Put a 2 on the HCl.
2HCl+Ba-->BaCl2+H2
Answer:
NaCl + AgF → NaF + AgCl
Explanation:
A double replacement reaction is a type of chemical reaction that occurs when two reactants exchange cations or anions to yield two new products.
From all the reactions given ,
- 2Na + Cl₂ → 2NaCl is an example of combination reaction because two or more reactants (Na & Cl₂) react with each other to form a single product (NaCl)
- H₂SO₃ → H₂O + SO₂ is an example of decomposition reaction because a single reactant (H₂SO₃) breaks down into two or more products (H₂O & SO₂).
- 2K + 2H₂O → 2KOH + H₂ is an example of displacement reaction because a highly reactive element (K) displaces a least reactive element (H) from its compound (H₂O).
- NaCl + AgF → NaF + AgCl is an example of double replacement reaction because there's an exchange between Cations (
&
) and Anions (
&
).
Electrons uniting with electrons of another atom is the cause in this relationship. The effect is a chemical change.
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
Answer:
0.01836 M
Explanation:
Again the reaction equation is;
Fe(s) + Mn2+(aq) → Fe2+(aq) + Mn(s)
E°cell= 0.77 V
Ecell= 0.78 V
[Mn2+] = 0.040 M
[Fe2+] = the unknown
n=2
From Nernst's equation;
Ecell= E°cell- 0.0592/n log Q
0.78= 0.77 - 0.0592/2 log [Fe2+] /[0.040]
0.78-0.77= - 0.0592/2 log [Fe2+] /[0.040]
0.01/ -0.0296= log [Fe2+] /[0.040]
-0.3378= log [Fe2+] /[0.040]
Antilog(-0.3378) = [Fe2+] /[0.040]
0.459= [Fe2+] /[0.040]
[Fe2+] = 0.459 × 0.040
[Fe2+] = 0.01836 M