Answer:
Explanation:
2C₄H₁₀ + 13O₂ ⟶ 8CO₂ + 10H₂O
n/mol: 4.3
13 mol of O₂ react with 2 mol of 2C₄H₁₀
Assume there is 100g of the substance at first
Answer:
C. Its oxidation number increases.
Explanation:
- <em><u>Oxidation is defined as the loss of electrons by an atom while reduction is the gain of electrons by an atom</u></em>.
- Atoms of elements have an oxidation number of Zero in their elemental state.
- When an atom looses electrons it undergoes oxidation and its oxidation number increases.
- For example, <em><u>an atom of sodium (Na) at its elemental state has an oxidation number of 0. When the sodium atom looses an electrons it becomes a cation, Na+, with an oxidation number of +1 , the loss of electron shows an increase in oxidation number from 0 to +1.</u></em>
Answer:
K = 3.37
Explanation:
2 NH₃(g) → N₂(g) + 3H₂(g)
Initially we have 4 mol of ammonia, and in equilibrium we have 2 moles, so we have to think, that 2 moles have been reacted (4-2).
2 NH₃(g) → N₂(g) + 3H₂(g)
Initally 4moles - -
React 2moles 2m + 3m
Eq 2 moles 2m 3m
We had produced 2 moles of nitrogen and 3 mol of H₂ (ratio is 2:3)
The expression for K is: ( [H₂]³ . [N₂] ) / [NH₃]²
We have to divide the concentration /2L, cause we need MOLARITY to calculate K (mol/L)
K = ( (2m/2L) . (3m/2L)³ ) / (2m/2L)²
K = 27/8 / 1 → 3.37
