Answer is: D. 6.02 x 1023.
Because this is Avogadro constant<span> (the number of </span>constituent particles, in this example atoms of gold<span> that are contained in the </span>amount of substance<span> given by one </span>mole). <span>The </span>mole<span> is the </span>unit of measurement<span> for </span>amount of substance, t<span>he mole is an </span>SI base unit<span>, with the unit symbol </span>mol<span>.</span>
First, we write the half equations for the reduction of the chemical species present:
Cu⁺² + 2e → Cu; E° = 0.34 V
Ni⁺² + 2e → Ni; E° = - 0.23 V
In order to determine the potential of the cell, we find the difference between the two values. For this:
E(cell) = 0.34 - (-0.23)
E(cell) = 0.57 V
The second option is correct. (The difference in values is due to different values in literature, and it is negligible)
Answer:
solution given:
Explanation:
4Fe +3O2------------>2Fe2O3
4mole. 3mole. 2 mole
224g. 96g. 320g
we have
224 g of fe needed to react completely with 96 g of O2
500g of fe needed to react completely with
96 ×500/224=214.3g of O2
214.3g of O2 is a required answer.
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
Soil covered, saturated, submerged, flooded w water, standing water
