Explanation:
The chemical reaction given in the question is as follows -
MnO₄⁻ (aq) + 8H⁺ (aq) + 5e⁻ → Mn²⁺ (aq) + 4H₂O (l)
NO₃⁻ (aq) + 4H⁺ (aq) + 3e⁻ → NO (g) + 2H₂O (l)
As we know , the value for reduction potential are -
Mn²⁺ = + 1.51 V
NO₃⁻ = +0.96 V
From , the data given above , the value of the reduction potential of NO₃⁻ is less than the reduction potential of Mn²⁺ .
Hence ,
NO₃⁻ can not oxidize Mn²⁺ .
Answer:
In Fahrenheit it is 294.8°F and in Celsius it is 146°C.
Answer:
The Net reaction is
-
-
-
Explanation:
From the Question we are told that the buffers are
and 
When NaOH is added the Net ionic reaction would be
-
-
-
Answer:
0.4 moles of water produced by 6.25 g of oxygen.
Explanation:
Given data:
Mass of oxygen = 6.25 g
Moles of water produced = ?
Solution:
Chemical equation;
2H₂ + O₂ → 2H₂O
Number of moles of oxygen:
Number of moles = mass/ molar mass
Number of moles = 6.35 g/ 32 g/mol
Number of moles = 0.2 mol
Now we will compare the moles of oxygen with water:
O₂ : H₂O
1 : 2
0.2 : 2×0.2 = 0.4 mol
0.4 moles of water produced by 6.25 g of oxygen.
