Question

Can you explain it please?

Complete the table: Grams of Molar element mass element 32g Manganese Moles of element Atoms of element 2.3 x 10-5 moles Zr 2.2 x 1022 Atoms of gold 53 mg carbon 25.8 mol W 4.984 x 1028 atoms Ar 500 kg He

Answer 1

SolutioN Equati on Molarmass Mass Mole Conversion MoleMulybyN Divideby N tpybyMolar mass Divide byMolarmass Mass Atoms Avogadro's Number, N4 =6.023x102particles /mol For 32 g Manganese (Mn) Mass of Mn 32 g Molarmass of Mn = 54.9380 g /mol Mass of Mn 32 g Moles of Mn=olarmass of Mn 54.9380g/mo0.58247 mol Moles of Mn = 0.58 mol Atoms of Mn Moles of Mnx N = 0.58molx6.023 x 105 atoms / mol = 3.493x 103 atoms Atomsof Mn 3.5 x 1023 atoms For 2.3x10 mol es Zr Moles of Zr 2.3x10 mol Molarmass of Zr = 91.224 g /mol Atoms of Zr Moles of Zr x N = 2.3 x105 molx6.023 x 10 atoms/mol 1.385 x1013 atoms Atomsof Zr 1.4 x 1019 atoms Mass of Zr Molarmass of Zr x Moles of Zr 91.224 g/mol x 2.3 x10 mol = 0.002098g Mass of Zr 2.1x10g For 2.2x10 atoms of gold: Atoms of Au 2.2 x102 atoms Molarmass of Au 196.96657 g /mol 2.2x10 atoms Atoms of Au Moles of Au =0.03652mol 6.023x1023 atoms/mol NA Moles of Au 0.037 mol Mass of Au Molarmass of Au xMoles of Au = 196.96657 g /molx 0.037 mol 7.288 g Mass of Au 7.3g