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194.5 g of BCl₃ is present in 1 × 10²⁴ molecules of BCl₃.
Explanation:
In order to convert the given number of molecules of BCl₃ to grams, first we have to convert the molecules to moles.
It is known that 1 moles of any element has 6.022×10²³ molecules.
Then 1 molecule will have moles.
So
Thus, 1.66 moles are included in BCl₃.
Then in order to convert it from moles to grams, we have to multiply it with the molecular mass of the compound.
As it is known as 1 mole contains molecular mass of the compound.
As the molecular mass of BCl₃ will be
Mass of boron is 10.811 g and the mass of chlorine is 35.453 g.
Molar mass of BCl₃ = 10.811+(3×35.453)=117.17 g.
So, 194.5 g of BCl₃ is present in 1 × 10²⁴ molecules of BCl₃.
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