Answer:
Explanation:
1. Write the skeleton equation for the half-reaction
NO₃⁻ ⟶ N₂O
2. Balance all atoms other than H and O
2NO₃⁻ ⟶ N₂O
3. Balance O by adding H₂O molecules to the deficient side.
2NO₃⁻ ⟶ N₂O + 5H₂O
4. Balance H by adding H⁺ ions to the deficient side.
2NO₃⁻ + 10H⁺ ⟶ N₂O + 5H₂O
5. Balance charge by adding electrons to the deficient side.
2NO₃⁻ + 10H⁺ + 8e⁻ ⟶ N₂O + 5H₂O
The amount of charge required to reduce 2 mol of NO₃⁻ is 8 F
Answer:
the range should be 2.2 to 4.3
Explanation:
I think so because the numbers at the left side of the scale from 1 are more acidic so as it increases it's still acidic but lesser so 1 is more acidic than 2 so I used 2.2 as the beginning of the range because it's less acidic than A even though its a greater number and 4.3 is lesser than 4.4 but its still greater on the scale. frankly speaking I don't feel so correct because it's in decimal so try and compare facts thank you
Answer:
This is an oxidation-reduction (redox) reaction:
2 Ni0 - 4 e- → 2 NiII
(oxidation)
2 O0 + 4 e- → 2 O-II
(reduction)
Ni is a reducing agent, O2 is an oxidizing agent.
Answer:
120g
Explanation:
We'll begin by writing the balanced equation for the reaction. This is given below:
Sn + 2HF —> SnF2 + H2
Next, we shall determine the number of mole of HF needed to react with 3 moles of Sn.
From the balanced equation above, 1 mole of Sn reacted with 2 moles of HF.
Therefore, 3 moles of Sn will react with = 3 x 2 = 6 moles of HF.
Finally, we shall convert 6moles of HF to grams
This is illustrated below:
Number of mole of HF = 6moles
Molar Mass of HF = 1 + 19 = 20g/mol
Mass of HF =..?
Mass = number of mole x molar Mass
Mass of HF = 6 x 20
Mass of HF = 120g
Therefore, 120g of HF is needed to react with 3 moles of Sn
