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alexandr1967 [171]
3 years ago
13

Based on the molecular formula, determine whether each compound is an alkane, alkene, or alkyne. (Assume that the hydrocarbons a

re noncyclical and there is no more than one multiple bond.)
a. C6H10
b. C4H10
c. C8H14
d. C8H16
Chemistry
2 answers:
topjm [15]3 years ago
5 0

Answer:

a. alkyne

b. alkane

c. alkyne

d. alkene

Explanation:

The general formula for each class of compound is given below

Alkane: C_nH_{2n+2}

Alkene: C_nH_{2n}

Alkyne: C_nH_{2n-2} (assuming single multiple bonds)

Now let us classify according to the above formulas:

a. It has two hydrogen atoms less than the two times of carbon atoms hence, it's alkyne

b. It has two hydrogen atoms more than the two times of carbon atoms hence, it's alkane

c. It has two hydrogen atoms less than the two times of carbon atoms hence, it's alkyne

d. It has hydrogen atoms two times of carbon atoms hence, it's alkene

Lisa [10]3 years ago
5 0

Answer:Based on the molecular formula, the following compounds belongs to the following groups:

a. C6H10 ( alkyne)

b. C4H10 ( alkane)

c. C8H14 (alkyne)

d. C8H16 (alkene)

Explanation: Hydrocarbon is an organic chemical compound that contains Hydrogen and carbons. There are three main types of Hydrocarbons which includes:

- Saturated Hydrocarbons.( They are composed entirely of single bonds and are saturated with hydrogen.). They are the alkane.

- Unsaturated Hydrocarbons: They are composed of either a double or triple bonds. The double bonds are the alkene and the triple bonds are the alkynes.

- Aromatic Hydrocarbons: they contain an aromatic ring. Example is the Benzene.

I hope this helps. Thanks.

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In this case, since silver is initially hot as it cools down, the heat it loses is gained by the liquid, which can be thermodynamically represented by:

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Since no phase change is happening. Thus, solving for the heat capacity of the liquid we obtain:

Cp_{liquid}=\frac{m_{Ag}Cp_{Ag}(T_2-T_{Ag})}{-m_{liquid}(T_2-T_{liquid})} \\\\Cp_{liquid}=\frac{31.2g*0.237\frac{J}{g\°C}*(28.3-227.2)\°C}{185.8g*(28.3-24.4)\°C}\\ \\Cp_{liquid}=2.54\frac{J}{g\°C}

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