The matter is going to have to be in a <em>plasma </em>state! =)
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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We know that the number of moles HCl in 14.3mL of 0.1M HCl can be found by multiplying the volume (in L) by the concentration (in M).
(0.0143L HCl)x(0.1M HCl)=0.00143 moles HCl
Since HCl reacts with KOH in a one to one molar ratio (KOH+HCl⇒H₂O+KCl), the number of moles HCl used to neutralize KOH is the number of moles KOH. Therefore the 25mL solution had to contain 0.00143mol KOH.
To find the mass of KOH in the original mixture you have to divide the number of moles of KOH by the 0.025L to find the molarity of the KOH solution..
(0.00143mol KOH)/(0.025L)=0.0572M KOH
Since the morality does not change when you take some of the solution away, we know that the 250mL solution also had a molarity of 0.0572. That being said you can find the number of moles the mixture had by multiplying 0.0572M KOH by 0.250L to get the number of moles of KOH.
(0.0572M KOH)x(0.250L)=0.0143mol KOH
Now you can find the mass of the KOH by multiplying it by its molar mass of 56.1g/mol.
0.0143molx56.1g/mol=0.802g KOH
Finally you can calulate the percent KOH of the original mixture by dividing the mass of the KOH by 5g.
0.802g/5g=0.1604
the original mixture was 16% KOH
I hope this helps.
Answer:
What Structures??
Explanation:
Can you please explain the question more
Answer:
Ne has the larger ionization energy
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