It depends, for example, it is quite important to know the Kelvin scale (i.e 0 degrees Celsius is 273 K and -273 degrees Celsius is 0 K ) when dealing gases. But I don't know other situations where you would need to know other temperature scales.
Hope this helps and also if you are using Fahrenheit 1 Fahrenheit is -17.22 degrees Celsius
Baking powder is a mixture of baking soda and a mild edible acid such as tartaric acid
16 g. The mass of 0.60 mol Al is 16 g.
Molar mass of Al = 26.98 g/mol
Mass of Al = 0.60 mol Al x (26.98 g Al/1 mol Al) = 16 g Al
From the reduction standard potentials;
The emf of Zinc = -0.76 V
and the emf of Aluminium = -1.66 V
In a galvanic cell the component with lower standard reduction potential gets oxidized and that it is added to the anode compartment.
Therefore. the voltage of a galvanic cell made with zinc and aluminium will be;
Voltage =Ered- Eoxd
= -0.76 - (-1.66)
= 0.9 V
Answer:
2MnO₄⁻ + 5Zn + 16H⁺ → 2Mn²⁺ + 8H₂O + 5Zn²⁺
Explanation:
To balance a redox reaction in an acidic medium, we simply follow some rules:
- Split the reaction into an oxidation and reduction half.
- By inspecting, balance the half equations with respect to the charges and atoms.
- In acidic medium, one atom of H₂O is used to balance up each oxygen atom and one H⁺ balances up each hydrogen atom on the deficient side of the equation.
- Use electrons to balance the charges. Add the appropriate numbers of electrons the side with more charge and obtain a uniform charge on both sides.
- Multiply both equations with appropriate factors to balance the electrons in the two half equations.
- Add up the balanced half equations and cancel out any specie that occur on both sides.
- Check to see if the charge and atoms are balanced.
Solution
Zn + MnO₄⁻ → Zn²⁺ + Mn²⁺
The half equations:
Zn → Zn²⁺ Oxidation half
MnO₄⁻ → Mn²⁺ Reduction half
Balancing of atoms(in acidic medium)
Zn → Zn²⁺
MnO₄⁻ + 8H⁺ → Mn²⁺ + 4H₂O
Balancing of charge
Zn → Zn²⁺ + 2e⁻
MnO₄⁻ + 8H⁺ + 5e⁻→ Mn²⁺ + 4H₂O
Balancing of electrons
Multiply the oxidation half by 5 and reduction half by 2:
5Zn → 5Zn²⁺ + 10e⁻
2MnO₄⁻ + 16H⁺ + 10e⁻→ 2Mn²⁺ + 8H₂O
Adding up the two equations gives:
5Zn + 2MnO₄⁻ + 16H⁺ + 10e⁻ → 5Zn²⁺ + 10e⁻ + 2Mn²⁺ + 8H₂O
The net equation gives:
5Zn + 2MnO₄⁻ + 16H⁺ → 5Zn²⁺ + 2Mn²⁺ + 8H₂O
