The empirical formula for the given compound has been
. Thus, option C is correct.
The empirical formula has been the whole unit ratio of the elements in the formula unit.
<h3>Computation for the Empirical formula</h3>
The given mass of Sulfur has been, 50.05 g
The given mass of oxygen has been 49.95 g.
The moles of elements in the sample has been given by:


The moles of sulfur in the unit has been 1.56 mol.

The moles of oxygen in the unit has been 3.12 mol.
The empirical formula unit has been given as:

Thus, the empirical formula for the given compound has been
. Thus, option C is correct.
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Answer:
When the concentration of all the reactants increases, more molecules or ions interact to form new compounds, and the rate of reaction increases. When the concentration of a reactant decreases, there are fewer of that molecule or ion present, and the rate of reaction decreases.
(a) The nature of bond between A and B is an ionic bond.
(b) The two main properties of the ionic compounds are:
- Ionic Compounds have high boiling and melting points as they're very strong and require a lot of energy to break.
- The electrostatic forces of attraction between oppositely charged ions lead to the formation of ions.
(c) If the ionic compound is dissolved in water, the ions in the solid separate and disperse uniformly
<h3>What is an ionic compound?</h3>
Ionic compounds contain ions and are held together by the attractive forces among the oppositely charged ions.
An ionic bond is formed by the complete transfer of some electrons from one atom to another. The atom losing one or more electrons becomes a cation—a positively charged ion.
In ionic bonds, the metal loses electrons to become a positively charged cation, whereas the nonmetal accepts those electrons to become a negatively charged anion.
When ionic compounds dissolve in water, the ions in the solid separate and disperse uniformly throughout the solution.
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The % yield if 500 g of sulfur trioxide reacted with excess water to produce 575 g of sulfuric acid is calculated using the below formula
% yield = actual yield/ theoretical yield x100
actual yield =575 grams
to calculate theoretical yield
find the moles of SO3 used =mass/molar mass
= 500g/ 80 g/mol =6.25 moles
SO3+H2O=H2SO4
by use of mole ratio of SO3 : H2SO4 which is 1:1 the moles of H2SO4 is also= 6.25 moles
the theoretical yield of H2SO4 is therefore = moles /molar mass
= 6.25 x98= 612.5 grams
%yield is therefore= 575 g/612 g x100= 93.9 %