For a hydrocarbon, the combustion reactions are the following:
C + O₂ --> CO₂
H₂ + 1/2 O₂ --> H₂O
The molar mass of CO₂ is 44 g/mol while C is 12 g/mol. Let's solve for amount of C in hydrocarbon.
Mass of C = (14.1 g CO₂)(1mol/44g)(1 mol C/1 mol CO₂)(12 g/mol) = 3.845 g C
So, that means that the mass of hydrogen is:
Mass of H = 4.4 - 3.845 = 0.555 g
Moles C = 3.845/12 = 0.32042
Moles H = 0.555/1 = 0.555
Divide both by the smaller value, 0.32042.
C: 0.32042/0.32042 = 1
H: 0.555/0.32042 = 1.732
We have to get an answer that is closest to a whole number. Let's try multiplying both with 4.
C: 1*4 = 4
H: 1.732*4 = 6.93≈7
<em>Thus, the empirical formula is C₄H₇.</em>
Answer:
Explanation:
Oxidation state of Cr in CrPO₄
As a general rule, the algebraic sum of all the oxidation numbers of all atoms in a neutral compound is zero.
The compound above is in its neutral state and we sum all the oxidation numbers and equate to zero:
The oxidation number of P is -2
O is -2
Let the oxidation number of Cr be x:
x + (-2) + 4(-2) = 0
x -2-8 = 0
x -10 = 0
x = +10
For Cr in Cr₃(PO₄)₂
Using the same rule:
2(x) + 2[-2 + 4(-2)] = 0
2x + 2(-2-8) = 0
2x -20 = 0
x = +10
<span>This problem uses the
relationship between Ka and the concentrations of the ions. Calculations are as follows:</span>
<span>
</span><span>1.9 x 10-5</span>= x^2 / (0.25 - x)
<span>x is very small and the denominator is approximately equal to 0.25. Thus, x is 2.2 x 10^-3
</span><span>pH = -log (2.2 x 10^-3)</span> = 2.66
