium. These factors shall change the rate of successful collisions and hence the reaction rate.

  • Changes in concentration influence the number of particles per unit space.
  • Changes in temperature influence both the rate of collision and the percentage of particles with sufficient energy of reaction.

For reactions that involve gases,

  • Changing the volume of the container will change the concentration of gases and change the reaction rate.

However, there are cases where the number of gases particles on the reactant side and the product side are equal. Rates of the forward and backward reaction will change by the same extent. In such cases, there will not be a change in the final concentrations. Similarly, catalysts change the two rates by the same extent and will not change the final concentrations. Adding noble gases will also change the pressure. However, concentrations stay the same and the equilibrium position will not change.

8 0
3 years ago
Pleassssdseee help meee
SpyIntel [72]
I believe it’s Non polar molecule
8 0
3 years ago
How many moles of potassium nitrate, KNO3 are present in a sample with a mass of 85.2 g?
Bingel [31]
0.843 moles would be the answer
5 0
3 years ago
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