The change in Gibbs Free Energy<span> </span><span> for any reaction is related to the equilibrium constant </span><span> by the simple equation- thats all i know</span>
Answer: Nail polish and Lithium batteries are Limited quantity item.
Explanation:
Answer:
See the answer below
Explanation:
Organic molecules in the form of carbohydrates are created during photosynthesis or chemosynthesis and these processes cause carbon to be removed from the atmosphere or other environment into the biosphere - the living organisms of the ecosystem.
During photosynthesis, atmospheric carbon dioxide is fixed as carbohydrates into green plants in a two-stage process that requires sunlight energy and water. The entire process can be summarized as an equation below:
Instead of using water as a reactant, chemosynthetic organisms make use of high energy chemicals such as hydrogen sulfide to fix carbon dioxide in order to generate carbohydrates.
Both processes ensure that carbon dioxide moves from the environment into living organisms.
Answer:
1. 15.71 g CO2
2. 38.19 % of efficiency
Explanation:
According to the balanced reaction (2 CO(g) + O2(g) → 2 CO2(g)), it is clear that the CO is the limitant reagent, because for every 2 moles of CO we are using only 1 mole of O2, so even if we have the same quantity for both reagents, not all of the O2 will be consumed. This means that we can just use the stoichiometric ratios of the CO and the CO2 to solve this question, and for that we need to convert the gram units into moles:
For CO:
C = 12.01 g/mol
O = 16 g/mol
CO = 28.01 g/mol
(10.0g CO) x (1 mol CO/28.01 g) = 0.3570 mol CO
For CO2:
C = 12.01 g/mol
O = 16 x 2 = 32 g/mol
CO2 = 44.01 g/mol
We now that for every 2 moles of CO we are going to get 2 moles of CO2, so we resolve as follows:
(0.3570 mol CO) x (2 mol CO2/2 mol CO) = 0.3570 moles CO2
We are obtaining 0.3570 moles of CO2 with the 10g of CO, now lets convert the CO2 moles into grams:
(0.3570 moles CO2) x (44.01 g/1 mol CO2) = 15.71 g CO2
Now for the efficiency question:
From the previous result, we know that if we produce 15.71 CO2 with all the 10g of CO used, we would have an efficiency of 100%. So to know what would that efficiency be if we would only produce 6g of CO2, we resolve as follows,
(6g / 15.71g) x 100 = 38.19 % of efficiency
