Answer:
The Ka value used for the buffer system is :- Ka2 = pKa2 = 7.2
pH = 7.1
Explanation:
The buffer system given in the question is :-
and
The reaction taking place is:-
The Ka value used for the buffer system is :- Ka2 = pKa2 = 7.2
The pH can be calculated as:-
pH = 7.2 + log 0.40/0.50 = 7.1
Formula for water is H20 and there are 5 mol water. 2 mol of hydrogen per 1 mol water so 5 mol of water times 2 mol hydrogen will give you 10 mol of hydrogen.
Same process with oxygen except 1 mol of oxygen per mol of water. So 5 mol water times 1 mol of oxygen will give you 5 mol of oxygen.
Answer:
Octet Rule is defined as the chemical rule in which atoms tend to form bonds in such a way that it completes its valence shell or tend to achieve the configuration of a noble gas.
Explanation:
The lithium is chemical element, which has atomic number 3. It has one valence electron in its outermost shell, therefore, it will require 7 more electrons to complete its octet.
The element nitrogen has the atomic number 7, such that it has 5 valence electrons in its outer most shell. Therefore, it will require 3 more electrons to complete its octet.
The noble gas Argon have complete octet, such that it does not require any electrons to complete its octet. the gas is inert and non-reactive.
The chemical element Boron has atomic number 5, such that it has three electrons in its outer most shell. Therefore, it will require 5 more electrons to complete its octet.
Hence, the correct option is Nitrogen.
The chemical formula for the compound can be written as,
CxHyOz
where x is the number of C atoms, y is the number of H atoms, and z is the number of O atoms. The combustion reaction for this compound is,
CxHyOz + O2 --> CO2 + H2O
number of moles of C:
(0.7191 g)(1 mol CO2/44 g of CO2) = 0.0163 mol CO2
This signifies that 0.0163 mole of C and the mass of carbon in the compound,
(0.0163 mols C)(12 g C/ 1 mol C) = 0.196 g C
number of moles H:
(0.1472 g H2O)(1 mol H2O/18 g H2O) = 0.00818 mol H2O
This signifies that there are 0.01635 atoms of H in the compound.
mass of H in the compound = (0.01635 mols H)(1 g of H) = 0.01635 g H
Mass of oxygen in the compound,
0.3870 - (0.196 g C + 0.01635 g H) = 0.1746 g
Moles O in the compound = (0.1746 g O)(1 mol O/16 g O) = 0.0109 mols O
The formula of the compound is,
C0.0163H0.01635O0.0109
Dividing the numbers by the least number,
C3/2H3/2O
The empirical formula of the compound is therefore,
<em> C₃H₃O₂</em>