Answer: Option D) covalent bonds between water molecules
In water, hydrogen bonds are best described as covalent bonds between water molecules
Explanation:
The hydrogen bonds between water molecules are covalent bonds because they are formed when oxygen attract the lone electron in hydrogen, thus resulting in the formation of a partially negative charge on the oxygen atom and a partially positive charge on two hydrogen atoms
Thus, the sharing of electrons between oxygen and hydrogen atoms is responsible for the covalent bonds between water molecules
Answer:
1.. Lanthanides and actinides are kept separately in the modern periodic table Why?
Lanthanides and actinides are kept separately in the modern periodic table because of following reason:
- If they are placed in the body of the periodic table, the periodic table will become extremely long and hard to recognize
- more scientific and technical reason why they chose to keep the lanthanides and actinides away from the rest of the periodic table was to help in the grouping of elements in blocks
Explanation:
2. Differentiate between long period and short period in any two points.
Long period:
- The 4th and 5th periods of the periodic table are called long periods.
- The long periods coontain 18 elements in them
short period:
- The 2nd and 3rd periods of the periodic table are called short periods.
- The short periods contain 8 elements in them
Explanation:
The given data is as follows.
mass (m) = 107 g,
= 
Specific heat of water = 4.18 
Since, the relation between heat energy, mass and temperature change is as follows.
q = 
Hence, putting the given values into the above formula to calculate the heat energy as follows.
q = 
= 
= 1654.86 J
Therefore, calculate the enthalpy of this reaction for 2.00 mol of a compound as follows.
= 827.43 J/mol
or, =
(as 1 kJ = 1000 J)
= 0.827 kJ/mol
Therefore, we can conclude that enthalpy of this reaction is 0.827 kJ/mol.
Given: weight of solute (propane) = 52.6 g
weight of solvent (benzene) = 196 g = 0.196 kg
We know that, molecular weight of propane = 44.1 g/mol
∴ Molality of solution =

=

= 6.085 m
Now, Depression of freezing point = Kf m
where Kf = cryoscopic constant = <span>5.12 oC/m
</span>∴ Depression of freezing point = 5.12 X 6.085
= 31.15 oC
Therefore,
freezing point of solution = 5.50 - 31.15 = -25.65 oC