Covalent bonds do not involve electron transfer because in covalent bonds, the electrons are shared between the atoms to form the molecule rather than being transferred to form ions as it is in the case of ionic bonds.<span />
The balanced equation for the above reaction is as follows;
C + H₂O ---> H₂ + CO
stoichiometry of C to H₂O is 1:1
1 mol of C reacts with 1 mol of H₂O
we need to find which is the limiting reactant
2 mol of C and 3.1 mol of H₂O
therefore C is the limiting reactant and H₂O is in excess.
stoichiometry of C to H₂ is 1:1
then number of H₂ moles formed are equal to C moles reacted
number of H₂ moles formed = 2 mol
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1. 0.33 M
2. 0.278 M
<h3>Further explanation</h3>
Molarity is a way to express the concentration of the solution
Molarity shows the number of moles of solute in every 1 liter of solute or mmol in each ml of solution
Where
M = Molarity
n = Number of moles of solute
V = Volume of solution
1. 0.350 mol of NaOH in 1.05 L of solution.
n=0.35
V=1.05 L
Molarity :
2. 14.3 g of NaCl in 879 mL of solution.
mol NaCl(MW=58.5 g/mol) :
Molarity :
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
