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STatiana [176]
3 years ago
15

A carbon-carbon triple bond is found in a molecule of

Chemistry
2 answers:
posledela3 years ago
5 0
The answer is (4) butyne. The carbon-carbon triple bond is found in alkynes. And the alkynes has -yne at the end of the name. "-ene" end means that it is alkene and has carbon-carbon double bonds.
jolli1 [7]3 years ago
5 0

Answer: Option (4) is the correct answer.

Explanation:

A hydrocarbon in which all atoms are bonded by single bond and have a general formula C_{n}H_{2n+2} is known as an alkane.

A suffix "ane" adds at the end of the name of this hydrocarbon.

A hydrocarbon in which an  atom is bonded by a double bond and have a general formula C_{n}H_{2n} is known as an alkene.

A suffix "ene" adds at the end of the name of this hydrocarbon.

A hydrocarbon in which an atom is bonded by triple bond and have a general formula C_{n}H_{2n-2} is known as an alkene.

A suffix "yne" adds at the end of the name of this hydrocarbon.

Thus, we can conclude that a carbon-carbon triple bond is found in a molecule of butyne.

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At 25∘C, the decomposition of dinitrogen pentoxide, N2O5(g), into NO2(g) and O2(g) follows first-order kinetics with k=3.4×10−5
Annette [7]

Answer:

4600s

Explanation:

2N_{2}O_{5}(g) - - -> 4NO_{2}(g) +O_{2}

For a first order reaction the rate of reaction just depends on the concentration of one specie [B] and it’s expressed as:

-\frac{d[B]}{dt}=k[B] - - -  -\frac{d[B]}{[B]}=k*dt

If we have an ideal gas in an isothermal (T=constant) and isocoric (v=constant) process.

PV=nRT we can say that P = n so we can express the reaction order as a function of the Partial pressure of one component.  

-\frac{d[P(B)]}{P(B)}=k*dt  

-\frac{d[P(N_{2}O_{5})]}{P(N_{2}O_{5})}=k*dt

Integrating we get:

\int\limits^p \,-\frac{d[P(N_{2}O_{5})]}{P(N_{2}O_{5})}=\int\limits^ t k*dt

-(ln[P(N_{2}O_{5})]-ln[P(N_{2}O_{5})_{o})])=k(t_{2}-t_{1})

Clearing for t2:

\frac{-(ln[P(N_{2}O_{5})]-ln[P(N_{2}O_{5})_{o})])}{k}+t_{1}=t_{2}

ln[P(N_{2}O_{5})]=ln(650)=6.4769

ln[P(N_{2}O_{5})_{o}]=ln(760)=6.6333

t_{2}=\frac{-(6.4769-6.6333)}{3.4*10^{-5}}+0= 4598.414s

4 0
3 years ago
I need help with this ASAP PLEASE
Artemon [7]

439.3 g CO2

Explanation:

First find the # of moles of CO2 that results from the combustion of 3.327 mol C3H6:

3.227 mol C3H6 × (6 mol CO2/2 mol C3H6)

= 9.981 mol CO2

Use the molar mass of CO2 to determine the # of grams of CO2:

9.981 mol CO2 x (44.01 g CO2/1 mol CO2)

= 439.3 g CO2

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2 years ago
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natali 33 [55]

Answer:

the advantage of using Microsoft Excel to create a graph, as opposed to pen and paper is it will help you get the right answer

Explanation:

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3 years ago
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TiliK225 [7]

Answer:

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Explanation:

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Answer:

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Explanation:

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