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2 years ago

How do distinctive rock layers support the theory of continental drift

1 answer:

8 0

The lithosphere (under the tectonic plates) is semifluid, allowing for the plates above to move over it, causing earthquakes, and essentially continental drift.

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6. 100 ml of gaseous hydrocarbon consumes 300

mario62 [17]

Answer:

a. C₂H₄

Explanation:

At constant pressure and temperature, the mole ratio of the gases is equal to their volume ratio (a consequence of Avogadro's law).

Hence, the complete combustion reaction that has a ratio of 100 ml of gaseous hydrocarbon to 300 ml of oxygen, is that whose mole ratio is 1 mol hydrocarbon : 3 mol of oxygen.

Then, you must write the balanced chemical equations for the complete combustion of the four hydrocarbons in the list of choices, and conclude which has such mole ratio (1 mol hydrocarbon : 3 mol oxygen).

A complete combustion reaction of a hydrocarbon is the reaction with oxygen that produces CO₂ and H₂O, along with the release of heat and light.

a. C₂H₄:

C₂H₄ (g) + 3O₂ (g) → 2CO₂(g) + 2H₂O (g)

Precisely, for this reaction the mole ratio is 1 mol C₂H₄: 2 mol O₂, hence, this is the right choice.

The following analysis just shows that the other options are not right.

b. C₂H₂:

2C₂H₂ (g) + 5O₂ (g) → 4CO₂(g) + 2H₂O (g)

The mole ratio for this reaction is 2 mol C₂H₂ :5 mol O₂.

с. С₃Н₈

C₃H₈ (g) + 5O₂ (g) → 3CO₂(g) + 4H₂O (g)

The mole ratio is 1 mol C₃H₈ : 5 mol O₂

d. C₂H₆

2C₂H₆ (g) +7 O₂ (g) → 4CO₂(g) + 6H₂O (g)

The mole ratio is 2 mol C₂H₆ : 7 mol O₂

7 0

3 years ago

Mg3N2(s)+6H2O(l)→3Mg(OH)2(s)+2NH3(g) When 36.0 g of H2O react, how many grams of NH3 are produced? When 36.0 g of H2O react, how

Afina-wow [57]

Answer:

11.3 g of $NH_{3}$ are produced from 36.0 g of $H_{2}O$

Explanation:

1. The balanced chemical equation is the following:

$Mg_{3}N_{2}(s)+6H_{2}O(l)=3Mg(OH)_{2}(s)+2NH_{3}(g)$

2. Use the molar mass of the $H_{2}O$ , the molar mass of the $NH_{3}$ and the stoichiometry of the balanced chemical reaction to find how many grams of $NH_{3}$ are produced:

Molar mass $H_{2}O$ = 18 $\frac{g}{mol}$

Molar mass $NH_{3}$ = 17 $\frac{g}{mol}$

$36.0gH_{2}O*\frac{1molH_{2}O}{18gH_{2}O}*\frac{2molesNH_{3}}{6molesH_{2}O}*\frac{17gNH_{3}}{1molNH_{3}}=11.3gNH_{3}$

Therefore 11.3 g of $NH_{3}$ are produced from 36.0 g of $H_{2}O$

3 0

2 years ago

Landfills are one source of CO2. As garbage decomposes, it releases CO2 into the air. Can you think of at least

Margaret [11]

Answer:

Recycling and reuse of materials

Explanation:

One of the greatest problems facing the human population is the problem of solid waste disposal. The menace of solid waste disposal has led to the idea of landfills. Land fills are depressions on the earth surface prepared for the purpose of solid waste disposal.

The most important approach towards solid waste disposal is the idea of recycling of materials. A material can be collected after use and processed into the same material or serve as a precursor in another manufacturing process. This means that no waste is generated as the materials which are supposed to be disposed of as solid waste are processed into other useful materials. This will reduce the volume of solid wastes generated that may need to be disposed in a landfill.

7 0

2 years ago

Researchers used a combustion method to analyze a compound used as an antiknock additive in gasoline. A 9.394 mg sample of the c

LuckyWell [14K]

Answer:

The percent composition of the compound is 90.5 % C and 9.5 % H

Explanation:

Step 1: Data given

Mass of compound = 9.394 mg

Mass of CO2 yielded = 31.154 mg

Mass of H2O yielded = 7.977 mg

Molar mass of CO2 = 44.01 g/mol

Molar mass of H2O = 18.02 g/mol

Step 2: Calculate moles CO2

moles of CO2 = (0.031154 g / 44.01 g/mol) = 7.08 * 10^-4 mol CO2

Step 3: Calculate moles C

moles of C = moles of CO2 * (1 mol C / 1 mol CO2)

moles of C = 7.08 * 10^-4 mol

Step 4: Calculate moles H2O

moles of H2O = (0.007977 g / 18.02 g/mol) = 4.43 * 10^-4 mol H2O

Step 5: Calculate moles of H

moles of H = moles of H2O * (2 mol H / 1 mol H2O)

moles of H = 4.43* 10^-4 *2 = 8.86 * 10^-4 mol H

Step 6: Calculate mass of C

mass C = moles C * molar mass C

mass C = 7.08 * 10^-4 mol*12.01 g/mol

mass C = 0.0085 grams

Step 7: Calculate mass of H

mass H = moles H * molar mass H

mass H = 8.86 * 10^-4 mol*1.01 g/mol

mass H = 0.000894 grams

Step 8: Calculate total mass of compound =

0.0085 grams + 0.000894 grams = 0.009394 grams = 9.394 mg

Step 9: Calculate the percent composition:

% C = (8.50 mg / 9.394 mg) x 100 = 90.5%

% H = (0.894 mg / 9.394 mg) x 100 = 9.5%

The percent composition of the compound is 90.5 % C and 9.5 % H

6 0

3 years ago

2. Draw a structural formula for 3,4-hexene (C6H12), which has a double bond between the number 3 and 4 carbons in the chain. Wh

Rufina [12.5K]

The structure will be:
H₃C-CH₂-CH=CH-CH₂-CH₃
This class of compounds is known or referred to as alkenes. Alkenes are unsaturated hydrocarbons that contain a carbon-carbon double bond. The present of this double bond alters the properties of alkenes rom alkanes.

8 0

3 years ago

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