Answer:
One gallon of octane produces approximately 7000 L of carbon dioxide.
Note:
I believe that the mass of octane should have been given as 2661 g. However, I understand that your instructor probably gave you this problem, so I will use 4000 g for the approximate mass of one gallon of octane. You can rework the problem on your own, substituting the correct masses of octane if you wish.
Step1. You must first determine the number of moles that are in 4000 g of octane, using the molar mass of octane. Step 2. Then you must determine the number of moles of carbon dioxide that can be produced by that number of moles of octane, based on the mole ratio between octane and carbon dioxide in the balanced equation. Step 3. Then use the ideal gas law to determine the volume in liters of carbon dioxide that can be formed.
Answer: The mass is 980.6g of Gold.
Explanation:
We begin by looking for the number of moles equivalent to 3.0 x 10^24 gold atoms.
Using the Avogadro's number,
6.02 x 10^23 atoms of gold make up 1 mole of gold.
3.0 x 10^24 atoms would make up: 1 / 6.02 x 10^23 x 3.0 x 10^24 = 4.98moles.
Now that we know the number of moles, we can then look for the mass using the formular:
Moles = mass/ molar mass
4.98 = mass / 196.9 (atomic mass of gold)
Making "mass" the subject of formula : mass = 4.98 x 196.9= 980.6g
Answer:
False
Explanation:
The first reaction is;
NO(g) + 1/2O2(g) ---->NO2(g)
K= [NO2]/[NO] [ O2]^1/2
The second reaction is;
2NO(g) + O2(g) ---->2NO2(g)
K'= [NO2]^2/[NO]^2 [O2]
It now follows that;
K'= K^2
Hence the statement in the question is false
The percentage by mass of oxygen in the compound
find the total mass=( 1.900+ 0250 +0.850) = 3
the percentage mass mass of oxgyen/total mass x100
that is (0.850/3) x100=28.33%
The little dipper is located in Ursa Minor you would also get a clue because... Minor and little
