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fredd [130]
3 years ago
6

What is the physical state of a straight-chain hydrocarbon containing three carbon atoms at room temperature?

Chemistry
2 answers:
WARRIOR [948]3 years ago
8 0

Answer: Gaseous

Explanation: Straight Chain hydrocarbon having three carbon atoms is known as Propane. It molecular formula is

                                CH_{3}CH_{2}CH_{3}

Inter molecular forces of attraction that is London Dispersion Forces existing between its molecules are weaker that is why the propane is gaseous in nature at room temperature.

It is also used in LPG. Boiling Point of propane is -42°  C.

madam [21]3 years ago
6 0
<span>The straight-chain hydrocarbon containing three carbon atoms is propane. It's molecular formula is C3H8. 

Following is the structure of propane
                                                         CH3-CH2-CH3

Boiling point of propane is -42 oC, this suggest that propane is gas at room temperature. </span>
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An industrial chemist is studying a sample of an unknown metal. Describe two ways he could change the metal physically and two w
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Answer:

Some of the physical changes used by the industrial chemist in order to identify it is by scratching it with other metals in order to find the hardness of it. Trying to deform it in order to find the malleability, and to heat it and measure the temperature in order to find the melting point.  

Some of the chemical changes used by the industrial chemist in order to identify it is by inserting it in water to observe that whether it reacts with it or not, if the reaction is violent, then the metal belongs to either group I or group II. The other method is to insert it in acids of distinct strength and to observe its reaction. The metals belonging to the second group react briskly with acids. The other metals react gradually with acids and others are almost inert.  

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  1. Compounds are formed when two or more <u>elements</u> are chemically combined.
  2. When these elements are <u>chemically</u> combined, a new substance is formed with new chemical and physical <u>properties</u>.
  3. An element is a <u>pure</u> substance that cannot be separated into simpler substances by <u>physical</u> or chemical means.
  4. A compound is two or more elements <u>combined</u> chemically to produce a new substance.
  5. When two or more elements chemically combine, the compound has new properties, different from the chemical and physical properties of the <u>original</u> elements.

<h3>What is a chemical element?</h3>

A chemical element can be defined as a pure substance which comprises atoms that have the same atomic number (number of protons) in its nuclei and as such it is the primary constituent of matter.

Generally, some examples of a chemical element include the following:

  1. Argon.
  2. Sodium.
  3. Carbon.
  4. Oxygen.
  5. Hydrogen.
  6. Phosphorus
  7. Copper
  8. Aluminum
  9. Potassium
  10. Magnesium

<h3>What is a pure substance?</h3>

A pure substance can be defined as a single sample of matter that cannot be separated into other kinds of matter through the use of any physical or chemical separating technique because it has distinct chemical properties and a definite and constant composition.

Read more on pure substances here: brainly.com/question/2056940

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5 0
2 years ago
A 3.4 g sample of an unknown monoprotic organic acid composed of C,H, and O is burned in air to produce 8.58 grams of carbon dio
Pavlova-9 [17]

Answer:

C_7H_6O_2

Explanation:

Hello there!

In this case, we can divide the problem in three stages: (1) determine the empirical formula with the combustion analysis, (2) compute the molar mass of acid via the moles of the acid in the neutralization and (3) determine the molecular formula.

(1) In this case, since 8.58 g of carbon dioxide are released, we can first compute the moles of carbon in the compound:

n_C=8.58gCO_2*\frac{1molCO_2}{44.01gCO_2}*\frac{1molC}{1molCO_2}=0.195molC

And the moles of hydrogen due to the produced 1.50 grams of water:

n_H=1.50gH_2O*\frac{1molH_2O}{18.02gH_2O}*\frac{2molH}{1molH_2O}  =0.166molH

Next, to compute the mass and moles of oxygen, we need to use the initial 3.4 g of the acid:

m_O=3.4g-0.195molC*\frac{12.01gC}{1molC}-0.166molH*\frac{1.01gH}{1molH} =0.89gO\\\\n_O=0.89gO*\frac{1molO}{16.0gO}=0.0556molO

Thus, the subscripts in the empirical formula are:

C=\frac{0.195}{0.0556}=3.5 \\\\H=\frac{0.166}{0.0556}=3\\\\O=\frac{0.0556}{0.0556}=1\\\\C_7H_6O_2

As they cannot be fractions.

(2) In this case, since the acid is monoprotic, we can compute the moles by multiplying the concentration and volume of KOH:

n_{KOH}=0.279L*0.1mol/L\\\\n_{KOH}=0.0279mol

Which are equal to the moles of the acid:

n_{acid}=0.0279mol

And the molar mass:

MM_{acid}=\frac{3.4g}{0.0279mol} =121.86g/mol

(3) Finally, since the molar mass of the empirical formula is:

7*12.01 + 6*1.01 + 2*16.00 = 122.13 g/mol

Thus, since the ratio of molar masses is 122.86/122.13 = 1, we infer that the empirical formula equals the molecular one:

C_7H_6O_2

Best regards!

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