Gold is found in pure state in nature but not the iron because gold is un-reactive metal, so it does not react with other element in normal condition but iron is reactive metal so it reacts with atmospheric oxygen in presence of moist (water) to form oxide.
Explanation:
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Answer:
Plants will absorb water through their roots and release water as vapor into the air through these stomata. To survive in drought conditions, plants need to decrease transpiration to limit their water loss. Some plants that live in dry conditions have evolved to have smaller leaves and therefore fewer stomata.
Explanation:
Plants will absorb water through their roots and release water as vapor into the air through these stomata. To survive in drought conditions, plants need to decrease transpiration to limit their water loss. Some plants that live in dry conditions have evolved to have smaller leaves and therefore fewer stomata.
Answer is: the nature of the initial nickel sulfide mixture is a suspension.
Suspension<span> is a </span>heterogeneous mixture (solute<span> particles do not </span>dissolve), <span>that contains </span>solid<span> particles (in this example nickel sulfide or NiS) sufficiently large for </span>sedimentation. <span> The internal phase (solid nickel sulfide) is dispersed throughout the external phase (water).</span>
Answer:
T° freezing solution → -11.3°C
T° boiling solution → 103.1 °C
Explanation:
Assuming 100 % dissociation, we must find the i, Van't Hoff factor which means "the ions that are dissolved in solution"
This salt dissociates as this:
SnCl₄ (aq) → 1Sn⁴⁺ (aq) + 4Cl⁻ (aq) (so i =5)
The formula for the colligative property of freezing point depression and boiling point elevation are:
ΔT = Kf . m . i
where ΔT = T° freezing pure solvent - T° freezing solution
ΔT = Kb . m . i
where ΔT = T° boiling solution - T° boiling pure solvent
Freezing point depression:
0° - T° freezing solution = 1.86°C/m . 1.22 m . 5
T° freezing solution = - (1.86°C/m . 1.22 m . 5) → -11.3°C
Boiling point elevation:
T° boiling solution - 100°C = 0.512 °C/m . 1.22 m . 5
T° boiling solution = (0.512 °C/m . 1.22 m . 5) + 100°C → 103.1 °C
