Answer:
Here's what I get.
Explanation:
An acid is a proton donor; a base is a proton acceptor.
Thus, HF is the acid and H₂O is the base.
The conjugate base is what's left after the acid has given up its proton.
The conjugate acid is what's formed when the base has accepted a proton.
HF/F⁻ make one conjugate acid/base pair, and H₂O/H₃O⁺ are the other conjugate acid/base pair.
<em>Answer:</em>
- Organic compounds are those which are derivatives of hydrocarbons. They are classified into following functional groups.
<em>Alkane: </em>
- Alkane are simplest hydrocarbons.
- They have general formula CnH2n+2. These hydrocarbons contain single bond.
- For example ethane , H3C----CH3.
<em>Alkene:</em>
- Alkene are most reactive.
- They have general formula CnH2n.
- These contain double bond in their structure.
- For example , ethene, H2C=CH2
<em>Alkyne:</em>
- These are less less reactive as compare to alkenes.
- They have general formula CnH2n-2.
- They contain triple bonds in their structure.
- For example Acetylene HC≡CH
<em>Alcohol:</em>
Alcohol have functional group OH. They have general formual R---OH, R may be alkyl group.For example Ethanol H3CH2C---OH
<em>Amine:</em>
- Amine contain NH2 F.G.
- They have general formula R---NH2.
- There are three types of amine like primary, secondary and tertiary amine.
- For example H3CH2C---NH2
<em>Aldehyde:</em>
- Aldehydes have CHO F.G .
- They have general formula R--CHO.
- For example H3CH2C---CHO
<em>Ketone:</em>
- Ketones have R--CO--R functional group.
- For example acetone H3C---CO---CH3
<em>Carboxylic acid:</em>
- They functional group COOH.
- Their general formula is R---COOH.
- For example Acetic acid H3C---COOH
Exothermic is when heat escapes an object. So the answer should be the last one.
Answer:
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Explanation:
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<u>Answer: </u>
The pressure of a sample of argon gas was increased from 2.12atm to 6.96atm at constant temperature. If the final volume of the argon sample was 16.9L
. The initial volume of the argon sample was 55.5L
<u>Explanation:</u>
Considering the initial volume, initial pressure to be V1 and P1 respectively
Assuming the final volume and final temperature of the gas to be V2 and P2 respectively
Given,
Pressure of the gas was increased from 2.12 atm to 6.96 atm
Therefore, P1 = 2.12 atm
P2 = 6.96 atm
Final volume, V2 = 16.9 L
Applying Boyle’s law,
Substituting the value
V1 = 55.5 L
Therefore, the initial volume was 55.5L