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⁺ .
Carbon dioxide (CO2)
Argon (Ar)
Hydrogen (H)
Helium (He)
Gas i took the test plz mark brainliest!
Answer:
The specific rotation of D is 11.60° mL/g dm
Explanation:
Given that:
The path length (l) = 1 dm
Observed rotation (∝) = + 0.27°
Molarity = 0.175 M
Molar mass = 133.0 g/mol
Concentration in (g/mL) = 0.175 mol/L × 133.0 g/mol
Concentration in (g/mL) = 23.275 g/L
Since 1 L = 1000 mL
Concentration in (g/mL) = 0.023275 g/mL
The specific rotation [∝] = ∝/(1×c)
= 0.27°/( 1 dm × 0.023275 g/mL
)
= 11.60° mL/g dm
Thus, the specific rotation of D is 11.60° mL/g dm
