Sagot :
Solution (1A)
[tex]\text{moles of Au = 95.0 g Au} × \frac{\text{1 mol Au}}{\text{196.97 g Au}}[/tex]
[tex]\boxed{\text{moles of Au = 0.4823 mol}}[/tex]
Solution (1B)
[tex]\text{number of Au atoms = 95.0 g Au} × \frac{\text{1 mol Au}}{\text{196.97 g Au}} × \frac{\text{6.022 × 10²³ Au atoms}}{\text{1 mol Au}}[/tex]
[tex]\boxed{\text{number of Au atoms = 2.904 × 10²³ atoms}}[/tex]
Solution (2)
Step 1: Calculate the molar mass of C₃H₆O₃.
molar mass = (12.0 g/mol × 3) + (1.008 g/mol × 6) + (16.0 g/mol × 3)
molar mass = 90.05 g/mol
Step 2: Calculate the number of molecules of C₃H₆O₃.
[tex]\text{number of C₃H₆O₃ molecules = 5.0 g C₃H₆O₃} × \frac{\text{1 mol C₃H₆O₃}}{\text{90.05 g C₃H₆O₃}} × \frac{\text{6.022 × 10²³ C₃H₆O₃ molecules}}{\text{1 mol C₃H₆O₃}}[/tex]
[tex]\boxed{\text{number of C₃H₆O₃ molecules = 3.34 × 10²² molecules}}[/tex]
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