<span><span>N2</span><span>O5</span></span>
Explanation!
When given %, assume you have 100 g of the substance. Find moles, divide by lowest count. In this case you'll end up with
<span><span>25.92 g N<span>14.01 g N/mol N</span></span>=1.850 mol N</span>
<span><span>74.07 g O<span>16.00 g O/mol O</span></span>=4.629 mol O</span>
The ratio between these is <span>2.502 mol O/mol N</span>, which corresponds closely with <span><span>N2</span><span>O5</span></span>.
We will use this formlula: Mass in grams = Number of moles x Molecular mass of 1 mole.
Since, we know the avagadro number is 6.02 x 10²³, we only have two unknown values left which are the molecular mass of CH3OH and its mole.
Molecular Mass: C = 12, H= 1, O = 16, since we have C=12, H4 = 4, O = 16, we will add them up: 12 + 4 + 16 =32
We know that one mole of anything = 6.02 x 10²³.
So we will use this formula to find the mole of methanol: Number of moles = Number of molecules / Avagadro number
Number of moles of CH3OH = (9.79 x 10^24)/6.02 x 10²³) = 16.263 moles.
Now we know that the molecular mass = 32 and the mole is = 16.263.
Now we can find its mass by using this formula: <span>Mass in grams = Number of moles x Molecular mass of 1 mole.
</span>
Mass in grams = 16.263 x 32 = 520g
Answer:
Explanation:
the elements are arranged according to their atomic number - not their relative atomic mass . In the periodic table the elements are arranged into: rows, called periods , in order of increasing atomic number. vertical columns, called groups , where the elements have similar properties
Answer:
Keep it simple. If all the oxygen contained in the 200 grams of potassium chlorate is produced in the decomposition, then all we have to do is find out how many grams of oxygen are there in the 200 grams. This we can do by calculating the ratio of oxygen mass to the whole. Using 39.1 for potassium, 35.45 for chlorine and 3 times 16, or 48 for the oxygen, we get a total of 122.55 grams per mole for potassium chlorate, of which 48 grams are oxygen. This ratio is 48/122.55. This ratio times the original 200 grams of the compound, gives us 78.34 grams of oxygen produced.
Explanation:
