The molecular mass of fluorine gas, F2 is 2(19.0) = 38.0Amount of F2 = mass of sample/molar mass = 9.5/38.0 = 0.25mol
There are 0.25 moles of fluorine in a 9.5g pure sample.
45 g is equal to 1,184 moles fluorine.
The answer is 1,184 moles.
2.35 grams water (1 mole H2O/18.016 grams) = 0.130 moles water ===============
45g = 0.045kg
Boron trifluoride is BF3. So each mole of BF3 contains 1 moles of boron (B) and 3 moles of fluorine (F). Thus, 3 moles of BF3 contains NINE moles of fluorine.
1 mole F atoms = 6.022 x 1023 F atoms 1.01 x 1023 F atoms x (1mol F atoms/6.022 x 1023 Fe atoms) = 0.168mol Fe
16 grams of oxygen how many moles is 0,5 moles.
2.35 grams water (1 mole H2O/18.016 grams) = 0.130 moles water ===============
(76 g F) * (1mol F/ 19g F) = 4 moles of F
45.
45g = 0.045kg
The answer is two moles of F atoms or 1 mole of F2 molecules.
45g = 1.587oz
45g = 1.59oz
0.8 moles Explanation: from the equation we can see, 2 mole A l is needed to react completely with 3 mole F e O so, 3 moles of F e O needs 2 moles A l so, 1 mole F e O needs 2 3 moles A l so, 1.2 mol F e O needs 2 × 1.2 3 moles A l = 0.8 moles A l
45g = 1.59oz
45g is about 25.39 drams.
1.59 oz.
Boron trifluoride is BF3. So each mole of BF3 contains 1 moles of boron (B) and 3 moles of fluorine (F). Thus, 3 moles of BF3 contains NINE moles of fluorine.