Remember that a cation will be smaller than its neutral atom, and an anion will be larger than its neutral atom. This would automatically eliminate answer choices A and D.
Also keep in mind that atomic radii decreases from left to right as you move along a periodic table. It also decreases from bottom up.
Atomic radii increases as you move from right to left and as you go from up to down.
As bromine is higher up in the periodic table than Iodine, it would have a smaller radius. Iodine would have a larger radius.
The correct answer is B. Br
Answer: The ion that contribute to water hardness are:
--> a. Ca2+
--> b. (HCO)3^- and
--> c. Mg2+
While K+ DOES NOT contribute to water hardness.
Explanation:
WATER in chemistry is known as a universal solvent. This is so because it is polar in nature and dissolves most inorganic solutes and some polar organic solutes to form aqueous solutions. It is composed of elements such as hydrogen and oxygen in the combined ratio of 2:1.
Water is said to be HARD if it does not lather readily with soap. There are two types of water hardness:
--> Permanent hardness: This is mainly due to the presence of CALCIUM and MAGNESIUM ions in the form of soluble tetraoxosulphate(VI) and chlorides. These ions are removed by adding washing soda or caustic soda.
--> Temporary hardness: This is due to the presence of calcium HYDROGENTRIOXOCARBONATES. It can be removed by boiling and using slaked lime.
Therefore from the above given ions, Ca2+,(HCO)3^- and Mg2+ contributes to water hardness.
Answer:
Kindly check the explanation section.
Explanation:
PS: kindly check the attachment below for the required diagram that is the diagram showing solid sodium chloride looks like at the atomic level.
The chemical compound known as sodium chloride, NaCl has Molar mass: 58.44 g/mol, Melting point: 801 °C and
Boiling point: 1,465 °C. The structure of the solid sodium chloride is FACE CENTRED CUBIC STRUCTURE. Also, solid sodium chloride has a coordination number of 6: 6.
In the diagram below, the positive sign shows the sodium ion while the thick full stop sign represent the chlorine ion.
H₂CO₃ ⇔ HCO₃⁻ + H⁺
I 0.160 0 0
C -x +x +x
E 0.160-x +x +x
Ka1 = [HCO₃⁻][H⁺] / [H₂CO₃]
4.3 x 10⁻⁷ = x² / (0.160-x) (x is neglected in 0.160-x = 0.160)
x² = 6.88 x 10⁻⁸
x = 2.62 x 10⁻⁴
HCO₃⁻ ⇔ CO₃⁻² + H⁺
I 2.62 x 10⁻⁴ 0 2.62 x 10⁻⁴
C -x +x +x
E 2.62 x 10⁻⁴ - x +x 2.62 x 10⁻⁴ + x
Ka2 = [CO₃⁻²][H⁺] / [HCO₃⁻]
5.6 x 10⁻¹¹ = x(2.62 x 10⁻⁴ + x) / (2.62 x 10⁻⁴ - x)
x = 5.6 x 10⁻¹¹
Thus,
[H₂CO₃] = 0.160 - (2.62 x 10⁻⁴) = 0.16 M
[HCO₃⁻] = 2.62 x 10⁻⁴ - ( 5.6 x 10⁻¹¹) = 2.6 x 10⁻⁴ M
[CO₃⁻²] = 5.6 x 10⁻¹¹ M
[H₃O⁺] = 2.62 x 10⁻⁴ + 5.6 x 10⁻¹¹ = 2.6 x 10⁻⁴ M
[OH⁻] = 3.8 x 10⁻¹¹