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Darya [45]
3 years ago
11

A chemist decomposes several samples of water into hydrogen and oxygen and weighs or more correctly measures the mass of the hyd

rogen and the oxygen obtained. The results are as follows sample 1 has 1.5grams of hydrogen and 12grams of oxygen sample 2 has 2grams of hydrogen and 16 grams of oxygen sample 3 has 2.5grams and 20 grams of oxygen
a. Can you summarize these observations in a short statement? Next, the chemist decomposes several samples of carbon dioxide into carbon and oxygen. The results are as follows: sample 1 has 0.5grams of carbon and 1.3grams of oxygen sample 2 has 1.0grams of carbon and 2.7grams of oxygen sample 3 has 1.5grams of carbon and 4.0grams of oxygen
b. Summarize these observations in a short statement?
c. Formulate a law from the observations in (a) and (b)?
d. Formulate a theory that might explain your law in (c)?
Chemistry
1 answer:
MariettaO [177]3 years ago
3 0

Answer:

A. Hydrogen and oxygen combines in a constant mole ratio in all the three water samples

B. Carbon and oxygen are combined in a constant mole ratio in all the three samples of carbon dioxide

C. All pure samples of a compound contain the same elements combined in the same proportion by mass

D. Since the same elements are always combining to form a certain compound, and the atoms of each element has its own mass, the atoms of each element combining to form the compound do so in a fixed proportion by mass.

Explanation:

A. molar mass of hydrogen = 1.0 g/mol; molar mass of oxygen = 16 g/mol

The molar ratio of elements in a compound gives the proportion each element is combined with the other by mass.

molar ratio of hydrogen to oxygen in the samples:

Sample 1: Hydrogen; 1.5/1 = 1.5: Oxygen; 12/16 = 0.75

molar ratio = 1.5/0.75 = 2 : 0.75./0.75 = 1

molar ratio = 2:1

Sample 2: Hydrogen; 2/1 = 2: Oxygen; 16/16 = 1

molar ratio = 2/1 = 2 : 1/1 = 1

molar ratio = 2:1

Sample 3: Hydrogen; 2.5/1 = 2.5: Oxygen; 20/16 = 1.25

molar ratio = 2.5/1.25 = 2 : 1.25/1.25 = 1

molar ratio = 2:1

Hydrogen and oxygen combines in a constant mole ratio in all the three water samples

B. molar mass of carbon = 12 g/mol; molar mass of oxygen = 16 g/mol

molar ratio of carbon to oxygen in the samples:

Sample 1: Carbon; 0.5/12 = 0.04 : Oxygen; 1.3/16 = 0.08

molar ratio = 0.04/0.04 = 1 : 0.08/0.04 = 2

molar ratio = 1:2

Sample 2: Carbon; 1.0/12 = 0.083 : Oxygen; 2.7/16 = 0.168

molar ratio = 0.083/0.083 = 1 : 0.168/0.083 = 2

molar ratio = 1:2

Sample 3: Carbon; 1.5/12 = 0.125 : Oxygen; 4.0/16 = 0.250

molar ratio = 0.125/0.125 = 1: 0.250/0.125 = 2

molar ratio = 1:2

Carbon and oxygen are combined in a constant mole ratio in all the three samples of carbon dioxide

C. All pure samples of a compound contain the same elements combined in the same proportion by mass

D. Since the same elements are always combining to form a certain compound, and the atoms of each element has its own mass, the atoms of each element combining to form the compound do so in a fixed proportion by mass.

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Calculate the total amount of energy required to change 10.0 g of water from 35.0 degrees Celsius to 110. degrees Celsius.
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Answer:

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Explanation:

Calorimetry is the measurement and calculation of the amounts of heat exchanged by a body or a system.

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To calculate the latent heat the formula is used:

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In this case, you have in the first place a heat to raise the temp of the water from 35.0 C to 100 C, where the specific heat value for water is  4.184 \frac{J}{g*C}:

q1 = m*c*(Tfinal-Tinitial)

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Now you have the heat to vaporize the water, where the heat of vaporization is 2259.36 \frac{J}{g}:

q2 = m*(heat of vaporization)

q2 = 10.0 g*(2259.36 \frac{J}{g}) = 22593.6 J

Finally, you have the heat to raise temp of steam to 110 C, where the specific heat value for steam is  2.02 \frac{J}{g*C} :

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The total amount of energy can be calculated as:

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Since, density is mass divided by volume and it will decrease at higher temperature.

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At 20^{o}C, density of water=0.9982071 g/ml  

Therefore, \frac{concentration}{density} will be calculated as follows.

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Also,             \frac{C_{1}}{d_{1}} = \frac{C_{2}}{d_{2}}

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