Answer: evaporation of water is equal to precipitation of water
Explanation:
An hydrological cycle or water cycle is the cyclic process in which the water circulates between different spheres of the earth in solid, liquid and gaseous forms. It circulates in earth's ocean or any other water body, then the vapors reach upto the atmosphere due to evaporation, where these vapors condense and precipitate into rain or snow, again it becomes part of the water body.
On the basis of the above explanation, evaporation of water is equal to precipitation of water is the correct option for the maintenance of the balance of the hydrological cycle on earth.
Explanation:
<em>We </em><em>know </em><em>that </em><em>metals </em><em>are </em><em>good </em><em>conductor </em><em>of </em><em>heat</em><em>. </em><em>so </em><em>it </em><em>allows </em><em>the </em><em>heat </em><em>to </em><em>flow </em><em>through </em><em>them </em>
<em>.</em><em> </em><em>So </em><em>as </em><em>we </em><em>know </em><em>the </em><em>spoon </em><em>is </em><em>a </em><em>metal </em><em>which </em><em>is </em><em>left </em><em>on </em><em>a </em><em>bowl </em><em>of </em><em>hot </em><em>soup </em><em>allows </em><em>heat </em><em>to </em><em>flow </em><em>through </em><em>it </em><em>easily </em><em>so </em><em>if </em><em>becomes </em><em>warm</em><em>. </em>
Answer:
0.59 mol O₂
Explanation:
The balanced chemical equation for the decomposition of potassium chlorate (KClO₃) to produce potassium chloride (KCl) and oxygen gas (O₂) is the following:
2 KClO₃ → 2 KCl + 3 O₂
According to the equation, 3 moles of O₂ are produced from 2 moles of KClO ⇒ conversion factor: 3 mol O₂/2 mol KClO₃
Now, we calculate the number of moles of KClO₃ there is in 48.1 g, by dividing the mass into the molecular weight (Mw) of O₂:
Mw(KClO₃) = 39.1 g/mol + 35.4 g/mol + (16 g/mol x 3) = 122.5 g/mol
moles KClO₃ = mass KClO₃/Mw(KClO₃) = 48.1 g/(122.5 g/mol) = 0.3926 mol KClO₃
Finally, we multiply the moles of KClO₃ by the conversion factor to calculate the moles of O₂ produced:
0.3926 mol KClO₃ x 3 mol O₂/2 mol KClO₃ = 0.59 mol O₂
Answer:
-285.4 J/K
Explanation:
Let's consider the following balanced equation.
HCl(g) + NH₃(g) ⇒ NH₄Cl(s)
We can calculate the standard entropy change for the reaction (ΔS°r) using the following expression.
ΔS°r = 1 mol × S°(NH₄Cl(s)) - 1 mol × S°(HCl(g)) - 1 mol × S°(NH₃(g))
ΔS°r = 1 mol × 94.6 J/K.mol - 1 mol × 187 J/K.mol - 1 mol × 193 J/K.mol
ΔS°r = -285.4 J/K