potassium chromate - k2cro4. ,,silver nitrate-agno3. ,, potassium nitrate-kno3 ,,Silver chromate- ag2cro4
Particles of gas are more scarcely placed as compared to that of liquid.
the intermolecular forces will be less in gaseous state and hence is less stable
The Relative Formula Mass of NaH2PO4 is 120 g/mol
Therefore, the number of moles = 6.6/120
= 0.055 moles of NaH2PO4 which is also equal to the number of moles of H2PO4.
[H2PO4-] = Number of moles oof H2PO4-/Volume of the solution in L
= 0.055/ ( 355 ×10^-3)
= 0.155 M
Na2HPO4 undergoes complete dissociation as follows;
Na2HPO4 (aq)= 2Na+ (aq) + HPO4^2- (aq)
1 mole of Na2HPO4 = 142 g/mol
Therefore; number of moles = 8.0/142
= 0.0563 moles
[HPO4 ^-2] is given by no of moles HPO4^2- /volume of the solution in L
= 0.0563/(355×10^-3)
= 0.1586 M
Both H2PO4^2- and HPO4^2- are weak acids the undergoes partial dissociation
Ka of H2PO4- = 6.20 × 10^-8
[H+] =Ka*([H2PO4-]/[HPO4(2-)]
= (6.20 ×10^-8)×(0.155/0.1586)
= 6.059 ×10^-8 M
pH = - log[H+]
= - log (6.059×10^-8)
= 7.218
The balanced reaction
is:
4NH3 + 3O2 --> 2N2 + 6H2O
<span>We
are given the amount of reactants to be used for the reaction. This
will be the starting point of our calculation.</span>
83.7g of O2 ( 1 mol / 32 g) = 2.62 mol O2
2.81 moles of NH3
From the balanced reaction, we have a 4:3 ratio of the reactants. The limiting reactant would be oxygen. We will use the amount for oxygen for further calculations.
<span>2.62 mol O2</span><span> (6 mol H2O / 3 mol O2) (18.02 g H2O / 1 mol H2O) = 94.42 g H2O</span>
