the greatest amount of work is required if the process is adiabatic.The correct option is adiabatic.
The process in which heat is constant is called adiabatic process.
The The process in which temperature is constant is called isothermal process.
The process in which pressure is constant is called isobaric process.
The P-V diagram for adiabatic , isothermal and isobaric process is given below.
Work done in process = area encloses by P-V diagram axis . Since area under the curve is maximum for adiabatic process which is shown in the above diagram. So, work done by the gas will be maximum for adiabatic process.
learn more about adiabatic process.
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Answer:
1 mole of Al2O3 = 102 grams
1 mole of Al2 = 54 grams
102 grams of Al2O3 contains = 54 gram of Al2
10kg of Al2O3 contains = (54/102)*10000g Al2
= 5294.11 g Al2 or 5.29411 kg
<h3>Answer:</h3>
The Alkane formed is 5,5-dibromo-2,2,3-trimethylhexane. as shown below in attached scheme (Green Color).
<h3>Explanation:</h3>
Alkynes like Alkenes undergo <em>Electrophillic Addition Reactions</em>. The reaction given is a two step reaction. In step 1, the Alkyne adds first equivalent of HBr obeying <em>Markovnikov's rule</em> (i.e. Bromine will add to carbon containing less number of hydrogen atoms) and forms <em>2-bromo-4,5,5-trimethylhex-1-ene</em>. In step 2, the alkene formed in first step (2-bromo-4,5,5-trimethylhex-1-ene) undergoes addition reaction with the second equivalent of HBr via Markovnikov's rule to produce <em>5,5-dibromo-2,2,3-trimethylhexane</em>.
The scheme is attached below, Blue color is assigned to starting Alkyne, Red color is assigned to intermediate Alkene and Green color is assigned to product Alkane respectively.
You can use P1V1/T1 = P2V2/T2 but since pressure is constant is becomes V1/T1=V2/T2
V1=0.5 L
T1=203 K
T2=273 K
V2=unknown
0.5L/203 = V2/273
V2= 0.67 L so C
Hope this helps :)
Answer:
Adding heat.
Decreasing and increasing temperatures.
Explanation:
