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AfilCa [17]
3 years ago
7

You have a 1.153 g sample of an unknown solid acid, HA, dissolved in enough water to make 20.00 mL of solution. HA reacts with K

OH(aq) according to the following balanced chemical equation:?
Chemistry
1 answer:
seropon [69]3 years ago
7 0

Answer:

HA +  KOH  →  KA  +  H₂O

Explanation:

The unknown solid acid in water can release its proton as this:

HA  +  H₂O  →  H₃O⁺  +  A⁻

As we have the anion A⁻, when it bonded to the cation K⁺, salt can be generated, so the reaction of HA and KOH must be a neutralization one, where you form water and a salt

HA +  KOH  →  KA  +  H₂O

It is a neutralization reaction because H⁺ from the acid and OH⁻ from the base can be neutralized as water

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If a gas occupies 4600 mL at 0.9 atm and 195°C, what is the new volume in ml
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Answer:

The new volume is 2415 mL

Explanation:

The STP conditions refer to the standard temperature and pressure. Pressure values at 1 atmosphere and temperature at 0 ° C are used and are reference values for gases.

Boyle's law says that the volume occupied by a given gas mass at constant temperature is inversely proportional to the pressure and is expressed mathematically as:

P * V = k

Charles's law is a law that says that when the amount of gas and pressure are kept constant, the ratio between volume and temperature will always have the same value:

\frac{V}{T} =k

Gay-Lussac's law indicates that when there is a constant volume, as the temperature increases, the gas pressure increases. And when the temperature is decreased, the gas pressure decreases. This can be expressed mathematically in the following way:

\frac{P}{T} =k

Combined law equation is the combination of three gas laws called Boyle's, Charlie's and Gay-Lusac's law:

\frac{P*V}{T} =k

Having two different states, an initial state and an final state, it is true:

\frac{P1*V1}{T1} =\frac{P2*V2}{T2}

In this case:

  • P1= 0.9 atm
  • V1=4,600 mL= 4.6 L (being 1 L=1,000 mL)
  • T1= 195 °C= 468 °K (being 0°C=273°K)

The final state 2 is in STP conditions:

  • P2= 1 atm
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Replacing:

\frac{0.9 atm*4.6L}{468K} =\frac{1 atm*V2}{273K}

Solving:

V2=\frac{0.9 atm*4.6L}{468K}*\frac{273K}{1 atm}

V2= 2.415 L =2,415 mL

<u><em>The new volume is 2415 mL</em></u>

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