Element at Extreme Left In Periodic Table:
The elements of Group I-A (1) are present at extreme left of the periodic table. They are called as Alkali Metals. Alkali Metals are strong metals. These elements can easily loose their valence electron. The valence shell electronic configuration of these elements is,
ns¹
where n is principle quantum number, which shows main energy level or shell. These metals can gain Noble gas configuration (stable configuration) either by loosing one electron or by gaining seven or more electrons. As it is quite reasonable to loose one electron instead of gaining seven or more electrons so these element easily loose one electron to gain noble as configuration. The Metallic character decreases along the period from left to right. So Group II-A (2) are second most metallic elements and so on. These metals at extreme left mainly exist in solid form.
Element at Extreme Right In Periodic Table:
Elements present at extreme right of the periodic table lacks the properties of metallic character and act as non-Metals. They have almost complete outermost shell or have the deficiency of one or two electrons. They are not as hard as metallic elements and they exist with complete octet like in Noble gases, or deficient with one electron (Halogens) or two electrons (oxygen group). These elements tend to gain or accept electron if their valence shell is deficient with required number of elements. Like the valence electronic configuration of Halogens is,
ns², np⁵
So, Halogens readily accept one electron and attain noble gas configuration. Elements at extreme left exist mainly in gas phase.
the formula we is as follows:-
M1V1= M2V2
where
M1=1.2
V1=0.133l
V2=41l
M2=?
1.2 × 0.133 = 41 × M2
0.1596 = 41 × M2
M2 = 0.15960/41
M2 = 0.0038926829
<h3>
Answer:</h3>
1.85 M
<h3>
Explanation:</h3>
<u>We are given;</u>
- Number of moles as 0.50 mol
- Volume of the solution is 270 ml
But, 1000 mL = 1 L
- Thus, volume of the solution is 0.27 L
We are required to calculate the molarity of the solution;
- Molarity refers to the concentration of a solution in moles per liter.
- It is calculated by dividing number of moles with the volume.
Molarity = Moles ÷ Volume
In this case;
Molarity = 0.50 moles ÷ 0.27 L
= 1.85 Mol/L or 1.85 M
Therefore, molarity of the solution is 1.85 M
Answer : Amoxicillin Suspension 125 mg/ 5 ml is 125 mg of Amoxicillin per 5 ml of suspension is an example of weight to volume.
Explanation :
Weight by volume (w/v) means that the mass of solute present in 100 mL volume of solution.
Weight by weight (w/w) means that the mass of solute present in 100 gram of solution.
Volume by volume (v/v) means that the volume of solute present in 100 mL volume of solution.
As per question, amoxicillin suspension is, 125 mg/ 5 ml that means 125 mg of Amoxicillin present in 5 mL of suspension. So, it is an example of weight to volume.
Hence, it is an example of weight to volume.
