<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:
look at it is the correct answer because if a new gas has formed bubbles will how or change of colour so look at it
Explanation:
Answer:
0.726 mol·L⁻¹
Step-by-step explanation:
c = moles/litres
=====
Moles = 29.8 × 1/342.30
Moles = 0.087 06 mol
=====
Litres = 120 × 1/1000
Litres = 0.120 L
=====
c = 0.087 06/0.120
c = 0.725 mol·L⁻¹
The type of reaction in the example below is a double replacement.
A double replacement can be identified by the switch between two reactants. This can be verified because both a Br and Cl are being switched to a new position in the reaction.