Cesium-131 has a half life of 9.7 days. what percent of a cesium-131 sample remains after 60 days?

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.

Light waves are as they pass through the

Consider the reaction: 2al(s) + fe2o3(s) → al2o3(s) + 2fe(s) the δhf for fe2o3(s) = -824.3 kj/mole. the δhf for al2o3(s) = -1675.7 kj/mole. finish the equation. δhrxn = [(1)( kj/mole) + (2)( kj/mole)] - [(1)( kj/mole) + (2) ( kj/mole)]