Answer:
2.14 × 10⁻³ molecules/RSP
3.31 × 10⁻³ molecules/ESP
Explanation:
Step 1: Calculate the number of moles of Acetaminophen per Regular Strength Pill (RSP)
A Regular Strength Pill has 1.29 × 10²¹ molecules of Acetaminophen per pill. To convert molecules to moles we will use Avogadro's number: there are 6.02 × 10²³ molecules in 1 mole of molecules.
1.29 × 10²¹ molecules/RSP × 1 mol/6.02 × 10²³ molecules = 2.14 × 10⁻³ molecules/RSP
Step 2: Calculate the number of moles of Acetaminophen per Extra Strength Pill (ESP)
An Extra Strength Pill has 1.99 × 10²¹ molecules of Acetaminophen per pill. To convert molecules to moles we will use Avogadro's number: there are 6.02 × 10²³ molecules in 1 mole of molecules.
1.99 × 10²¹ molecules/ESP × 1 mol/6.02 × 10²³ molecules = 3.31 × 10⁻³ molecules/ESP
Its would be fluorine as its the most electronegative
hope that helps <span />
Answer:
.371 mole of NaCl
Explanation:
Na Cl Mole weight = 22.989 + 35.45 = 58.439 g/mole
21.7 g / 58.439 g/mole = .371 mole
<span>D) recycling ;)
Waste Management's Aerobic-Anaerobic Bioreactor* is designed to accelerate waste degradation by combining attributes of the aerobic and anaerobic bioreactors. The objective of the sequential aerobic-anaerobic treatment is to cause the rapid biodegradation of food and other easily degradable waste in the aerobic stage in order to reduce the production of organic acids in the anaerobic stage resulting in the earlier onset of methanogenesis. In this system the uppermost lift or layer of waste is aerated, while the lift immediately below it receives liquids. Landfill gas is extracted from each lift below the lift receiving liquids. Horizontal wells that are installed in each lift during landfill construction are used convey the air, liquids, and landfill gas. The principle advantage of the hybrid approach is that it combines the operational simplicity of the anaerobic process with the treatment efficiency of the aerobic process. Added benefits include an expanded potential for destruction of volatile organic compounds in the waste mass. (*US Patent 6,283,676 B1)</span>
Answer:
MoClBr₂
Explanation:
First we calculate the mass of bromine in the compound:
- 300.00 g - (82.46224 g + 45.741 g) = 171.79676 g
Then we<u> calculate the number of moles of each element</u>, using their <em>respective molar masses</em>:
- 82.46224 g Mo ÷ 95.95 g/mol = 0.9594 mol Mo
- 45.741 g Cl ÷ 35.45 g/mol = 1.290 mol Cl
- 171.79676 g Br ÷79.9 g/mol = 2.150 mol Br
Now we <u>divide those numbers of moles by the lowest number among them</u>:
- 0.9594 mol Mo / 0.9594 = 1
- 1.290 mol Cl / 0.9594 = 1.34 ≅ 1
- 2.150 mol Br / 0.9594 = 2.24 ≅ 2
Meaning the empirical formula is MoClBr₂.
