Explain why CH3CO2H has a lower pK than CH3CH2OH in terms of their conjugate bases

A6.10 m nacl can be made by adding [x]g of nacl to a container and making the volume of water up to the 1.00 l line

Predict the products of the following acid-base reactions, and predict whether the equilibrium lies to the left or to the right of the reaction arrow. part ao2-(aq)+h2o(l)< => express your answer as part of a chemical equation. identify all of the phases in your answer. o2-(aq)+h2o(l) < => oh-(aq)+oh-(aq)part bpredict whether the equilibrium lies to the left or to the right of the equation in previous part. h2o is a stronger acid than oh–, so the equilibrium lies to the right. h2o is a weaker acid than oh–, so the equilibrium lies to the left. h2o is a stronger acid than oh–, so the equilibrium lies to the left. h2o is a weaker acid than oh–, so the equilibrium lies to the right. part cch3cooh(aq)+hs? (aq) < => express your answer as part of a chemical equation. identify all of the phases in your answer. ch3cooh(aq)+hs-(aq) < => h2s(aq)+c2h3o2-(aq)h2s(aq)+c2h3o2-( aq)part dpredict whether the equilibrium lies to the left or to the right of the equation in previous part. ch3cooh is a weaker acid than h2s, so the equilibrium lies to the right. ch3cooh is a weaker acid than h2s, so the equilibrium lies to the left. ch3cooh is a stronger acid than h2s, so the equilibrium lies to the right. ch3cooh is a stronger acid than h2s, so the equilibrium lies to the left. part eno2-(aq)+h2o(l) < => express your answer as part of a chemical equation. identify all of the phases in your answer. no2-(aq)+h2o(l) < => part fpredict whether the equilibrium lies to the left or to the right of the equation in previous part. hno2 is a stronger acid than h2o, so the equilibrium lies to the right. hno2 is a weaker acid than h2o, so the equilibrium lies to the left. hno2 is a stronger acid than h2o, so the equilibrium lies to the left. hno2 is a weaker acid than h2o, so the equilibrium lies to the right.

All forces cause motion? true or false