A. True would be the best answer
The noble gas notation is as follows- you must start with the noble gas that is before the element which in this case is Krypton and then from there you continue the electron configuration as follows- [Kr] 5s^2 4d^10 5p^2
For the compound B the following statement is correct-
B. It is an ether because it is unable to form a hydrogen bond, so it is less soluble in water.
The solubility of alcohol in water depends upon the capability of formation of hydrogen bond in the solute. Now in alcohol the -OH group is polar in nature which enhance the possibility of hydrogen bond formation and it is more soluble in water.
On the other hand although there presence a -O- functional group in ether. It is less soluble in water due to non polarity of the functional group.
From the given data it is seen that compound A is more soluble in water than compound B. Thus it may be predicted that compound A is alcohol and B is ether.
Henceforth, for the compound B the following statement is correct-
B. It is an ether because it is unable to form a hydrogen bond, so it is less soluble in water.
The reason of incorrect options:
A. compound B cannot be an alcohol as it is less soluble in water.
C. In ether the functional group is -O-, thus electronegative atom (O) is present.
D. As both the compound (alcohol and ether) has equal molecular mass thus the organic chain will be same in alcohol and the hydrogen bond interaction will be more prominent than the dispersion force between the -OH group.
25 drops of acid is required to neutralize the 50.0 ml of 0.010m of NaOH in the experiment.
The equation of the reaction is;
NaOH(aq) + HCl(aq) ---------> NaCl(aq) + H2O(l)
We can use the titration formula;
CAVA/CBVB = NA/NB
CA= concentration of acid
VA = volume of acid
CB = concentration of base
VB = volume of base
NA = number of moles of acid
NB = number of moles of base
CB = 0.010 M
VB = 50.0 ml
CA = 0.50 M
VA = ?
NA = 1
NB = 1
Substituting values;
CAVANB = CBVBNA
VA = 0.010 × 50.0 × 1/ 0.50 × 1
VA = 1 ml
Since the total volume of acid used is 1 ml and each drop contains 0.040 ml
The number of drops required is 1ml/0.040 ml = 25 drops
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