1.45 g Cr x (1 mol Cr/52.0 g Cr) x (6.022x 1023 atoms Cr/1 mol Cr) = 1.68 x 1022 atoms
1.45 grams Chromium (1mol Cr/52.00g )(6.022 X 10^23/1mol Cr ) = 1.68 X 10^22 atoms Chromium.
Which sample contains the greatest number of atoms. A sample of Mn that contains 3.29E+24 atoms or a 5.18 mole sample of I?The sample of _____ contains the greatest number of atoms.Answer:In order to compare the two samples, it is necessary to express both quantities in the same units. Since the question was phrased in terms of atoms, it is convenient to convert moles of I to atoms of I.The conversion factor between atoms and moles is Avogadro's number: 6.02 x 1023 "things" / molTo convert 5.18 moles of I to atoms of I:atoms I= 5.18 mol I6.02 x 1023 atoms I = 3.12E+24 atoms I1 mol IMultiply by atoms per mole. Moles cancel out.The sample of Mn contains 3.29E+24 atoms.Since 3.12E+24 is smaller than 3.29E+24, the sample of Mn contains the greatest number of atoms.
both. the chromium metal would form an ionic bond, and chromium nitrate is molecular because it contains atoms of different elements.
A "mole" of chromium is generally regarded as consisting of Avogadro's Number of atoms. Therefore, 4.37 X 1023 atoms constitutes (4.37 X 1023)/(6.022 X 1023) or 0.726 moles.
This depends on the mass of the gold sample.
Two Chromium Atoms, and 3 Oxygen Atoms. (Cr2O3)
Take the actual sample weight of 13grams, and divide it by the atomic weight of chromium. This gives you your molar percentage of atoms. Now multiply this molar percentage by Avogadro's constant, the number of atoms in one mole, and this will give you your number of atoms in the sample.
Which sample contains the greatest number of atoms. A sample of Mn that contains 3.29E+24 atoms or a 5.18 mole sample of I?The sample of _____ contains the greatest number of atoms.Answer:In order to compare the two samples, it is necessary to express both quantities in the same units. Since the question was phrased in terms of atoms, it is convenient to convert moles of I to atoms of I.The conversion factor between atoms and moles is Avogadro's number: 6.02 x 1023 "things" / molTo convert 5.18 moles of I to atoms of I:atoms I= 5.18 mol I6.02 x 1023 atoms I = 3.12E+24 atoms I1 mol IMultiply by atoms per mole. Moles cancel out.The sample of Mn contains 3.29E+24 atoms.Since 3.12E+24 is smaller than 3.29E+24, the sample of Mn contains the greatest number of atoms.
both. the chromium metal would form an ionic bond, and chromium nitrate is molecular because it contains atoms of different elements.
A "mole" of chromium is generally regarded as consisting of Avogadro's Number of atoms. Therefore, 4.37 X 1023 atoms constitutes (4.37 X 1023)/(6.022 X 1023) or 0.726 moles.
Chromate is an anion which contains 4 atoms of oxygen, and 1 of Chromium
Which sample contains the greatest number of atoms. A sample of Al that contains 8.18E+23 atoms or a 5.16 mole sample of S?The sample of______ contains the greatest number of atoms.Answer:In order to compare the two samples, it is necessary to express both quantities in the same units. Since the question was phrased in terms of atoms, it is convenient to convert moles of S to atoms of S.The conversion factor between atoms and moles is Avogadro's number: 6.02 x 1023 "things" / molTo convert 5.16 moles of S to atoms of S:atoms S=5.16 mol S6.02 x 1023 atoms S= 3.11E+24 atoms S1 mol SMultiply by atoms per mole. Moles cancel out.The sample of Al contains 8.18E+23 atoms.Since 3.11E+24 is larger than 8.18E+23, the sample of S contains the greatest number of atoms.
This depends on the mass of the gold sample.
Two Chromium Atoms, and 3 Oxygen Atoms. (Cr2O3)
150 (50 x 3)
a chromium atom is a type of atom! it is also known as (Cr)
Whatever be the substance the one gram mole of that substance would have 6.023 x 1023 atoms or molecules or ions in it. Hence to get the mole just divide the number given by 6.023 x 1023
This is some easy math: chromium (Cr, #24) has an atomic mass of 52 grams per mole (from the periodic table). According to the Avogadro constant, 1 mole of any substance contains 6.02x1023 representative particles. So, 52g of Cr is 1 mole, and contains 6.02x1023 atoms.