Suppose that 6.2 mL of 2.1 M KOH(aq) is transferred to a 250 mL volumetric flask and diluted to the mark. It was found that 36.9
1 answer:
Answer:
0.928 M
Explanation:
The concentration of acid can be determined by using the volume used and the concentration and volume used of base.
We will use the law of equivalence of moles.
M₁V₁=M₂V₂
M₁ = concentration of base used
V₁ = volume of base used
M₂ = concentration of acid used =? (to be determined)
V₂ = volume of acid used
The initial concentration of KOH used is diluted so let us find the final concentration of KOH after dilution
initial moles = final moles
initial concentration X initial volume = final concentration X final volume
6.2 X 2.1 = 250 X final concentration
final concentration = 0.052 M = M₁
V₁ = 36.9 mL
V₂ = 6.2 mL
Here with each mole of phosphoric acid three moles of KOH are used.
Therefore
3 M₁V₁ = M₂V₂
M₂ =
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REASON:
Elements which have 4 valence electrons are generally metalloids.
The metalloids show the properties of both metals and non-metals.
We know that,
no. of protons= Atomic number
So,
Atomic no.= 32
Hence,
The element is Germanium which is a metalloid with 4 valence electrons and has 32 protons in nucleus of each atom because it has atomic no. 32
HOPE IT IS USEFUL
REASON:
Elements which have 4 valence electrons are generally metalloids.
The metalloids show the properties of both metals and non-metals.
We know that,
no. of protons= Atomic number
So,
Atomic no.= 32
Hence,
The element is Germanium which is a metalloid with 4 valence electrons and has 32 protons in nucleus of each atom because it has atomic no. 32
HOPE IT IS USEFUL
E0,correctly written as E
, is the <span> standard electrode potential of the cell. This is usually used in the subject, electrochemistry. This is used to measure the difference in potential between two half cells used in an electrochemical cell. The o above signifies that it is at its standard state. </span>