Answer:
ΔS° = 180.5 J/mol.K
Explanation:
Let's consider the following reaction.
4 NH₃(g) + 5 O₂(g) → 4 NO(g) + 6 H₂O(g)
The standard molar entropy of the reaction (ΔS°) can be calculated using the following expression.
ΔS° = ∑np × S°p - ∑nr × S°r
where,
ni are the moles of reactants and products
S°i are the standard molar entropies of reactants and products
ΔS° = 4 mol × S°(NO(g)) + 6 × S°(H₂O(g)) - 4 mol × S°(NH₃(g)) - 5 mol × S°(O₂(g))
ΔS° = 4 mol × 210.8 J/K.mol + 6 × 188.8 j/K.mol - 4 mol × 192.5 J/K.mol - 5 mol × 205.1 J/K.mol
ΔS° = 180.5 J/K
This is the change in the entropy per mole of reaction.
B) Tellurium is a metalloid. Metalloids in some classifications are also or alternatively called “semi-metals.”
Answer:
second order
Explanation:
units of reaction and their order.
Zero order --> M^1 s^-1 = M/s
First order --> M^0 s^-1 = 1/s
Second order --> M^-1 s^-1 = L/mol s
In the question rate constant k = 4.65 L mol-1s-1. = 4.65 L/mol s
Hence, the reaction is a second order reaction
Oxidation of D -Ribose in presence of hypobromous acid gives D-Ribonic acid
N₂H₄ + 2H₂O₂ → N₂ + 4H₂O
mol = mass ÷ molar mass
If mass of hydrazine (N₂H₄) = 5.29 g
then mol of hydrazine = 5.29 g ÷ ((14 ×2) + (1 × 4))
= 0.165 mol
mole ratio of hydrazine to Nitogen is 1 : 1
∴ if moles of hydrazine = 0.165 mol
then moles of nitrogen = 0.165 mol
Mass = mol × molar mass
Since mol of nitrogen (N₂) = 0.165
then mass of hydrazine = 0.165 × (14 × 2)
= 4.62 g