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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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MrRissso [65]

The question is incomplete, complete question is :

The frequency factors for these two reactions are very close to each other in value. Assuming that they are the same, compute the ratio of the reaction rate constants for these two reactions at 25°C.

\frac{K_1}{K_2}=?

Activation energy of the reaction 1 ,Ea_1 = 14.0 kJ/mol

Activation energy of the reaction 2,Ea_1  = 11.9 kJ/mol

Answer:

0.4284 is the ratio of the rate constants.

Explanation:

According to the Arrhenius equation,

K=A\times e^{\frac{-Ea}{RT}}

The expression used with catalyst and without catalyst is,

\frac{K_2}{K_1}=\frac{A\times e^{\frac{-Ea_2}{RT}}}{A\times e^{\frac{-Ea_1}{RT}}}

\frac{K_2}{K_1}=e^{\frac{Ea_1-Ea_2}{RT}}

where,

K_2 = rate constant reaction -1

K_1 = rate constant reaction -2

Activation energy of the reaction 1 ,Ea_1 = 14.0 kJ/mol = 14,000 J

Activation energy of the reaction 2,Ea_1  = 11.9 kJ/mol = 11,900 J

R = gas constant = 8.314 J/ mol K

T = temperature = 25^oC=273+25=298 K

Now put all the given values in this formula, we get

\frac{K_1}{K_2}=e^{\frac{11,900- 14,000Jl}{8.314 J/mol K\times 298 K}}=2.3340

0.4284 is the ratio of the rate constants.

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ira [324]

Answer:

35.9 ml

Explanation:

Start with the balanced equation:

3CuCl2(aq)+2Na3PO4(aq)→Cu3(PO4)2(s)+6NaCl(aq)

This tells us that 3 moles of CuCI2 react with 2 moles Na3PO4-

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We know that concentration = moles/volume i.e:

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v=nc=6.491×10−30.181=35.86×10−3L

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