Page. 5. What is the position of alkali metals in the Modern periodec table.
2 answers:
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<u>Answer:</u> Yes, metals and non-metals forms ionic bonds.
<u>Explanation:</u>
Ionic bond is defined as the bond which is formed by complete transfer of electrons from one atom to another atom.
The atom which looses the electron is known as electropositive atom and the atom which gains the electron is known as electronegative atom. This bond is usually formed between a metal and a non-metal.
<u>For Example:</u> Formation of sodium chloride
Sodium is a metal and is the 11th element of periodic table having electronic configuration of
To form ion, this element will loose 1 electron.
Chlorine is a non-metal and is the 17th element of periodic table having electronic configuration of .
To form ion, this element will gain 1 electron.
By criss-cross method, the oxidation state of the ions gets exchanged and they form the subscripts of the other ions. This results in the formation of a neutral compound.
So, the compound formed between sodium and chlorine atom is NaCl
Hence, metals and non-metals forms ionic bonds.
Given that:
- The heat of vaporization = 22.44 kJ/mol = 22440 J/mol
- normal boiling point which is the initial temperature = 0.4° C = (273 + (-0.4))K = 272.6 K
- volume = 250 mL = 0.250 L
- Mass of butane = 0.8 g
- the final temperature = -22° C = (273 + (-22)) K = 251 K
The first step is to determine the vapor pressure at the final temperature of 251K by using the Clausius-Clapeyron equation. This is following by using the ideal gas equation to determine the numbers of moles of butane gas. After that, the mass of butane present in the liquid is determined by using the relation for the number of moles.
Using Clausius-Clapeyron Equation:
where;
P1 and P2 correspond to the temperature at T1 and T2.
∴
replacing the values into the given equation, we have;
As such, at -22° C; the vapor pressure = 0.427 atm
Now, using the ideal gas equation:
PV = nRT
where:
- P = Pressure
- V = volume
- n = number of moles of butane
- R = universal gas constant
- T = temperature
∴
Making (n) the subject of the formula:
We all know that the standard molecular weight of butane = 58.12 g/mol
∴
Using the relation for the number of moles which is:
mass = 0.00518 mole × 58.12 g/mol
mass = 0.301 g
∴
The mass of butane in the flask = 0.301 g
But the mass of the butane present as a liquid in the flask is
= 0.8 g - 0.301 g
= 0.499 g
In conclusion, the mass of the butane present as a liquid in the flask is 0.499 g
Learn more about vapourization here: