Answer:82.7% C and 17.3% H
Explanation:
C4 = 12.0107g/mol x 4 = 48.0428 g/mol
H10 = 1.00794 g/mol x 10 = 10.0794 g/mol
Molar Mass of Butane, C4H10 = 48.0428 g/mol + 10.0794 g/mol= 58.1222 g/mol
percent composition of Carbon is = ( mass of carbon contained in butane / molar mass of Butane) x 100
=(48.0428 /58.1222) x 100% = 0.8265 x 100
=82.65% =82.7% of Carbon.
Percent composition of Hydrogen = (mass of Hydrogen contained in Butane / molar mass of Butane )x 100
( 10.0794/58.122 ) x 100% =0.1734 x 100
= 17.3% OF Hydrogen.
Answer:
A. your observations in writing
Explanation:
Descriptive investigation is one of the mode of inquiries in science. It is devoid of the usual collection of data in the field or setting up laboratory experiments to compared variables.
- In descriptive investigation, observations are usually done in writing.
- They are usually common in Astrology.
- Such investigation involves a question without a hypothesis ensuing.
- Tests are usually not carried out in descriptive investigation.
Answer:
Reducing the amount of water you use, by having a 5-minute shower or not running the water when washing up the dishes, can help protect vital wetlands. Plant scientists are also working to help conserve by developing crop varieties that use less water.
Explanation:
You must use 1880 mL of O₂ to react with 4.03 g Mg.
A_r: 24.305
2Mg + O₂ ⟶ 2MgO
<em>Moles of Mg</em> = 4.03 g Mg × (1 mol Mg/24.305 g Mg) = 0.1658 mol Mg
<em>Moles of O₂</em> = 0.1658 mol Mg × (1 mol O₂/2 mol Mg) = 0.082 90 mol O₂
STP is 25 °C and 1 bar. At STP, 1 mol of an ideal gas has a volume of <em>22.71 L</em>.
<em>Volume of O₂</em> = 0.082 90 mol O₂ × (22.71 L O₂/1 mol O₂) = 1.88 L = 1880 mL
