2. What pressure is required to compress 196.0 L of air at 1.00 atmosphere into a cylinder
1 answer:
Answer:
<h2>7.54 atm </h2>Explanation:
The required pressure can be found by using the formula for Boyle's law which is
where
P1 is the initial pressure
P2 is the final pressure
V1 is the initial volume
V2 is the final volume
From the question we have
We have the final answer as
<h3>7.54 atm </h3>Hope this helps you
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= 4.8 mole * 98 g/mole = 470g
m(H2SO4) which is =<span>470g
</span><span>m(solution) = m(H2SO4) + m(solvent) = 470 + 1000 = 1470 g
d(solution) = m(solution) / V(solution) =>
=> 1.249 g/mL = 1470 g / V(solution) =></span>
According to that: </span>
= 4.8 mole * 98 g/mole = 470g
</span><span>m(solution) = m(H2SO4) + m(solvent) = 470 + 1000 = 1470 g
d(solution) = m(solution) / V(solution) =>
=> 1.249 g/mL = 1470 g / V(solution) =></span>