True, if you have rough surface the more friction you have. The More slipery, less friction.
The periodic table of the elements are describe the electronic configuration of the elements on which the properties of the elements depends. Among the given groups only metal, non-metal and semi-metal group are the part of periodic table. The metallic property depends upon the binding energy of the electrons with the nucleus. Thus the elements which have the valence electrons more near to the nucleus that is s-block elements are more metallic in nature. On the other hand the elements which have the valence electrons far from the nucleus are more non-metallic in nature like p-block elements. However the binding energy or the attraction of the outermost electrons to the nucleus depends not only its valence electrons position but also some other factors like shielding effect, effective nuclear charge etc.
The elements which are in between the metals and non-metals can be classified as semi-metals.
Although the conductivity of a material is an inherent property of the metals but sometime the nonmetals or semi-metals are also behave like a conductor due to presence of the other elements, thus it cannot be a p[property of the periodic table. Similarly acidity, flammable gases are not part of the periodic table.
Ca(OH)2 is a strong base.
It’s name is calcium hydroxide and often it is easy to tell if something is a strong base if it has (OH) in the name.
Answer:
Measure 2.47 mL of the stock solution (i.e 0.0405 mM) and dilute it to the 100 mL mark with water
Explanation:
To make 100 mL of 0.001 mM solution from 0.0405mM solution, we need to determine the volume of 0.0405mM solution needed. This can be obtained as follow:
Molarity of stock (M₁) = 0.0405 mM
Volume of diluted (V₂) = 100 mL
Molarity of diluted solution (M₂) = 0.001 mM
Volume of stock solution needed (V₁) =?
M₁V₁ = M₂V₂
0.0405 × V₁ = 0.001 × 100
0.0405 × V₁ = 0.1
Divide both side by 0.0405
V₁ = 0.1 / 0.0405
V₁ = 2.47 mL
Therefore, to make 100 mL of 0.001 mM solution from 0.0405mM solution, measure 2.47 mL of the stock solution (i.e 0.0405 mM) and dilute it to the 100 mL mark with water.
