The answer is: It expands when it freezes.
When molecule frezes, it lose energy. When molecules are far apart, it means the volume is greater and it expands.
For example, ice expands when water is freezing.
Hydrogen bond is an electrostatic attraction between two polar groups that occurs when a hydrogen atom (H), covalently bound to a highly electronegative atom such as flourine (F), oxygen (O) and nitrogen (N) atoms.
Answer:
Water is a unique molecule and thus consists of unique properties. It is considered polar because of its charges and its bent shape. The two oxygen are negative and the hydrogen is positive. This creates two poles, making the molecule polar. This is an example of polar covalent chemical bonding.
Answer:
0.095 moles of Calcium is there in 5.74 x 1022 atoms of calcium.
Explanation:
- As we know, 6.023*10^23 atoms of an element is equal to its atomic weight.
And, 6.023*10^23 atoms of an element is also equal to 1 mole of the element.
We have,
- 6.023*10^23 atoms of element calcium equals to 1 mole of Calcium
- 5.74*10^22 atoms of element calcium equals to
(1/(6.023*10^.23)) * 5.74*10^22 moles of calcium
Therefore,
- 5.74 x 1022 atoms of calcium= 0.095 moles of calcium.
Answer:
<u><em>neurons</em></u>
Explanation:
The long-axoned cells, called principal neurons, transmit information over long distances from one brain region to another (Sheperd,1979). Principal neurons provide the pathways of communication within the nervous system.
Explanation:
It is known that 
value of acetic acid is 4.74. And, relation between pH and is as follows.
pH = pK_{a} + log ![\frac{[CH_{3}COOH]}{[CH_{3}COONa]}](https://tex.z-dn.net/?f=%5Cfrac%7B%5BCH_%7B3%7DCOOH%5D%7D%7B%5BCH_%7B3%7DCOONa%5D%7D)
= 4.74 + log 
So, number of moles of NaOH = Volume × Molarity
= 71.0 ml × 0.760 M
= 0.05396 mol
Also, moles of 
= moles of 
= Molarity × Volume
= 1.00 M × 1.00 L
= 1.00 mol
Hence, addition of sodium acetate in NaOH will lead to the formation of acetic acid as follows.
Initial : 1.00 mol 1.00 mol
NaoH addition: 0.05396 mol
Equilibrium : (1 - 0.05396 mol) 0 (1.00 + 0.05396 mol)
= 0.94604 mol = 1.05396 mol
As, pH = pK_{a} + log ![\frac{[CH_{3}COONa]}{[CH_{3}COOH]}](https://tex.z-dn.net/?f=%5Cfrac%7B%5BCH_%7B3%7DCOONa%5D%7D%7B%5BCH_%7B3%7DCOOH%5D%7D)
= 4.74 + log 
= 4.69
Therefore, change in pH will be calculated as follows.
pH = 4.74 - 4.69
= 0.05
Thus, we can conclude that change in pH of the given solution is 0.05.
