uating learning Activity #2: Choose the letter of the best answer. Write your answers in your notebook 1. Where is the best place to start helping your community? a. At home c. In your barangay b. In school d. wherever you are 2. Which is the best step a school can do to help prevent tobacco smoking among its pupils? a. Include the bad effects of tobacco smoking in classroom lessons b. Invite resource persons to talk about the dangers of smoking c. Invite parents to a meeting about the dangers of smoking d. Advertise on television against the ill-effects of smoking 3. Which of the following is NOT a way to build a healthy communitys a. Throw wastes and garbage in trash bins and segregation boxes b. Plant vegetables and fruit bearing trees in your community c. Remove stagnant water in old tires, plastic bottles, and other containers d. Burn solid and brown wastes 4. Which of the following DOES NOT help your school? a. Vandalizing on walls c. Helping clean the room b. Disposing of garbage properly d. volunteering to start a garden 5. What SHOULD NOT be done with empty bottles? a. Sell them to junk peddlers c. Use them as plant pots b. Make art out of them d. Bury them
Answers & Comments
Answer:
Solution (No. 1A)
\text{moles of Au = 95.0 g Au} × \frac{\text{1 mol Au}}{\text{196.97 g Au}}moles of Au = 95.0 g Au×
196.97 g Au
1 mol Au
\boxed{\text{moles of Au = 0.482 mol}}
moles of Au = 0.482 mol
Solution (No. 1B)
\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}}number of Au atoms = 95.0 g Au×
196.97 g Au
1 mol Au
×
1 mol Au
6.022 × 10²³ Au atoms
\boxed{\text{number of Au atoms = 2.90 × 10²³ atoms}}
number of Au atoms = 2.90 × 10²³ atoms
Solution (No. 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 = 90.048 g/mol
Step 2: Calculate the number of molecules of C₃H₆O₃.
\text{number of C₃H₆O₃ molecules = 5.0 g C₃H₆O₃} × \frac{\text{1 mol C₃H₆O₃}}{\text{90.048 g C₃H₆O₃}} × \frac{\text{6.022 × 10²³ C₃H₆O₃ molecules}}{\text{1 mol C₃H₆O₃}}number of C₃H₆O₃ molecules = 5.0 g C₃H₆O₃×
90.048 g C₃H₆O₃
1 mol C₃H₆O₃
×
1 mol C₃H₆O₃
6.022 × 10²³ C₃H₆O₃ molecules
\boxed{\text{number of C₃H₆O₃ molecules = 3.3 × 10²² molecules}}
number of C₃H₆O₃ molecules = 3.3 × 10²² molecules
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