Answer:
its false ...............
Answer:
camphor sublimates salt is soluble in water while sand does not sublime and does not dissolve in water you first heat the mixture in a beaker covered with a watch glass camphor will then accumulate on the watch glass then you dissolve the remaining mixture of sand and salt salt will dissolve forming a salt solution then you filter using a filter paper and a beaker the residue on the filter paper is sand while the filtrate is salt solution you then heat the salt solution so that it can evaporate leaving salt particles thus you will have obtained salt sand and camphor
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.
P1: 741 mmHg
V1: 3.49 L P1 x V1 / P2= (741 mmHg) (3.49 L) / 760 mmHg = 3.40 L
P2: 760 mmHg
V2: ? L