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aniked [119]
3 years ago
3

Check all correct statements concerning pentyne.

Chemistry
1 answer:
Ipatiy [6.2K]3 years ago
8 0
1. <span>Pentyne is a hydrocarbon. It is classified as an alkyne. The general formula for the alkyne is CnH2n-2. The functional group of pentyne is a triple bond which is made from one </span>σ bond and two π bonds<span>. </span>Since, it has a triple bond, pentyne is an unsaturated hydrocarbon. The formula of pentyne is C5H8 and it only consists of H and C atoms.

2. Alkenes are also hydrocarbons. But they have at least one double bond. The general formula for the alkenes is CnH2n. Since, it has double bond, alkenes are unsaturated hydrocarbons. The hybridization of C which has double bond is sp². Due to the unsaturation it is a reactive hydrocarbon but less stable than alkanes.

Hence, the false statement is "<span>They are more stable than alkanes."

</span>

3. <span>Butene would have 4 carbon atoms and a </span>double bond.

Butene is an alkene. Since the general formula of alkene is CnH2n, formula for the butene is C4H4 because n = 4. It is an unsaturated hydrocarbon due to the presence of the double bond. The double bond is made from one σ bond and one π bond.


<span>4. Alkenes generally undergo addition reactions.</span>

<span>
</span>

Alkenes are unsaturated hydrocarbons which have double bonds. The hybridization of the carbon atom which has the double bond is sp². Hence, that double bond has one σ bond and one π bond. In addition reactions, the π bond breaks down and another atom is connected to the carbon atom by forming a σ bond. By undergoing addition reaction, the hybridization of c atom which has double bond changes from sp² to sp³.


5. If the carbon atom of hydrocarbon is bonded to less than 4 other atoms, the hydrocarbon is considered unsaturated.


Maximum number of bonds that a carbon atom can form is 4. Those bonds can be formed by forming single bonds with four another atoms of by forming one double bond and two single bond with three atoms or by forming one single bond and one triple bond with to another atoms. The single bonds are saturated and double bonds and triple bonds are considered as unsaturated.


6. <span>Unsaturated hydrocarbons </span>have double bond and triple bonds<span> as their functional groups but saturated hydrocarbons have only single bonds. Due to the unsaturation, </span>they will react readily with other elements and compounds. Although the hydrocarbons are unsaturated, they also have isomers. But the stability of the unsaturated hydrocarbons is less than saturated ones.


7. Alkenes have the general formula as CnH2n. Hence C5H10 is an alkene which has 5 carbon atoms. Since, it is an alkene, that hydrocarbon has a double C-C bond. Since it is an unsaturated one, it is highly reactive than alkanes and unstable. It is less inert than alkanes.


8. The general formula of alkenes is CnH2n.


Alkanes have the general formula as CnH2n+2 and for the alkynes, it is CnH2n-2. The general formula of alkenes and alkynes are given when there is only one double bond or triple bond. Due to the presence of double bonds and triple bonds, alkenes and alkynes are less stable and unsaturated.


9. Benzene is a cyclic closed ring and aromatic alkene. The general formula is CnH2n. Benzene has three double bonds. Hence, three C-C have double bonds which share 2 pairs of electrons. Since, those C-C bonds are double bonds, they have π bonds. The all of C atoms in benzene are sp² hybridized. Hence, all C atoms are actually identical. 


10. 

1. H-C≡C-H is a alkyne which has two carbon atoms. Hence, the name of this is  ethyne.

2. Benzene is a cyclic compound which has both saturated and unsaturated characteristics.

3. The general formula of the alkenes is CnH2n.

4. addition is the typical reaction type for the unsaturated hydrocarbons.

5. The general formula of the alkynes is CnH2n-2.

6. series name of hydrocarbons with triple bond is alkyne.

7. The general formula of the alkenes is CnH2n+2.

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A man with normal lungs and an arterial PO2 of 40 mmHg takes an overdose of barbiturates that halves his alveolar ventilation wi
Colt1911 [192]

Answer:

\frac{4}{5} = \frac{40}{x}

Where x represent the value of interest on this case. And solving for the value of x we have:

x = 40 * \frac{5}{4}= 50 mm Hg

So then the new arterial pressure needs to be now 50 mm Hg to mantain the original level.

And in order to find the concentration we can use a figure called the "O2

-CO2  diagram showing a ventilation-perfusion ratio line. " and when we use this graph to calculate the pressure of Co2 for PO2= 40 mmHg and for Po2=50 mm Hg we got and increase of 0.07 or 7%  

So then the final answer for this case would be an increase of 7%

Explanation:

For this case we know that a man with normal lungs have an arterial Po2 os 40 mm Hg.

Then we know that this man take an overdose of barbiturates thats halves his aveolar ventilation without changing his metabolism

We also know that the respiratory exchange ratio is 0.8 or 8/10

0.8 = \frac{8}{10}= \frac{4}{5}

And we want to find hor much does his inspired oxyden concentration % have to increased to return his avelolar Po2 to the original level.

On this case we can apply a proportion rule and we have this:

\frac{4}{5} = \frac{40}{x}

Where x represent the value of interest on this case. And solving for the value of x we have:

x = 40 * \frac{5}{4}= 50 mm Hg

So then the new arterial pressure needs to be now 50 mm Hg to mantain the original level.

And in order to find the concentration we can use a figure called the "O2

-CO2  diagram showing a ventilation-perfusion ratio line. " and when we use this graph to calculate the pressure of Co2 for PO2= 40 mmHg and for Po2=50 mm Hg we got and increase of 0.07 or 7%  

So then the final answer for this case would be an increase of 7%

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Write the acid-base reaction where hydrogen carbonate is the acid and water is the base.
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HI

I found a link that will direct you to the answer of this questions
<span>http://click.dji.com/ANbvbbP7bwUWtSACp6U_?pm=link&as=0004</span>
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Shkiper50 [21]

Answer:

3

Explanation:

third answer might be right

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If I react 6 units of AB with 10 units of CD in the equation below, what is the limiting reactant? 3AB + 4CD --&gt; 2AD + 6CB
irga5000 [103]

If 6 units of AB were reacted with 10 units of CD in the described equation, the limiting reactant would AB

<h3>Limiting reactants</h3>

They are reactants that determine how far a reaction can go in terms of yield.

From the equation: 3AB + 4CD --> 2AD + 6CB

The mole ratio of AB to CD is 3:4

Thus, 6 units of AB will require 8 units of CD.

But 10 units of CD were reacted with only 6 units of AB. This means that CD is in excess by 2 units while AB will limit the yield of the reaction.

More on limiting reactants can be found here: brainly.com/question/14225536

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