How an air mattress supports the weight of a person lying on it

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.

Answer asap need to be answered by wednesday morning explain how a buffer works, using an ethanoic acid / sodium ethanoate system including how the system resists changes in ph upon addition of a small amount of base and upon addition of a small amount of acid respectively. include the following calculations in your i. calculate the ph of a solution made by mixing 25cm3 0.1m ch3cooh and 40cm3 0.1m ch3coo-na+. [ka = 1.74 x 10-5 m] ii. calculate the ph following the addition of a 10cm3 portion of 0.08 m naoh to 500cm3 of this buffer solution. iii. calculate the ph following the addition of a 10cm3 portion of 0.08 m hcl to 200cm3 of the original buffer solution.

What conclusion can you draw from this experiment about the components of the black ink?