1.3 Question 8 What is the equilibrium constant expression K, for the following reaction? CH4 (aq) + H₂O (8) CO (aq) + 3

Business, Finance, Economics, Accounting, Operations Management, Computer Science, Electrical Engineering, Mechanical Engineering, Civil Engineering, Chemical Engineering, Algebra, Precalculus, Statistics and Probabilty, Advanced Math, Physics, Chemistry, Biology, Nursing, Psychology, Certifications, Tests, Prep, and more.
Post Reply
answerhappygod
Site Admin
Posts: 899603
Joined: Mon Aug 02, 2021 8:13 am

1.3 Question 8 What is the equilibrium constant expression K, for the following reaction? CH4 (aq) + H₂O (8) CO (aq) + 3

Post by answerhappygod »

1 3 Question 8 What Is The Equilibrium Constant Expression K For The Following Reaction Ch4 Aq H O 8 Co Aq 3 1
1 3 Question 8 What Is The Equilibrium Constant Expression K For The Following Reaction Ch4 Aq H O 8 Co Aq 3 1 (41.8 KiB) Viewed 38 times
1 3 Question 8 What Is The Equilibrium Constant Expression K For The Following Reaction Ch4 Aq H O 8 Co Aq 3 2
1 3 Question 8 What Is The Equilibrium Constant Expression K For The Following Reaction Ch4 Aq H O 8 Co Aq 3 2 (51.86 KiB) Viewed 38 times
1.3 Question 8 What is the equilibrium constant expression K, for the following reaction? CH4 (aq) + H₂O (8) CO (aq) + 3 H₂ (8) OKP О Кр О КР O = = Kp= = O Kp = (Poo) (P₂) (Pcn₂) (PH₂o) (Poo) (PH₂) (Pon) (P₂o) (PCR) (PH₂O) (Poo)(PH₂)² (P₂) PH₂0 (PCH₂) (PH₂O) (Poo)(P₂) 1 pts
E D Question 2 1 pts The decomposition of a certain insecticide in water is a first order reaction. If it takes the insecticide 150 days to decay from a concentration of 0.300 M to a concentration of 0.200 M, what is the rate constant (in yr¹) for this reaction? 365 days 1 yr Question 3 1 pts The decomposition of SO₂Cl₂ is a first order process with a half-life of 2000.0 seconds. If a reaction starts with 0.750 M SO₂Cl₂, what will be the concentration (in M) after 3750 seconds?
Join a community of subject matter experts. Register for FREE to view solutions, replies, and use search function. Request answer by replying!
Post Reply