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.
(a) The displacement of point P at time t=0.10s is determined as +2cm.
(b) The displacement of point P at time t=0.20s is determined as -2cm.
<h3>
What is displacement?</h3>
Displacement is the change in position of an object. It is obtained from the product of velocity and time of motion.
x = vt
<h3>Displacement of the waves after 0.1 s</h3>
x = 10 m/s x 0.1 s = 1 m
Each wave will travel 1 m to the right or to the left, depending on the initial direction.
- wave B from left will stop at point 0 m
- wave A from left will stop at point -1 m
- wave C from right will stop at point 0 m
- wave D from right will stop at point + 1 m
wave B and C superimposed and the displacement will be between A and D.
Amplitude of A = - 2cm
Amplitude of D = + 4cm
Displacement of point P = 4 cm - 2 cm = 2cm
<h3>Displacement of the waves after 0.2 s</h3>
x = 10 m/s x 0.1 s = 2 m
Each wave will travel 2 m to the right or to the left, depending on the initial direction.
- wave B from left will stop at point 1 m
- wave A from left will stop at point 0 m
- wave C from right will stop at point -1 m
- wave D from right will stop at point 0 m
Displacement of point P = (amplitude B + amplitude C) + (amplitude A + amplitude D)
Displacement of point P= (2cm - 2cm) + (2 cm - 4cm)= -2cm
Learn more about displacement here: brainly.com/question/2109763
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setup 1 : to the right
setup 2 : equilibrium
setup 3 : to the left
<h3>Further explanation</h3>
The reaction quotient (Q) : determine a reaction has reached equilibrium
For reaction :
aA+bB⇔cC+dD
![\tt Q=\dfrac{C]^c[D]^d}{[A]^a[B]^b}](https://tex.z-dn.net/?f=%5Ctt%20Q%3D%5Cdfrac%7BC%5D%5Ec%5BD%5D%5Ed%7D%7B%5BA%5D%5Ea%5BB%5D%5Eb%7D)
Comparing Q with K( the equilibrium constant) :
K is the product of ions in an equilibrium saturated state
Q is the product of the ion ions from the reacting substance
Q <K = solution has not occurred precipitation, the ratio of the products to reactants is less than the ratio at equilibrium. The reaction moved to the right (products)
Q = Ksp = saturated solution, exactly the precipitate will occur, the system at equilibrium
Q> K = sediment solution, the ratio of the products to reactants is greater than the ratio at equilibrium. The reaction moved to the left (reactants)
Keq = 6.16 x 10⁻³
Q for reaction N₂O₄(0) ⇒ 2NO₂(g)
![\tt Q=\dfrac{[NO_2]^2}{[N_2O_4]}](https://tex.z-dn.net/?f=%5Ctt%20Q%3D%5Cdfrac%7B%5BNO_2%5D%5E2%7D%7B%5BN_2O_4%5D%7D)
Setup 1 :

Q<K⇒The reaction moved to the right (products)
Setup 2 :

Q=K⇒the system at equilibrium
Setup 3 :

Q>K⇒The reaction moved to the left (reactants)