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attashe74 [19]
3 years ago
8

chemistry A basketball is inflated to a pressure of 1.10 atm in a 28.0°C garage. What is the pressure of the basketball outside

where the temperature is -2.00°C?
Chemistry
1 answer:
butalik [34]3 years ago
7 0

Answer : The final pressure of the basketball is, 0.990 atm

Explanation :

Gay-Lussac's Law : It is defined as the pressure of the gas is directly proportional to the temperature of the gas at constant volume and number of moles.

P\propto T

or,

\frac{P_1}{P_2}=\frac{T_1}{T_2}

where,

P_1 = initial pressure = 1.10 atm

P_2 = final pressure = ?

T_1 = initial temperature = 28.0^oC=273+28.0=301.0K

T_2 = initial temperature = -2.00^oC=273+(-2.00)=271.0K

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

\frac{1.10atm}{P_2}=\frac{301.0K}{271.0K}

P_2=0.990atm

Thus, the final pressure of the basketball is, 0.990 atm

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Na₃PO₄ is sodium phosphate.  

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The acid-dissociation constants of HC3H5O3 and CH3NH3+ are given in the table below. Which of the following mixtures is a buffer
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Explanation:

The pH of a buffer solution is calculated using following relation

pH=pKa+log(\frac{salt}{acid} )

Thus the pH of buffer solution will be near to the pKa of the acid used in making the buffer solution.

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A mixture of 100. mL of 0.1 M HC3H5O3 and 50. mL of NaOH

Half of the acid will get neutralized by the given base and thus will result in equal concentration of both the weak acid and the salt making the pH just equal to the pKa value.

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Answer:

2.7 x 10^-19 J

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