The reaction of removing CO2
using LiOH is the following:
2 LiOH + CO2 -----> Li2CO3
+ H2O
By solving the amount of CO2
the LiOH can scrub:
(3.50 × 10^4 g LiOH) (1 mol LiOH/
24 g LiOH) ( 1 mol CO2 / 2 mol LiOH) ( 44 g CO2 /1 mol CO2) = 32, 083.33 g CO2
it can scrub
<span>Since number of astronaut = 32,
083.33 g / 9 (8.8 × 10^2) = 4 astronaut</span>
Answer:
measuring the consumption of the products and the rate of conversion of products to reagents
Explanation:
It must be taken into account that the reaction rate also depends on the presence of a catalyst, since these advance the rate of the reaction.
Answer:
Mn(s)/Mn^2+(aq)//Co^2+(aq)/Co(s)
Explanation:
In writing the cell notation for an electrochemical cell, the anode is written on the left hand side while the cathode is written on the right hand side. The two half cells are separated by two thick lines which represents the salt bridge.
For the cell discussed in the question; the Mn(s)/Mn^2+(aq) is the anode while the Co^2+(aq)/Co(s) half cell is the cathode.
Hence I can write; Mn(s)/Mn^2+(aq)//Co^2+(aq)/Co(s)
Hello, I would like to help you, but I really don't understand the question
Answer:
I would say, what helps me is really paying attention in class and asking questions, also making sure you study for upcoming test's and quizzes and completely assingments on time
Explanation:
