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17.03.2020 •
Chemistry
10) How many grams are there in 1.00 x 10^24 molecules of BC13?
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Ответ:
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![1*10^{24} molecules = \frac{1*10^{24} }{6.022*10^{23} } =1.66 moles](/tpl/images/0550/9282/2259d.png)
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.
Ответ:
0,25 g/cm3 creo
Explanation:densidad es masa entre volumen