The name of the compound by using the <u>IUPAC nomenclature of organic compounds</u> is 1 -octene. The correct option is the last option - 1-octene.
<h3>Nomenclature of Organic compounds</h3>
From the question, we are to determine the name of the given molecule.
To name the compound, we will follow the IUPAC rules.
Some of IUPAC rules are
- Find the longest continuous carbon chain. Determine the root name for this parent chain.
- For Alkenes (organic compounds with double bond), number the chain of carbons that includes the C=C so that the C=C has the lower position number. Change “ane” to “ene” and assign a position number to the first carbon of the C=C.
The given compound has 8 carbons and a double bond. The root name of the compound is octane.
By <u>IUPAC rules</u>, the compound is an <u>Octene</u>.
Since the double bond is between carbon-1 and carbon-2. The compound becomes 1-octene.
Hence, the name of the compound by using the <u>IUPAC nomenclature of organic compounds</u> is 1 -octene. The correct option is the last option - 1-octene.
Learn more on Nomenclature of Organic compounds here: brainly.com/question/26754333
The diagram for the compound is attached below.
They are pretty much the cause and effect. The independent variable that is being changed in the experiment, so it is the cause. The dependant variable is the result of change the independant variable, so the effect.
It is made of polar molecules
There are 2 unpaired electrons in sulphur orbital...
Answer: hundred cubic meters of carbon dioxide initially at 150◦C and 50 bar is to beisothermally compressed in a frictionless piston-and-cylinder device to a final pressure of 300 bar.Calculatei The volume of the compressed gas.ii The work done to compress the gas.iii The heat flow on compression.assuming carbon dioxide(a) Is an ideal gas.(b) Obeys the principle of corresponding states of Sec. 6.6(c) Obeys the Peng-Robinson equation of state.SolutionWe haveT1= 150◦C,P1= 50 bar,T2= 150◦C,P2= 300 bar. (1 and 2 denote the initialand final conditions in this ’snapshot’ problem, respectively - we have sometimes called themt1andt2)(a) If CO2is an idea gas, we havePV=NRT.The number of moles can be calculated from theinitial conditions:N1=P 1 V 1 RT 1 = (6 × 10 6 Pa)(100 m 3 ) (8 . 314 J/(mol K))(150 + 273 K) = 142127 mol = 142 . 1 kmol i. Since we know N 1 = N 2 , T 2 , P 2 V 2 = N 2 RT 2 P 2 = (142127 mol)(8 . 314 J/(mol K))(150 + 273 K ) 30 × 10 6 Pa = 16 . 66 m 3 ii. Since there is no shaft work, and since the gas is isothermally compressed , we only have pressure-volume work: W = - Z V 2 V 1 PdV = - Z V 2 V 1 NRT V dV = - NRT ln V 2 V 1 W = - (142127 mol)(8 . 314 J/(mol K))(150 + 273 K ) ln 16 . 66 100 = 8 . 958 × 10 8 J iii. Energy balance (integral form) for the closed system is: U 2 - U 1 = Q + W Back from homework 2, for an ideal gas, stating from equation 6.2-21, dU = C V dT + " T ∂P ∂T V - P # dV reduces to: dU = C V dT However, the process is isothermal, so dT = 0 Which gives: dU = Δ U = 1 N Δ U Therefore 0 = Q + W → Q = - W Q = - 8 . 958 × 10 8 J 3
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Explanation:As revealed above, the stimuli connections are clearly stated
