The empirical formula of the given compound is 
.
The correct option is B.
<h3>What is an empirical formula?</h3>
The empirical formula of a chemical compound is the simplest whole-number ratio of atoms contained in the substance.
Given,
1.0 g of S
1.5 g of O
To calculate the empirical formula, we will divide the masses of the elements by their atomic weight.
For sulfur
For oxygen
Now, divide the greater value of mole came by the smaller value
Thus, the empirical formula for the given compound is 1 for S and 3 for O
Learn more about empirical formula, here:
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Answer:
Higher boiling point and lower freezing point
Explanation:
Boiling point is the temperature at which substances goes to gaseous phase form their liquid phase. Liquids boil when vapour pressure of the liquid becomes equal to atmospheric pressure. So boiling point of a liquid is affected by the change in its vapour pressure and vapour pressure is affected by adding a non-volatile solute in it.
Effect of adding non -volatile solute on vapour pressure:
When a non-volatile solute is added to a liquid, it alters the interaction between the liquid particles and prevents the liquid particles to going into gaseous phase which results in decrease in vapour pressure. More energy is needed for the vapour pressure of the liquid to becomes equal to atmospheric pressure. Therefore, boiling point increases.
CaCl2 is a non-volatile solute and hence increases the boiling point of the water.
Effect of adding non-volatile solute on freezing point:
When a non-volatile solute is added to a liquid, it alters the interaction between the liquid molecules. The new solute and solvent interaction prevents the liquid to soldify and requires more lowering in temperature. Therefore, freezing point decreases.
Answer:
entropy of reaction = 572-718= -146J/mol-k.
Explanation:
entropy of a reaction = entropy of product - entropy of reactant.
entropy of product = 2×entropy of SO2+2×entropy of NiO
=
=572 J/mol-k
entropy of reactant = 2×entropy if NiS+3×entropy of O2
=
=718 J/mol-k
therefore entropy of reaction = 572-718= -146J/mol-k.
There are two possible products from this elimination:
-2,3-dimethylbut-1-ene
-2,3-dimethylbut-2-ene
As the base is relatively unhindered, the reaction will form the Saytzeff product as the major product. The Saytzeff product is the most substituted alkene which is more stable due to hyperconjugation. In this reaction the Saytzeff product is 2,3-dimethylbut-2-ene.
