first we have to find the empirical formula of the compound
empirical formula is the simplest ratio of whole numbers of components making up a compound
for 100 g of the compound
N O
mass 46.7 g 53.3 g
number of 46.7 g/ 14 g/mol 53.3 g/ 16 g/mol
moles = 3.34 mol = 3.33 mol
divide by the least number of moles
3.34/3.33 = 1.00 3.33/ 3,33 = 1.00
therefore number of atoms are
N - 1
O - 1
empirical formula is - NO
mass of empirical unit - 14 g/mol + 16 g/mol = 30 g
molecular formula is actual composition of elements in the compound
molecular mass - 60.01 g/mol
number of empirical units = molecular mass / empirical unit mass
= 60.01 g/mol / 30 g = 2
there are 2 empirical units
2(NO)
molecular formula = N₂O₂
Answer:
Define Transportation in plants: Transportation in plants is when the plant transports water and other mineral throughout the whole plant from the roots to the stem and finally specific parts of a plant.
Define Respiration In Plants: Is when plants use photosynthesis to make their own food to make energy for the plant's growth
Answer: 1. Is2 2s2 2p3
2. Nitrogen
Explanation: The number of electron present In C = 6
But an extra electron is added since the charge on C is -1, this therefore makes the total electron 7.
1. By arrangement, the Electronic configuration is therefore;
Ans: 1s2 2s2 2p3
2. It is explained how C has 7 electrons, we can proceed then.
Neutral atom have atomic number of 7.
The element with atomic number of 7 is;
Ans: NITROGEN
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
