You can answer this question by only searching the element in the periodic table.
The atomic number of iodine, I, is 53. It is placed in the column 17 (this is the Group) and row 5 (this is the Period).
The conclusion is that the iodine is located in Period 5, Group 17, and is classified as a nonmetal.
One of the examples is radiation and chemistry of water. Environmental science requires the capacity to integrate data from the greater part of the significant fields of science, and in addition from arithmetic.
Geology is vital on the grounds that huge scale arrives forms make geology. The presence of mountains and valleys influences how much daylight and precipitation achieve the ground, how breezy an area is, the manner by which precipitation keeps running off, and numerous different variables that figure out what plants and creatures will have the capacity to occupy a district.
Answer:
[Cl-18]⁻ & [Cl-20]⁻
Explanation:
By definition isotopes are elements with the same number of protons by different number of neutrons. Elements X-18 & X-20 have 17 protons and represent Chlorine isotopes Cl-18 & Cl-20 each with 17 protons and 18 electrons to give the isotopes a -1 oxidation state. Both isotope of chlorine have 7 electron in its valence shell and 10 electrons in its core structure. Gaining 1 electron fills the valence octet and establishes a -1 oxidation state.
Chlorine is a halogen and all halogens and oxygen, nitrogen and hydrogen are diatomics
<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.
