Answer:
An input of heat energy Yr don't you know stupied
Answer:
By using renewable energy sources.
Explanation:
You need to satisfy a quota of energy for (whatever country you live in). People do this by using the cheapest way of producing the most energy, the most efficient. Sounds great right? Wrong! It stuffs our atmosphere with harmful gasses like carbon dioxide. You can reduce the use of these fossil fuels by using renewable energy sources such as windmills, watermills, and most notable, solar panels!
Answer:
ΔE = 150 J
Explanation:
From first law of thermodynamics, we know that;
ΔE = q + w
Where;
ΔE is change in internal energy
q is total amount of heat energy going in or coming out
w is total amount of work expended or received
From the question, the system receives 575 J of heat. Thus, q = +575 J
Also, we are told that the system delivered 425 J of work. Thus, w = -425 J since work was expended.
Thus;
ΔE = 575 + (-425)
ΔE = 575 - 425
ΔE = 150 J
Some of the salt would settle out. When the water was heated, it was able to absorb more salt than usual. This is known as super saturation. When the water is frozen it cannot hold as much salt, so some of it has to come out.
Answer:
30.4 g. NH3
Explanation:
This problem tells us that the hydrogen (H2) is the limiting reactant, as there is "an excess of nitrogen." Using stoichiometry (the relationship between the various species of the equation), we can see that for every 3 moles of H2 consumed, 2 moles of NH3 are produced.
But before we can use that relationship to find the number of grams of ammonia produced, we need to convert the given grams of hydrogen into moles:
5.4 g x [1 mol H2/(1.008x2 g.)] = 2.67857 mol H2 (not using significant figures yet; want to be as accurate as possible)
Now, we can use the relationship between H2 and NH3.
2.67857 mol H2 x (2 mol NH3/3 mol H2) = 1.7857 mol NH3
Now, we have the number of moles of ammonia produced, but the answer asks us for grams. Use the molar mass of ammonia to convert.
1.7857 mol NH3 x 17.034 g. NH3/mol NH3 = 30.4 g. NH3 (used a default # of 3 sig figs)
