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: 120.9 grams
Explanation:
Given the question : How many grams are in 3.67 moles of Lithium cyanide?
Chemical formula of lithium cyanide = LiCN
Molar mass of LiCN: Li = 6.941 ; C = 12, N = 14
LiCN = (6.941 + 12 + 14). = 32.941 g/mol
(Gram of substance / mole of substance) = (molar mass of substance / one mole)
(Gram of substance / 3.67) = (32.941 / 1)
= gram * 1 = 3.67 * 32.941
Gram of substance = 120.89347
= 120.9 grams
Answer:
How many moles of oxygen are in 2.71 x 1025 molecules of carbon dioxide (CO2)? 90 moles. How many moles are there in 125 grams of carbon? 10.42 moles of carbon.
Explanation:
