(a) In this section, give your answers to three decimal places.
(i)
Calculate the mass of carbon present in 0.352 g of CO
2
.
Use this value to calculate the amount, in moles, of carbon atoms present in 0.240 g
of
A
.
(ii)
Calculate the mass of hydrogen present in 0.144 g of H
2
O.
Use this value to calculate the amount, in moles, of hydrogen atoms present in 0.240 g
of
A
.
(iii)
Use your answers to calculate the mass of oxygen present in 0.240 g of
A
Use this value to calculate the amount, in moles, of oxygen atoms present in 0.240 g
of
A
(b)
Use your answers to
(a)
to calculate the empirical formula of
A
thank you
hope it helpsss
We can calculate the mass percent of an element by dividing its atomic mass by the mass of the compound and then multiply by 100:
% by mass of element = (mass of element/mass of compound) x100%
Impurities like n-eicosane with the molecular formula C20H42 could account for the low percent by mass of oxygen in the sample because it has a zero percent oxygen based on its compound formula which indicates that it does not have the element oxygen.
Answer:
194 g/mol.
Explanation:
Hello,
In this case, one first must compute the mass of each element as shown below:

Next, the corresponding moles:

Then, each element's subscripts is found to be:

Therefore, the empirical formula is:

Nonetheless, it has a molar mass of 97bg/mol, thereby, by multiplying such formula by 2 one gets:

Which has a molar mass of 194 g/mol being correctly contained in the given interval.
Best regards.
B is correct. As you move down group 1, the elements become more reactive with other elements because the electrons have a weaker attraction to their own atoms nucleus which means attraction with other elements is much stronger, making the atom more reactive.
Answer : The enthalpy change for the reaction is, 201.9 kJ
Explanation :
According to Hess’s law of constant heat summation, the heat absorbed or evolved in a given chemical equation is the same whether the process occurs in one step or several steps.
According to this law, the chemical equation can be treated as ordinary algebraic expression and can be added or subtracted to yield the required equation. That means the enthalpy change of the overall reaction is the sum of the enthalpy changes of the intermediate reactions.
The balanced reaction of
will be,

The intermediate balanced chemical reaction will be,
(1)

(2)

(3)

(4)

Now we will multiply the reaction 1 by 2, revere the reaction 2, reverse and half the reaction 3 and 4 then adding all the equations, we get :
(1)

(2)

(3)

(4)

The expression for enthalpy of the reaction will be,



Therefore, the enthalpy change for the reaction is, 201.9 kJ