CH3 is the empirical formula for the compound.
A sample of a compound is determined to have 1.17g of Carbon and 0.287 g of hydrogen.
The number of atom or moles in the compound is
1.17 g C X 1 mol of C / 12.011 g C = 0.097411 mol of C.
0.287 g H x 1 mol of H / 1 g H = 0.28474 mol H.
This compound contains 0.097411 mol of carbon and 0.28474 mol of Hydrogen.
So we can represent the compound with the formula C0.974H0.284.
Subscripts in formulas can be made into whole numbers by multiplying the smaller subscript by the larger subscript.
we can divide 0.284 by 0.0974.
0.284 / 0.0974 = 3.
So here, Carbon is one and hydrogen is 3.
We can write the above formula as a CH3.
Hence the empirical formula for the sample compound is CH3.
For a detailed study of the empirical formula refer given link brainly.com/question/13058832.
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Answer:
392 g
Explanation:
The given concentration tells us that<em> in 100 g of solution, there would be 15.3 g of 2-ethyltoluene</em>.
With that in mind we can<u> calculate how many grams of solution would contain 60.0 g of 2-ethyltoluene</u>:
- Mass of solution * 15.3 / 100 = 60.0 g 2-ethyltoluene
Explanation:
Galileo found Venus went through the entire phase cycle, proving that she revolved completely around the Sun, as predicted by Copernicus in his heliocentric model. ... The phases of Venus could not be explained by Ptolemy's Geocentric Model. Galileo knew that the Roman Catholic Church was incorrect.
Explanation:
For an isothermal process equation will be as follows.
W = nRT ln
It is given that mass is 10 kg/s or 10,000 g/s (as 1 kg = 1000 g). So, calculate number of moles of water as follows.
No. of moles = 
= 
= 555.55 mol/s
= 556 mol/s (approx)
As T = 
or (50 + 273.15) K = 323.15 K. Hence, putting the given values into the above formula as follows.
W = nRT ln[/tex]\frac{P_{1}}{P_{2}}[/tex]
= 
= 
= -3440193.809 J/s
Negative sign shows work is done by the pump. Since, 1 J = 0.001 kJ. Therefore, converting the calculated value into kJ as follows.
= 3440.193 kJ/s
= 3451 kJ/s (approx)
Thus, we can conclude that the pump work is 3451 kJ/s.
