Answer:
Electrons conducted heat
Explanation:
Iron is a metallic compound. One property of metallic compounds are that they have many loose electrons.
Consider the two rods to be a singular rod, since they touch:
An uncountable number of electrons at one end of the rod are heated, so they gain kinetic energy. Those electrons then collide with other electrons, which energize those as well (although less), which is equal to adding heat. This is a pattern that will flow all the way through the rod.
Answer:
Option D is correct.
The concentrations of both PCl₅ and PCl₃ are changing at equilibrium
Explanation:
Chemical equilibrium during a reversible chemical reaction, is characterised by an equal rate of forward reaction and backward reaction. It is better described as dynamic equilibrium.
This is because, the concentration of the elements and compounds involved in the reversible chemical reaction at equilibrium changes, but the rate of change of the reactants is always equal to the rate of change of products.
Hence, the concentration of reactants and products, such as PCl₅ and PCl₃ are allowed to change at equilibrium, but alas, the rate of forward reaction must always match the rate of backward reaction for the process to remain in a state of Chemical equilibrium.
Hope this Helps!!!
Answer:
Pentafluorobenzene: 11,92 min
Benzene: 12,14 min
Explanation:
<em>Retention time of pentafluorobenzene is 12,98 min and 13,20 min of benzene.</em>
The adjusted retention time is the time an analyte spends in the column not the stationary phase. As time of unretained solute is 1,06 min the adjusted retention time for an analyte is:
tr' = tr - 1,06min
For pentafluorobenzene:
tr' = 12,98min - 1,06min = <em>11,92 min</em>
For benzene:
tr' = 13,20 - 1,06min = <em>12,14 min</em>
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I hope it helps!
Answer:
0.33 mol/kg NH₃
Explanation:
Data:
b(NH₃) = 0.33 mol/kg
b(Na₂SO₄) = 0.10 mol/ kg
Calculations:
The formula for the boiling point elevation ΔTb is

i is the van’t Hoff factor — the number of moles of particles you get from a solute.
(a) For NH₃,
The ammonia is a weak electrolyte, so it exists almost entirely as molecules in solution.
1 mol NH₃ ⟶ 1 mol particles
i ≈ 1, and ib = 1 × 0.33 = 0.33 mol particles per kilogram of water
(b) For Na₂SO₄,
Na₂SO₄(aq) ⟶ 2Na⁺(aq) + 2SO₄²⁻(aq)
1 mol Na₂SO₄ ⟶ 3 mol particles
i = 1 and ib = 3 × 0.10 = 0.30 mol particles per kilogram of water
The NH₃ has more moles of particles, so it has the higher boiling point.