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How do you calculate mass fractions of calcium and fluorine in fluorite?
To calculate the mass fractions of calcium and fluorine in fluorite (CaF₂), first determine the molar masses of calcium (Ca, approximately 40.08 g/mol) and fluorine (F, approximately 19.00 g/mol). The formula for fluorite shows that one mole of calcium combines with two moles of fluorine, giving a total molar mass of CaF₂ as 40.08 g/mol + 2 × 19.00 g/mol = 78.08 g/mol. The mass fraction of calcium is calculated by dividing the molar mass of calcium by the total molar mass (40.08 g/mol / 78.08 g/mol), and the mass fraction of fluorine is calculated by dividing the total mass of fluorine (2 × 19.00 g/mol) by the total molar mass (38.00 g/mol / 78.08 g/mol).
What is fluorine's ionization energy?
The first ionization energy is 1681 kJ/mol.
Is fluorine lighter than helium?
NO, sure not. Helium (He) mass = 4 g/mol Fluorine gas (F2) mass = 2*19 = 38 g/mol (almost 10 times heavier), although fluorine gas hardly can be held stable: it is so VERY rective, even in air!!
What is the formula mass of barium flouride?
The formula mass of barium fluoride (BaF2) is calculated by adding the atomic masses of each element in the compound. Barium has an atomic mass of 137.33 g/mol, and fluorine has an atomic mass of 18.998 g/mol. Multiplying the atomic mass of barium by 1 and the atomic mass of fluorine by 2 (because there are two fluorine atoms in the compound) and adding them together gives a formula mass of 175.33 g/mol.
What is the formula weight of calcium fluoride?
To calculate the formula weight of calcium fluoride (CaF2), you would add the atomic weights of calcium (Ca) and two atoms of fluorine (F). The atomic weight of calcium is 40.08 g/mol and the atomic weight of fluorine is 18.998 g/mol. Therefore, the formula weight of calcium fluoride is 40.08 + 2(18.998) = 78.076 g/mol.
1 mole of fluorine has how many quantity?
1 mole is itself, a quantity. But, if you like, the quantity is 6.022140857*10^23 atoms of fluorine.
What is the weight of fluorine?
Fluorine (F) has an atomic weight of 18.9984032 g/mol
If you have 85.5 grams of fluorine which has a molar mass approxidmatley 19 g mol how many moles of fluorine do you have?
To calculate the number of moles of fluorine, divide the given mass (85.5 grams) by the molar mass of fluorine (19 g/mol). Number of moles of fluorine = 85.5 g / 19 g/mol ≈ 4.5 moles.
What is the cost of fluorine for 1 gram?
The cost of fluorine can vary depending on the supplier and quantity purchased. As of now, the price of fluorine is approximately $0.07 per gram.
How many atoms are in 25.9 grams of fluorine?
25.9 grams fluorine (1 mole F/19.0 grams)(6.022 X 1023/1 mole F) = 8.21 X 1023 atoms of fluorine ======================
How many moles in 5.67 grams of fluorine gas?
To find the number of moles in 5.67 grams of fluorine gas, you need to divide the mass by the molar mass of fluorine. The molar mass of fluorine is approximately 19 g/mol. Therefore, 5.67 grams of fluorine gas is equal to 5.67 g / 19 g/mol ≈ 0.298 mol.
How do you calculate mass fractions of calcium and fluorine in fluorite?
To calculate the mass fractions of calcium and fluorine in fluorite (CaF₂), first determine the molar masses of calcium (Ca, approximately 40.08 g/mol) and fluorine (F, approximately 19.00 g/mol). The formula for fluorite shows that one mole of calcium combines with two moles of fluorine, giving a total molar mass of CaF₂ as 40.08 g/mol + 2 × 19.00 g/mol = 78.08 g/mol. The mass fraction of calcium is calculated by dividing the molar mass of calcium by the total molar mass (40.08 g/mol / 78.08 g/mol), and the mass fraction of fluorine is calculated by dividing the total mass of fluorine (2 × 19.00 g/mol) by the total molar mass (38.00 g/mol / 78.08 g/mol).
How many grams of fluorine are needed to react with 6.20 g of phosphorus?
Find number of moles of phosphorusmolar mass of phosphorus is 30.97g/moldivide mass by molar mass to get the moles of phosphorus please that your molar mass mast be multiplied by 4 because you have four phosphorus atoms. 30.97*4= 123.88 g/moln= m/M = 62.0g/123.88g/mol= 0.50 molesTheoretically, there are 6 times the number of moles in fluorine than there are in phosphorus because of the ratio 1:6 but fluorine is diatomic gas so we multiply the number of moles by 12 instead of 6. 6 from the ratio times from fluorine being diatomic gas.So 0.5*12=6 moles of fluorine. we multiply that by the molar mass of fluorine multiplied by 12 because we 12fluorine atoms.So, 6= m/228g/mol so to find mass of fluorine we multiply moles times molar mass it gives us 1368g.
What is fluorine heat of fusion?
The heat of fusion of fluorine is approximately 2.62 kJ/mol. This value represents the amount of energy required to convert 1 mole of solid fluorine into liquid fluorine at its melting point while keeping the temperature constant.
How many fluorine atoms are present in 5.59 g of C2F4?
To find the number of fluorine atoms in 5.59 g of C2F4, start by calculating the molar mass of C2F4: 12.01 g/mol (C) + 2(19.00 g/mol) (F) = 70.01 g/mol. Then divide the given mass by the molar mass to get the number of moles: 5.59 g / 70.01 g/mol ≈ 0.08 moles. Since there are 4 fluorine atoms in 1 molecule of C2F4, multiply the number of moles by Avogadro's number (6.022 x 10^23) and then by 4 to get the number of fluorine atoms: 0.08 mol x 6.022 x 10^23 atoms/mol x 4 = approximately 1.93 x 10^23 fluorine atoms.
How many grams of iodine are produced when 0.72 mol of fluorine reacts with potassium iodide KI 2 KI F2 -- 2 KF I2?
From the balanced chemical equation, 2 moles of potassium iodide (KI) react with 1 mole of fluorine (F2) to produce 2 moles of iodine (I2). Therefore, when 0.72 mol of fluorine reacts, it produces 0.72 mol/2 = 0.36 mol of iodine. To convert this to grams, you would multiply the moles by the molar mass of iodine (I2), which is approximately 253.8 g/mol. So, 0.36 mol * 253.8 g/mol ≈ 91.4 grams of iodine are produced.
What is the molar mass of calcium fluoride?
The formula is CaF2 so, its three atoms, one calcium and two fluorine atoms. add the molar mass of calcium(40 g/mol) and fluorine(19 g/mol X 2 because there are two atoms of fluorine) and you get 40+(19x2)=78 78 grams per mol
