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EleoNora [17]
3 years ago
7

Hydrocarbons are compounds that are composed of carbon and hydrogen. Which statement does the law of definite proportions suppor

t?
A. Each sample of a single hydrocarbon has the same mass ratio of carbon to hydrogen
B. The mass ratio of carbon to hydrogen is the same for each hydrocarbon
C. No to hydrocarbons have the same mass ratio of carbon to hydrogen
Each hydrocarbon can have several mass ratio of carbon to hydrogen depending on its source
Chemistry
1 answer:
lora16 [44]3 years ago
6 0
The correct option is A.
The law of definite proportion states that a given chemical compound always contains its component elements in fixed ratio irrespective of the source they are obtain from.
For instance, water is made up of two atoms of hydrogen and one atom of oxygen. Water can not be formed in any other way. Combining the oxygen and the hydrogen in another ratio will give us another thing not water.
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Spontaneous combustion is characterized by: an explosion a self-starting fire rapid oxidation lightning
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Using the determined equivalence point from question 2 and the balanced reaction of acetic acid and sodium hydroxide, calculate
goldenfox [79]

Answer:

Molarity of the packet is 0.5M

Explanation:

In the reaction of acetic acid with NaOH:

CH₃COOH + NaOH → CH₃COO⁻ + H₂O + Na⁺

<em>1 mole of acetic acid reacts with 1 mole of NaOH.</em>

<em />

When you are titrating the acid with NaOH, you reach equivalence point when moles of acid = moles of NaOH.

Moles of NaOH are:

3.0mL = 3.0x10⁻³L ₓ (0.1 mol / L) =<em> 3.0x10⁻⁴ moles</em> of NaOH = moles of CH₃COOH.

Now, you find the moles of acetic acid in the hot sauce packet. But molarity is the ratio between moles of the acid and liters of solution.

As you don't know the volume of your packet, <em>you can assume its density as 1g/mL. </em>Thus, volume of 0.6g of hot sauce is 0.6mL = 6x10⁻⁴L.

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6 0
3 years ago
A 14.60g sample of an unknown compound, composed only of carbon, hydrogen, and oxygen, produced 28.6g of CO2 and 14.6g of H2O in
o-na [289]

Answer: The empirical formula for the given compound is C_{2}H_{5}O

Explanation:

The chemical equation for the combustion of compound having carbon, hydrogen and oxygen follows:

C_xH_yO_z+O_2\rightarrow CO_2+H_2O

where, 'x', 'y' and 'z' are the subscripts of carbon, hydrogen and oxygen respectively.

We are given:

Mass of CO_2=28.6g

Mass of H_2O=14.6g

We know that:

Molar mass of carbon dioxide = 44 g/mol

Molar mass of water = 18 g/mol

For calculating the mass of carbon:

In 44 g of carbon dioxide, 12 g of carbon is contained.

So, in 28.6 g of carbon dioxide, \frac{12}{44}\times 28.6=7.8g of carbon will be contained.

For calculating the mass of hydrogen:

In 18 g of water, 2 g of hydrogen is contained.

So, in 14.6 g of water, \frac{2}{18}\times 14.6=1.6 of hydrogen will be contained.

For calculating the mass of oxygen:

Mass of oxygen in the compound = (14.60) - (7.8 + 1.6) = 5.2 g

To formulate the empirical formula, we need to follow some steps:

Step 1: Converting the given masses into moles.

Moles of Carbon =\frac{\text{Given mass of Carbon}}{\text{Molar mass of Carbon}}=\frac{7.8g}{12g/mole}=0.65moles

Moles of Hydrogen = \frac{\text{Given mass of Hydrogen}}{\text{Molar mass of Hydrogen}}=\frac{1.6g}{1g/mole}=1.6moles

Moles of Oxygen = \frac{\text{Given mass of oxygen}}{\text{Molar mass of oxygen}}=\frac{5.2g}{16g/mole}=0.32moles

Step 2: Calculating the mole ratio of the given elements.

For the mole ratio, we divide each value of the moles by the smallest number of moles calculated which is 0.32 moles.

For Carbon = \frac{0.65}{0.32}=2.03\approx 2

For Hydrogen  = \frac{1.6}{0.32}=5

For Oxygen  = \frac{0.32}{0.32}=1

Step 3: Taking the mole ratio as their subscripts.

The ratio of C : H : O = 2 : 5 : 1

Hence, the empirical formula for the given compound is C_{2}H_{5}O_{1}=C_{2}H_{5}O

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