In an undisturbed sequence of layers of rocks, the younger layers lie on top of the older layers
<h3><u>Answer;</u></h3>
True
<h3><u>Explanation</u>;</h3>
- The molecule NH3 contains all single bonds.
- NH3 has a three single covalent bond among its nitrogen and hydrogen atoms,because one valence electron of each of three atom of hydrogen is shared with three electron.
- There are three covalent bonds are in NH3 . Each hydrogen make a single bond with nitrogen and there is also a pair of electron which is unpaired from nitrogen.
We can use the ideal gas equation:
PV = nRT
P = 202.6kPa = 202600 Pa (You have to
multiply by 1000)
n = 0.050 mole
R = 0.082 atm*l/(K*mol)
T = 400K
We will have to convert from Pa to atm or
viceversa.
101325 Pa________1 atm
202600 Pa________x = 2.00 atm
2atm*V = 0.050 mole*0.082 atm*l/(K*mol)* 400K
V = 0.050 mole*0.082 atm*l/(K*mol)* 400K/2atm
= 0.82 liters = 820 mililiters
Explanation:
Relation between pressure, latent heat of fusion, and change in volume is as follows.

Also, 
where,
is the difference in specific volumes.
Hence, 
As,
= 22.0 J/mol K
And,
...... (1)
where,
= density of water
= density of ice
M = molar mass of water =
Therefore, using formula in equation (1) we will calculate the volume of fusion as follows.
=
=
Therefore, calculate the required pressure as follows.

=
or, = 145 bar/K
Hence, for change of 1 degree pressure the decrease is 145 bar and for 4.7 degree change dP =
= 681.5 bar
Thus, we can conclude that pressure should be increased by 681.5 bar to cause 4.7 degree change in melting point.