11.0 kg = (11.0 kg)(1000 g/kg) = 11000 g
(11000 g)/(1400 cm3) = 7.857 g/cm3
Simplified = 7.86 g/cm3
Answer:
42.29 g of BrF₃.
Explanation:
Molar mass of OF2 = 16 + (2 * 19)
= 54 g/mol
Number of moles of OF₂ = mass / molar mass
= 25/54
= 0.46 mol.
Number of OF₂ molecules = number of moles * avogadros constant
= 0.46 * 6.022 x 10²³
= 2.8 x 10^23 molecules of OF2.
Since each OF₂ molecule has 2 Fluorine atoms,
Number of Fluorine atoms in 25.0 g of OF₂ = 2 x 2.8 x 10²³
= 5.576 x 10²³ atoms of Fluorine.
Since 1 BrF₃ molecule has 3 Fluorine atoms,
number of BrF₃ molecules = Number of Fluorine atoms / 3
= 5.576 x 10²³ / 3
= 1.8587 x 10²³ molecules of BrF₃.
Number of moles of BrF₃ = number of molecules of BrF₃ / Avogadros constant
= (1.8587 x 10²³) / (6.022 x 10²³ )
= 0.30865 moles of BrF₃.
Molar mass of BrF₃ = 80 + (19 * 3)
= 137 g/mol
mass of BrF₃ = number of moles * molar mass
= 0.30865 * 137
= 42.29 g of BrF₃.
Answer:
Explanation:
The relation between average velocity of a gas molecules and its molecular mass is as follows .
average velocity v
where k is a constant .
If v₁ and v₂ be the velocities corresponding to molecular masses 349 and 352 .
= 1.004
