Answer :
Answer is: there are 9,7·10²² atoms of oxygen.
m(Na₂Cr₂O₇) = 6,00 g.
n(Na₂Cr₂O₇) = m(Na₂Cr₂O₇) ÷ M(Na₂Cr₂O₇).
n(Na₂Cr₂O₇) = 6 g ÷ 262 g/mol.
n(Na₂Cr₂O₇) = 0,023 mol.
n(Na₂Cr₂O₇) : n(O) = 1 : 7.
n(O) = 7 · 0,023 mol.
n(O) = 0,161 mol.
N(O) = 0,161 mol · 6,022·10²³ 1/mol.
N(O) = 9,7·10²².
m(Na₂Cr₂O₇) = 6,00 g.
n(Na₂Cr₂O₇) = m(Na₂Cr₂O₇) ÷ M(Na₂Cr₂O₇).
n(Na₂Cr₂O₇) = 6 g ÷ 262 g/mol.
n(Na₂Cr₂O₇) = 0,023 mol.
n(Na₂Cr₂O₇) : n(O) = 1 : 7.
n(O) = 7 · 0,023 mol.
n(O) = 0,161 mol.
N(O) = 0,161 mol · 6,022·10²³ 1/mol.
N(O) = 9,7·10²².
Final answer:
approximately 9.69 × 10^23 oxygen atoms in 6.00 g of sodium dichromate.
Explanation:
to calculate the number of moles of Na2Cr2O7 in 6.00 g and then use this information to find the number of oxygen atoms. First, calculate the molar mass of Na2Cr2O7. Sodium (Na) has an atomic mass of about 23.0 g/mol, chromium (Cr) is roughly 52.0 g/mol, and oxygen (O) is 16.0 g/mol. Therefore, the molar mass of Na2Cr2O7 is (2×23.0) + (2×52.0) + (7×16.0) = 262.0 g/mol. To find the moles of Na2Cr2O7 in 6.00 g, use the formula: Moles = mass (g) / molar mass (g/mol), which gives us 6.00 g / 262.0 g/mol = 0.0229 mol.
Since Na2Cr2O7 contains 7 oxygen atoms per formula unit, the total number of oxygen atoms in 6.00 g of Na2Cr2O7 is 0.0229 mol × (7 atoms/mol) × (6.022 × 1023 atoms/mol) = 9.69 × 1023 oxygen atoms.
