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3 years ago

Which of the following is an example of a saturated hydrocarbon?

Chemistry

2 answers:

MakcuM [25]3 years ago

4 0

C. Ethyne. Hope this helped :)

KIM [24]3 years ago

3 0

Answer: B. Heptane

Explanation:

Saturated hydrocarbons are defined as the hydrocarbons in which a single bond is present between carbon and carbon atoms. they are known as alkanes. The general formula for these hydrocarbons is $C_nH_{2n+2}$

Unsaturated hydrocarbons are defined as the hydrocarbons which have double or triple covalent C-C bonds. They are known as alkenes and alkynes respectively. The general formula for these hydrocarbons is $C_nH_{2n}$ and $C_nH_{2n-2}$

A. Nonyne is an alkyne with 9 carbon atoms with molecular formula of $C_9H_{16}$

B. Heptane is an alkane with 7 carbon atoms with molecular formula of $C_7H_{16}$

C. Ethyne is an alkyne with 2 carbon atoms with molecular formula of $C_2H_{2}$ .

D. Hexene is an alkene with 6 carbon atoms with molecular formula of $C_6H_{12}$

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Question 6 the chemical formula ch2o can be classified as: empirical only. molecular only. empirical, possibly molecular. not en

frosja888 [35]

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3 years ago

chemistry A basketball is inflated to a pressure of 1.10 atm in a 28.0°C garage. What is the pressure of the basketball outside

butalik [34]

Answer : The final pressure of the basketball is, 0.990 atm

Explanation :

Gay-Lussac's Law : It is defined as the pressure of the gas is directly proportional to the temperature of the gas at constant volume and number of moles.

$P\propto T$

or,

$\frac{P_1}{P_2}=\frac{T_1}{T_2}$

where,

$P_1$ = initial pressure = 1.10 atm

$P_2$ = final pressure = ?

$T_1$ = initial temperature = $28.0^oC=273+28.0=301.0K$

$T_2$ = initial temperature = $-2.00^oC=273+(-2.00)=271.0K$

Now put all the given values in the above equation, we get:

$\frac{1.10atm}{P_2}=\frac{301.0K}{271.0K}$

$P_2=0.990atm$

Thus, the final pressure of the basketball is, 0.990 atm

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3 years ago

Name every block in mc

maks197457 [2]

Answer:

Andesite.
Bedrock.
Basalt. Smooth.
Blackstone.
Gilded.
Calcite.
Cobblestone.
Mossy.
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2 years ago

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Radium has a half-life of 1,620 years. In how many years will a 1 kg sample of radium decay and reduce to 0.125 kg of radium?

Tasya [4]

We can calculate years by using the half-life equation. It is expressed as:

A = Ao e^-kt

where A is the amount left at t years, Ao is the initial concentration, and k is a constant.

From the half-life data, we can calculate for k.

1/2(Ao) = Ao e^-k(1620)
k = 4.28 x 10^-4

0.125 = 1 e^-4.28 x 10^-4 (t)
t = 4259 years

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3 years ago

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makkiz [27]

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3 years ago

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