For the first question, salt is soluble while sand is insoluble or not dissolvable in water. The salt should have vanished or melted, but the sand stayed noticeable or visible, making a dark brown solution probably with some sand particles caught on the walls of the container when the boiling water was put in to the mixture of salt and sand. The solubility of a chemical can be disturbed by temperature, and in the case of salt in water, the hot temperature of the boiling water enhanced the salt's capability to melt in it.
For the second question, the melted or dissolved salt should have easily made its way through the filter paper and into the second container, while the undissolved and muddy sand particles is caught on the filter paper. The size of the pores of the filter paper didn’t change. On the contrary, the size of the salt became smaller because it has been dissolved which is also the reason why it was able to go through the filter paper, while the size of the sand may have doubled or even tripled which made it harder to pass through.
Variations in electronegativity prompt in the unequal halves of electrons in polar molecules because when one atom is more electronegative than the other, it becomes more polar than the other.
It results in the more electronegative atom to have a slightly negative (-ve) charges, and the other atom to have partial or slightly positive(+ve) charges.
Polar molecules have unequal sharing of electrons because the atoms have unequal attraction for electrons so the sharing is unequal.
The larger the difference in electronegativity between the two atoms, the more the polar the bond.
Hydrogen bonds are involved in unequal sharing of electrons between two atoms.
To know more about variations in electronegativity in polar molecules here :
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Answer:
Mass of oxygen in glucose = 29.3g
Explanation:
Mass of glucose given is 55grams.
We are to find the mass of oxygen in this compound.
In the compound we have 6 atoms of oxygen.
Solution
To find the mass of oxygen in glucose, we calculate the formula mass of glucose. We now divide the formula mass of the oxygen atom with that of the glucose and multiply by the given mass to find the unkown mass.
Atomic mass of C = 12g
H = 1g
O = 16g
Formula mass of C₆H₁₂O₆ = {(12x6) + (1x12) + (16x6)} = 180
Mass of O in glucose = 
x 55
= 
x 55
= 0.53 x 55
Mass of oxygen in glucose = 29.3g
