Answer:
The options are unclear, however, the correct option is:
Aqueous solutions of ionic compounds cause to dissociate, hence, ions are free to conduct electricity
Explanation:
Ionic compounds are compounds formed from ions (charged atoms). For example, NaCl is an ionic compound from the following ions; Na+ (cation) and Cl- (anion). One characteristics of ionic compounds is their ability to dissociate into the ions that form them when in an aqueous solution i.e. NaCl will dissociate into Na+ and Cl- when in an aqueous solution.
These disssociated ions are free to conduct electricity, hence, making ionic compounds good conductors of electricity.
Potential energy can be calculated by the formula Pe=mgh. Plug in your values:
Pe=mgh
Pe=(6 kg)(9.8m/s^2)(100 m)
Pe=5880 kg x m^2/s^2, or 5880 Joules
Answer: Option (b) is the correct answer.
Explanation:
Vitamins which are soluble in water are known as water-soluble vitamins. And, vitamins which remain insoluble in water are known as water-insoluble vitamins.
Water-soluble vitamins are able to be easily excreted by the human body through urine. For example, vitamin C and vitamin B-complex are water soluble vitamins.
Thus, we can conclude that water-soluble vitamins are easily excreted in the urine.
Answer:
0.17 lb
Explanation:
78 g * (1 lb/454 g)=0.17 lb
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Answer:
1.58x10⁻⁵
2.51x10⁻⁸
0.0126
63.10
Explanation:
Phenolphthalein acts like a weak acid, so in aqueous solution, it has an acid form HIn, and the conjugate base In-, and the pH of it can be calculated by the Handerson-Halsebach equation:
pH = pKa + log[In-]/[HIn]
pKa = -logKa, and Ka is the equilibrium constant of the dissociation of the acid. [X] is the concentrantion of X. Thus,
i) pH = 4.9
4.9 = 9.7 + log[In-]/[HIn]
log[In-]/[HIn] = - 4.8
[In-]/[HIn] = 
[In-]/[HIn] = 1.58x10⁻⁵
ii) pH = 2.1
2.1 = 9.7 + log[In-]/[HIn]
log[In-]/[HIn] = -7.6
[In-]/[HIn] = 
[In-]/[HIn] = 2.51x10⁻⁸
iii) pH = 7.8
7.8 = 9.7 + log[In-]/[HIn]
log[In-]/[HIn] = -1.9
[In-]/[HIn] = 
[In-]/[HIn] = 0.0126
iv) pH = 11.5
11.5 = 9.7 + log[In-]/[HIn]
log[In-]/[HIn] = 1.8
[In-]/[HIn] = 
[In-]/[HIn] = 63.10
