The number of particles in one mole is given be Avagadro's number <span>6.022×10^23
Multiply by number of moles.
3 ×10^-21 mol * 6.022 ×10^23 molecules/mol = </span><span>1,807 molecules
(rounded to nearest whole number)
</span>
In a combustion of a hydrocarbon compound, 2 reactions are happening per element:
C + O₂ → CO₂
2 H + 1/2 O₂ → H₂O
Thus, we can determine the amount of C and H from the masses of CO₂ and H₂O produced, respectively.
1.) Compute for the amount of C in the compound. The data you need to know are the following:
Molar mass of C = 12 g/mol
Molar mass of CO₂ = 44 g/mol
Solution:
0.5008 g CO₂*(1 mol CO₂/ 44 g)*(1 mol C/1 mol CO₂) = 0.01138 mol C
0.01138 mol C*(12 g/mol) = 0.13658 g C
Compute for the amount of H in the compound. The data you need to know are the following:
Molar mass of H = 1 g/mol
Molar mass of H₂O = 18 g/mol
Solution:
0.1282 g H₂O*(1 mol H₂O/ 18 g)*(2 mol H/1 mol H₂O) = 0.014244 mol H
0.014244 mol H*(1 g/mol) = 0.014244 g H
The percent composition of pure hydrocarbon would be:
Percent composition = (Mass of C + Mass of H)/(Mass of sample) * 100
Percent composition = (0.13658 g + 0.014244 g)/(<span>0.1510 g) * 100
</span>Percent composition = 99.88%
2. The empirical formula is determined by finding the ratio of the elements. From #1, the amounts of moles is:
Amount of C = 0.01138 mol
Amount of H = 0.014244 mol
Divide the least number between the two to each of their individual amounts:
C = 0.01138/0.01138 = 1
H = 0.014244/0.01138 = 1.25
The ratio should be a whole number. So, you multiple 4 to each of the ratios:
C = 1*4 = 4
H = 1.25*4 = 5
Thus, the empirical formula of the hydrocarbon is C₄H₅.
3. The molar mass of the empirical formula is
Molar mass = 4(12 g/mol) + 5(1 g/mol) = 53 g/mol
Divide this from the given molecular weight of 106 g/mol
106 g/mol / 53 g/mol = 2
Thus, you need to multiply 2 to the subscripts of the empirical formula.
Molecular Formula = C₈H₁₀
A.electrons are shared between two different nuclei
Arguably A, B and C are all correct. Atoms of elements can form compounds such as NaCl(table salt). Molecules form molecular compounds and minerals are just complex molecule compounds.
Answer:
Explanation:
Hello,
In this case, the law of mass action for the first reaction turns out:
![Kc=\frac{[AsH_3]^2}{[As]^2[H_2] ^3}=1.27](https://tex.z-dn.net/?f=Kc%3D%5Cfrac%7B%5BAsH_3%5D%5E2%7D%7B%5BAs%5D%5E2%5BH_2%5D%20%5E3%7D%3D1.27)
Now, for the second reaction is:
![Kc=\frac{[As][H_2] ^{3/2}}{[AsH_3]}](https://tex.z-dn.net/?f=Kc%3D%5Cfrac%7B%5BAs%5D%5BH_2%5D%20%5E%7B3%2F2%7D%7D%7B%5BAsH_3%5D%7D)
Therefore, by applying square root for the first reaction, one obtains:
![\sqrt{Kc} =\sqrt{\frac{[AsH_3]^2}{[As]^2[H_2] ^3}} =\sqrt{1.27}](https://tex.z-dn.net/?f=%5Csqrt%7BKc%7D%20%3D%5Csqrt%7B%5Cfrac%7B%5BAsH_3%5D%5E2%7D%7B%5BAs%5D%5E2%5BH_2%5D%20%5E3%7D%7D%20%3D%5Csqrt%7B1.27%7D)
![\frac{\sqrt{[AsH_3]^2} }{\sqrt{[As]^2} \sqrt{[H_2] ^3} } =\sqrt{1.27}](https://tex.z-dn.net/?f=%5Cfrac%7B%5Csqrt%7B%5BAsH_3%5D%5E2%7D%20%7D%7B%5Csqrt%7B%5BAs%5D%5E2%7D%20%5Csqrt%7B%5BH_2%5D%20%5E3%7D%20%7D%20%3D%5Csqrt%7B1.27%7D)
![\sqrt{1.27}=\frac{[AsH_3]}{[As][H_2] ^{3/2}}](https://tex.z-dn.net/?f=%5Csqrt%7B1.27%7D%3D%5Cfrac%7B%5BAsH_3%5D%7D%7B%5BAs%5D%5BH_2%5D%20%5E%7B3%2F2%7D%7D)
Finally, since Kc is asked for the inverse reaction, one modifies the previous equation as:
![Kc'=\frac{1}{\sqrt{1.27} }=\frac{[As][H_2] ^{3/2}}{[AsH_3]}=0.887](https://tex.z-dn.net/?f=Kc%27%3D%5Cfrac%7B1%7D%7B%5Csqrt%7B1.27%7D%20%7D%3D%5Cfrac%7B%5BAs%5D%5BH_2%5D%20%5E%7B3%2F2%7D%7D%7B%5BAsH_3%5D%7D%3D0.887)
Best regards.
