<em>Acetic acid, HC2H3O2</em>
First, calculate for the molar mass of acetic acid as shown below.
M = 1 + 2(12) + 3(1) + 2(16) = 60 g
Then, calculating for the percentages of each element.
<em> Hydrogen:</em>
P1 = ((4)(1)/60)(100%) = <em>6.67%</em>
<em> Carbon:</em>
P2 = ((2)(12)/60)(100%) = <em>40%</em>
<em>Oxygen</em>
P3 =((2)(16) / 60)(100%) = <em>53.33%</em>
<em>Glucose, C6H12O6</em>
The molar mass of glucose is as calculated below,
6(12) + 12(1) + 6(16) = 180
The percentages of the elements are as follow,
<em> Hydrogen:</em>
P1 = (12/180)(100%) = <em>6.67%</em>
<em>Carbon:</em>
P2 = ((6)(12) / 180)(100%) = <em>40%</em>
<em>Oxygen:</em>
P3 = ((6)(16) / 180)(100%) = <em>53.33%</em>
b. Since the empirical formula of the given substances are just the same and can be written as CH2O then, the percentages of each element composing them will just be equal.
Answer:
Deposition
Explanation:
It’s breaking down the rock :]
Answer:
Molarity of acid, Ca = Cb*Vb*A/Va*B
Explanation:
Using H2SO4 as acid, the reaction is as follow:
2NaOH + H2SO4 ⇒ Na2SO4 + 2H2O
Volume of acid = Va; Volume of base = Vb, Molar concentration of acid = Ca; Molar concentration of base = Cb; Molarity of acid = A and Molarity of base = B
Ca*Va/Cb*Vb =A/B
∴ Ca = Cb*Vb*A/Va*B
Therefore Chlorine is losing electrons and being oxidized. Hope it helps.
