first you find Mr of chlorine which is 17.
Then find moles of chlorine which is mass divided by Mr.. 35.5 divided by 17 equals 2.088 (4sf)
Finally avogadro constant.. 6x10^23 atoms per mole so multiply that by 2.088.
Havent got a calculator so you do it.
Think that's right but havent got calculator so check
the mol mass of chlorine atom is 35,453 g/mol. there are 6,022 141 79 × 1023 atoms in one mol. So one atom of chlorine would weigh around 5.83 x10-23 grams. Chlorine,the molecule, would be twice as heavy since the gas consists out of two chlorine atoms.
If the density of carbon tetrachloride is 1.59 g/L, the volume in L of 4.21 kg of carbon tetrachloride is about 2,647.8 L
When sulfur reacts with chlorine to produce disulfur dichloride, the name of the compound corresponds to a chemical formula of S2Cl2, which shows that the same numbers of atoms of each element are needed to form the compound. The gram atomic mass of sulfur is 32.06, and the gram atomic mass of chlorine is 35.453. 200.2 grams of sulfur corresponds to 200.2/32.06 or about 6.2445 gram atoms of sulfur, while 100.3 grams of chlorine corresponds to 100.3/35.453 or about 2.83 gram atoms of chlorine. Therefore, chlorine is the limiting reactant among these amounts of sulfur and chlorine.
1.26 mol of AlCl3
One chlorine molecule has an atomic mass of 35.453 g/mol. If it is diatomic, that means there are two atoms chemically bonded. Therefore: 2 * 35.453 = 70.906
6,687.1023 chlorine atoms
75,10 g of chlorine = 2,1183 moles
The answer is 6,176.10e23 atoms of chlorine.
two elements and two atoms (potassium and chlorine)In one molecule of KCl, there are two elements (potassium and chlorine).The molecular weight of KCl is 74.55 g / mol.So, 74.55 g of KCl will contain 6.023 x 1023 molecules or 12.046 x 1023 atoms.
The number of atoms of lead is 6,68.10e23.
55.0 g of Cl2 contains 55.0/35.45* or 1.551 gram atoms of chlorine. Each mole of PCl5 requires exactly 5 gram atoms of chlorine, as shown by the formula. Therefore, 1.551/5.000 or 0.310 moles of PCl5 can be formed, to the justified number of significant digits. *This number is the gram Atomic Mass of chlorine.
To determine the number of atoms of chlorine in 445g of lead chloride (PbCl2), you need to calculate the number of moles of PbCl2 in 445g and then multiply it by the number of chlorine atoms in one PbCl2 molecule. First, calculate the number of moles of PbCl2 using the formula: moles = mass / molar mass. The molar mass of PbCl2 is 278.1 g/mol, so moles = 445g / 278.1 g/mol = 1.6 moles. Since there are two chlorine atoms in one molecule of PbCl2, the total number of chlorine atoms is 2 * 1.6 moles = 3.2 moles of chlorine atoms. To convert moles to atoms, multiply by Avogadro's number. Therefore, there are 3.2 moles * 6.022 x 10^23 atoms/mole = 1.9264 x 10^24 atoms of chlorine in 445g of PbCl2.
2 Al + 3Cl2 --Ã 2AlCl3 so 2 moles of Aluminium atoms react with 3 moles of chlorine molecules or 1 atom of Al reacts with 3 atoms of chlorine. Aluminium has an atomic weight of 27 and Chlorine atoms 35.5. Multiplying by Avagadro's number 1 mole of Al atoms reacts with 3 moles of chlorine atoms. So 27 g of Al reacts with 106.5 g Chlorine (3 x 35.5) to produce 133.5 g AlCl3 15g Al = 15/27 = 0.5556 moles (4 decimal places) 2g Cl = 2/35.5 = 0.0563 moles of chlorine atoms but 3 needed per molecule of AlCl3 so 0.0563/3 = 0.0188 (4 decimal places) maximum number of moles of product possible. 0.0188 x 133.5 = 2.51 g maximum yield.
the mol mass of chlorine atom is 35,453 g/mol. there are 6,022 141 79 × 1023 atoms in one mol. So one atom of chlorine would weigh around 5.83 x10-23 grams. Chlorine,the molecule, would be twice as heavy since the gas consists out of two chlorine atoms.
the mol mass of chlorine atom is 35,453 g/mol. there are 6,022 141 79 × 1023 atoms in one mol. So one atom of chlorine would weigh around 5.83 x10-23 grams. Chlorine,the molecule, would be twice as heavy since the gas consists out of two chlorine atoms.
49.1740 g (6.02 x 1023 atoms) / (91.22 g) = 3.25 x 1023 atoms
If the density of carbon tetrachloride is 1.59 g/L, the volume in L of 4.21 kg of carbon tetrachloride is about 2,647.8 L