A sample of argon initially has a volume of 5.0 L and the pressure is 2 atm. If the final temperature is 30° C, the final volume
is 6 L, and the final pressure is 8atm, what was the initial temperature of the argon? Please give answer WITHOUT the unit of KELVIN and round to the tenth position (two places to the right of the decimal). Example: 25.14 not 25.4K
1 answer:
We have that every gas satisfies the fundamental gas equation, PV=nRT where P is the Pressure, V is the volume of the gas, n are the moles of the gas, R is a universal constant and T is the Temperature in Kelvin. We have that PV/T=nR and during our process, the moles of the gas do not change (no argon enters or escapes our sample). See attached.
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Answer:
See answer below
Explanation:
The picture below will show you the final product and mechanism.
In the first step, the NaNH₂ is a strong base, so, this base will substract the hydrogen from carbon 2, to generate a negative charge there, and then, carbon 2 becomes a nucleophyle.
As a nucleophyle it will attack to the CH₃I in the next step, and it will attach to the CH₃.
The second step is just a regular step to reduce the triple bond of the alkyne to alkane or alkene, this will depend on the quantity of the reactant. In this case, an alkene.
Hope this helps,,,,,,k
