Answer: The oxygen atom in a water molecule carries a negative electrical charge.
Explanation:
A polar covalent bond is defined as the bond which is formed when there is a difference of electronegativities between the atoms.
Hydrogen bonding (H-bonding) is an intermolecular force having partial ionic-covalent character. H-bonding takes place between a hydrogen atom (attached with an electronegative atom e.g. O, N and F) and an electronegative atom (O,N and F).
In , 
, O is a highly electronegative atom attached to a H atom through a covalent bond. Therefore O atom gets partial negative charge and H atoms get partial positive charge.
Answer:
Option D. 0.115 M
Explanation:
The following data were obtained from the question:
Mass of CuSO4 = 36.8 g
Volume of solution = 2 L
Molar mass of CuSO4 = 159.62 g/mol
Molarity of CuSO4 =..?
Next, we shall determine the number of mole in 36.8 g of CuSO4.
This can be obtained as shown below:
Mass of CuSO4 = 36.8 g
Molar mass of CuSO4 = 159.62 g/mol
Mole of CuSO4 =.?
Mole = mass /Molar mass
Mole of CuSO4 = 36.8 / 159.62
Mole of CuSO4 = 0.23 mole
Finally, we shall determine the molarity of the CuSO4 solution as illustrated below:
Mole of CuSO4 = 0.23 mole
Volume of solution = 2 L
Molarity of CuSO4 =..?
Molarity = mole /Volume
Molarity of CuSO4 = 0.23 / 2
Molarity of CuSO4 = 0.115 M
Therefore, the molarity of the CuSO4 solution is 0.115 M.
Answer:
D. 3220
Explanation:
1 km equals 1000 m, so 3.22 km is
3220 m
Answer:
Option C is false statement. The half life of a second order reaction is not dependent on concentration.
Explanation:
Half life of a reaction is defined as the amount of time which is required for a reactant concentration reduced by half comparison to its initial concentration.
Half life of a second order reaction is depend on the initial concentration of a reaction, in contrast to 1st order reaction.
Answer:
A. 85.6 g
= 0.0856 kg.
B. 0.00027 mol/g
= 0.27 mol/kg.
C. 8.39 %
Explanation:
Given:
Molar concentration = 0.25 M
Molar weight of sucrose = 342.296 g/mol
Density of solution = 1.02 g/mL
Mass of water = 934.4 g.
Density in g/l = 1.020 g/ml * 1000ml/1 l
= 1020 g/l
Mass of solution in 1 l of solution = 1020 g
Mass of solution = mass of solvent + mass of solute
Mass of sucrose = 1020 - 934.4
= 85.6 g of sucrose in 1 l of solution.
A.
Density of sucrose = mass/volume
= molar mass/molar concentration
= 342.296 * 0.25
= 85.6 g/l
Number of moles = mass/molar mass
= 85.6/342.296
= 0.25 mol
B.
Molality = number of moles of solute/mass of solvent
= 0.25/934.4
= 0.00027 mol/g
C.
% mass of sucrose = mass of sucrose/total mass of solution * 100
= 85.6/1020 * 100
= 8.39 %
