Practice: Solve each of the following problems using dimensional analysis (or another method). Show your work. (Use a periodic table to look up the necessary atomic masses.) A. 75 g KMnO4 = molecules KMnO4 B. 7.23 × 1024 Al atoms = grams Al C. 9.23 × 1023 Au atoms = moles Au D. 125 g H3PO4 = molecules H3PO4 E. 0.75 moles CO2 = total atoms

Explanation:

A )

molecular weight of KMnO₄ =   158

75 g KMnO4 = 75 / 158 moles

= ( 75 / 158 )x 6.02 x 10²³ molecules

= 2.857 x 10²³ molecules KMnO4

so 75g KMnO₄ = 2.857 x 10²³ molecules KMnO4

B )

7.23 × 10²⁴ Al atoms = grams Al

72.3 X 10²³ Al atoms = 72.3 X 10²³ / 6.02 X 10²³ moles of Al

= 12 moles of Al

= 12 x 27 gm of Al

= 324 gm of Al .

7.23 × 10²⁴ Al atoms = 324 grams Al

C )

9.23 × 10²³ Au atoms = moles Au

9.23 × 10²³Au atoms = 9.23 × 10²³ / 6.02 x 10²³ moles of Au

= 1.53 moles of Au .

D )

125 g H₃PO₄ =   molecules H₃PO₄

125 g H₃PO₄ = 125 / 98 moles of H₃PO₄

= (125 / 98) x 6.02 x 10²³ molecules

= 7.678 x 10²³ molecules of H₃PO₄

125 g H₃PO₄ = 7.678 x 10²³  molecules H₃PO₄

E )

.75 moles of CO₂

= .75  x 6.02 x 10²³ molecules of CO₂

= 4.515 x 10²³ molecules of CO₂.

one molecule of CO₂ = 3 atoms

4.515 x 10²³ molecules of CO₂ = 3 x 4.515 x 10²³ atoms

= 13.545 x 10²³ atoms .

o conteúdo de calor de um sistema é igual à entalpia apenas para um sistema que é constante