When finding empirical formulas decimals are fine so you wouldn't have to multiply all of them to turn into whole numbers. What you would do is divide each of those numbers by the smallest of those numbers and take the nearest whole number so if you had:
1.81 mol .6 mol and .3.61 mol you would divide each of those numbers by .6
and you would have 3, 1, and 6
for your problem you would divide them all by 1 and Carbon would be 2 (if your teacher normally has you round down at .5's then it would be 1) hydrogen would be 3 and oxygen would be 1.
C2H3O
Answer:
the most useful hydrocarbon is fuel.
Explanation:
some examples like: natural gases,coal,diesel fuel, kerosene etc.
Answer: 20.2
Explanation:
Mass of isotope 1 = 20
% abundance of isotope 1 = 90.5% = 
Mass of isotope 2 = 22
% abundance of isotope 2 = 8.0% = 
Mass of isotope 3 = 23
% abundance of isotope 3 = 1.5% = 
Formula used for average atomic mass of an element :
![A=\sum[(20\times 0.905+(22\times 0.08)+(23\times 0.015]](https://tex.z-dn.net/?f=A%3D%5Csum%5B%2820%5Ctimes%200.905%2B%2822%5Ctimes%200.08%29%2B%2823%5Ctimes%200.015%5D)
Therefore, the average atomic mass of the element is 20.2.
Answer:
Ingredients
2 ⅓ cups all-purpose flour.
1 tablespoon baking powder.
¾ teaspoon salt.
1 ½ cups white sugar.
½ cup shortening.
2 eggs.
1 cup milk.
1 teaspoon vanilla extract.
Explanation:
Answer:
0.56L
Explanation:
This question requires the Ideal Gas Law: 
where P is the pressure of the gas, V is the volume of the gas, n is the number of moles of the gas, R is the Ideal Gas constant, and T is the Temperature of the gas.
Since all of the answer choices are given in units of Liters, it will be convenient to use a value for R that contains "Liters" in its units:
Since the conditions are stated to be STP, we must remember that STP is Standard Temperature Pressure, which means 
and 
Lastly, we must calculate the number of moles of 
there are. Given 0.80g of , we will need to convert with the molar mass of . Noting that there are 2 oxygen atoms, we find the atomic mass of O from the periodic table (16g/mol) and multiply by 2: 
Thus, 
Isolating V in the Ideal Gas Law:
...substituting the known values, and simplifying...
So, 0.80g of would occupy 0.56L at STP.
