In one of the experiments we will use a spectrophotometer to determine the absorbance of ________________. In that same experime
2 answers:
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Answer:
Ethynylcyclopropane is the stable isomer for given alkyne.
Explanation:
In order to solve this problem we will first calculate the number of Hydrogen atoms. The general formula for alkynes is as,
CₙH₂ₙ₋₂
Putting value on n = 5,
C₅H₂.₅₋₂
C₅H₈
Also, the statement states that the compound contains one ring therefore, we will subtract 2 hydrogen atoms from the above formula i.e.
C₅H₈ ------------(-2 H) ----------> C₅H₆
Hence, the molecular formula for given compound is C₅H₆
Below, 4 different isomers with molecular formula C₅H₆ are attached.
The first compound i.e. ethynylcyclopropane is stable. As we know that alkynes are sp hybridized. The angle between C-C-H in alkynes is 180°. Hence, in this structure it can be seen that the alkyne part is linear and also the cyclopropane part is a well known moiety.
Compounds 3-ethylcycloprop-1-yne, <u>cyclopentyne </u>and 3-methylcyclobut-1-yne are highly unstable. The main reason for the instability is the presence of triple bond in three, five and four membered ring. As the alkynes are linear but the C-C-H bond in these compound is less than 180° which will make them highly unstable.
Answer:
i) 0.5071 (kg/s)
ii) -1407.1 kj/kg
iii) 204.05 Kw
iv) 5.881
v) 9.238
Explanation:
Given Data:
evaporation temperature ( T ) = 4°c = 277.15 K
Condensation Temperature ( T ) = 34°c = 307.15 K
<em>n</em> ( compressor efficiency ) = 0.76
refrigeration rate = 1200 kJ.s^-1
i) determine the circulation rate of the refrigerant
m = =
------- 1
Q = 1200 Kj.s^-1
H2 = entropy at step 2 = 2508.9 (kJ / kg ) ( gotten from Table F )
H4 = entropy at step 4 = 142.4 ( kJ/ kg )
back to equation 1
m ( circulation rate of refrigerant ) = 0.5071 (kg/s)
ii) heat transfer rate in the condenser
Q = m ( H4 - H3 )
= 0.5071 ( 142.4 - 2911.27 )
= -1407.1 kj/kg
where H3 = H2 + ΔH23 = 2911.27 (kj/kg) ( as calculated )
iii) power requirement
w = m * ΔH23
= 0.5071 (kg/s) * 402.37 (kj/kg) = 204.05 Kw
where: ΔH23 = =
= 402.37 (kj/kg)
iv) coefficient of performance of a cycle
W = Qc / w
= 1200 Kj.s^-1/ 204.05 kw
= 5.881
v) coefficient of performance of a Carnot refrigeration cycle
= 277.15 / ( 307.15 - 277.15 )
= 9.238
With the given problem you gave here, I can't answer the question because I need more details. Luckily, I found a similar problem that's provided with a diagram and a table shown in the attached picture.
This test is called the Benedict's test which is used as test for presence of sugars. If the solution contains sugar, like glucose, the solution would turn from blue to red. If not, it would stay blue. <em>Therefore, the correct results would be that in row 3.</em>
This test is called the Benedict's test which is used as test for presence of sugars. If the solution contains sugar, like glucose, the solution would turn from blue to red. If not, it would stay blue. <em>Therefore, the correct results would be that in row 3.</em>