To present his data on the <span>solubility of three different salts in water at 22°C, bar graph should be used since there are different salts and the only variable is the type of salt used. Line graph and scatter plot use two coordinates or variables and is common to comparing data using the same sample while using histogram to find out data distribution is irrelevant.</span>
Density is defined as mass/volume (or m/v).
So,
(126.0 g)/(12.5 cm^3)= 10.08 g/cm^3
If your teacher requires correct significant figures, the answer is 10.1 g/cm^3.
If not, the first answer is fine.
<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: 40 + 2x14 + 6x16 = 164g/mole
54.3g x [1mole / 164g] = 0.331moles
355mL x 1L / 1000mL = 0.355L
molarity = 0.331moles / 0.355L =
00
Explanation:
