The mass percent of carbon in a typical human is 18%, and the mass percent of 14c in natural carbon is 1.6 × 10-10%. assuming a 190.-lb person, how many decay events per second occur in this person due exclusively to the β-particle decay of 14c? for 14c, t1/2 = 5730 years

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.