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How many F atoms are in 12.15 mol of C2HBrClF3?
In one mole of C₂HBrClF₃, there are three fluorine (F) atoms. Therefore, to find the total number of F atoms in 12.15 mol of C₂HBrClF₃, you multiply the number of moles by the number of F atoms per mole: (12.15 , \text{mol} \times 3 , \text{F atoms/mol} = 36.45 , \text{F atoms}). Thus, there are 36.45 moles of F atoms in 12.15 mol of C₂HBrClF₃.
What is the mass in grams of 3.011 1023 atoms F?
To find the mass of 3.011 × 10²³ atoms of fluorine (F), first note that the molar mass of fluorine is approximately 19.00 g/mol. Since 1 mole contains Avogadro's number (approximately 6.022 × 10²³) of atoms, the number of moles in 3.011 × 10²³ atoms is 3.011 × 10²³ / 6.022 × 10²³ ≈ 0.5 moles. Therefore, the mass is 0.5 moles × 19.00 g/mol = 9.50 grams.
How many atoms of F are in 5.88 mg of ClF3?
To find the number of fluorine (F) atoms in 5.88 mg of ClF3, first calculate the molar mass of ClF3. ClF3 has a molar mass of 83.45 g/mol. Convert 5.88 mg to grams (0.00588 g) and then use the molar mass to find the number of moles of ClF3. Since there are 3 F atoms in each molecule of ClF3, multiply the number of moles by Avogadro's number (6.022 x 10^23) and then by 3 to find the number of F atoms. In this case, there are approximately 4.24 x 10^20 F atoms in 5.88 mg of ClF3.
How many moles of fluorine in 3 moles of borontrifluoride?
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.
What mass of phosphorus pentafluoride PF 5 has the same number of fluorine atoms as 25.0 g of oxygen difluoride OF2?
To find the mass of phosphorus pentafluoride (PF₅) that has the same number of fluorine atoms as 25.0 g of oxygen difluoride (OF₂), we first determine the number of fluorine atoms in OF₂. Each molecule of OF₂ contains 2 fluorine atoms, and the molar mass of OF₂ is approximately 42 g/mol, meaning 25.0 g of OF₂ contains about 1.19 moles (25.0 g / 42 g/mol). Since there are 2 fluorine atoms per molecule of OF₂, this equates to approximately 2.38 moles of fluorine atoms. PF₅ contains 5 fluorine atoms per molecule, so to match 2.38 moles of fluorine atoms, we need 0.476 moles of PF₅ (2.38 moles F / 5). The molar mass of PF₅ is about 130 g/mol, so the mass of PF₅ needed is approximately 61.9 g (0.476 moles × 130 g/mol).
How many F atoms are in 12.15 mol of C2HBrClF3?
In one mole of C₂HBrClF₃, there are three fluorine (F) atoms. Therefore, to find the total number of F atoms in 12.15 mol of C₂HBrClF₃, you multiply the number of moles by the number of F atoms per mole: (12.15 , \text{mol} \times 3 , \text{F atoms/mol} = 36.45 , \text{F atoms}). Thus, there are 36.45 moles of F atoms in 12.15 mol of C₂HBrClF₃.
How many moles are in 1.16 1023 F atoms?
There are 1.93 moles in 1.16 x 10^23 fluorine atoms. This calculation is done by dividing the number of atoms by Avogadro's number (6.022 x 10^23 atoms/mol).
What is the mass of 4 moles of fluorine atoms?
The mass of 4 moles of fluorine F atoms is 151,98 g (because fluorine is a diatomic element).
What is the amount in moles of 1.50 x 1023 atoms F?
Amount of F atoms = (1.50x1023)/(6.02x1023) = 0.249mol Note: F in elemental form exists as diatomic F2 so the amount of fluorine gas would be 0.125mol.
What is the mass in grams for 3.011 X 10 23 atoms F?
1 mole of any element is its atomic weight (from the periodic table) in grams.1 mole of atoms of an element is 6.022 x 1023 atoms (Avogadro's number).1 mole F = 18.9984032g F1 mole F = 6.022 x 1023 atoms FConvert atoms of F to moles F.3.011 x 1023 atoms F x (1mole F/6.022 x 1023 atoms F) = 0.5000 mole FConvert moles F to g F.0.5000 mole F x (18.9984032g F/1mole F) = 9.499g Fe
How many F atoms in 19g of CH2F2?
One mole of CH2F2 has a mass of (12.011)+2(1.0079)+2(18.9984) g = 52.0236 g 19 g of CH2F2 is equivalent to 19/52.0236 moles = 0.3652 moles For every mole of CH2F2 there are 2 F atoms, and 1 mole of a substance has 6.022 x 10^23 entities. So 19 g of CH2F2 has 2(0.3652)(6.022 x 10^23) F atoms = 4.40 x 10^23 F atoms.
How many atoms are present in 4.0 moles?
To determine the number of atoms present in 4.0 moles, you can use Avogadro's constant, which is 6.022 x 10^23 atoms/mol. Therefore, in 4.0 moles, there would be 4.0 x 6.022 x 10^23 atoms, which equals 2.409 x 10^24 atoms. This calculation is based on the concept that one mole of any substance contains Avogadro's number of atoms or molecules.
What is the mass in grams of 3.011 1023 atoms F?
To find the mass of 3.011 × 10²³ atoms of fluorine (F), first note that the molar mass of fluorine is approximately 19.00 g/mol. Since 1 mole contains Avogadro's number (approximately 6.022 × 10²³) of atoms, the number of moles in 3.011 × 10²³ atoms is 3.011 × 10²³ / 6.022 × 10²³ ≈ 0.5 moles. Therefore, the mass is 0.5 moles × 19.00 g/mol = 9.50 grams.
How many moles of molecules are there in 38g of fluorine F2?
To calculate the number of moles of F2 molecules in 38g, we first need to determine the molar mass of F2, which is 38 grams/mol. Next, we can use the formula: moles = mass / molar mass. Therefore, the number of moles in 38g of F2 is 1 mole.
How many atoms of F are in 5.88 mg of ClF3?
To find the number of fluorine (F) atoms in 5.88 mg of ClF3, first calculate the molar mass of ClF3. ClF3 has a molar mass of 83.45 g/mol. Convert 5.88 mg to grams (0.00588 g) and then use the molar mass to find the number of moles of ClF3. Since there are 3 F atoms in each molecule of ClF3, multiply the number of moles by Avogadro's number (6.022 x 10^23) and then by 3 to find the number of F atoms. In this case, there are approximately 4.24 x 10^20 F atoms in 5.88 mg of ClF3.
How many moles of H2SO4 are needed to react with 2.4 moles of Aluminum metal?
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
How many atoms of fluorine are there in 2.5 mol?
1 mole F = 6.022 x 1023 atoms F 2.5mol F x 6.022 x 1023 atoms F/1mol F = 1.5 x 1024 atoms F
