(easy)+(20 points)

Chemistry

2 answers:

Semmy [17]4 years ago

6 0

I am not sure but cold to hot

Vitek1552 [10]4 years ago

4 0

Liquid to solid and solid to liquid

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If the gas inside the flask is cooled so that its pressure is reduced from 797.3 torr to a value of 715.7 torr what will be the

nadezda [96]

The differential pressure is
797.9 - 715.7 = 82.2 torr = 138815.25 Pa
The height of the mercury in the open-end arm is calculated using
ΔP = ρgh
The density of mercury is 13560 kg/m3
The height of the mercury is
138815.25 Pa = 13560 kg/m3 (9.81 m/s2) h
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3 years ago

Importancia de las soluciones acuosas y su CONTAMINACIÓN

vova2212 [387]

En una solución acuosa donde el agua es el solvente, el soluto que será disuelto por el agua tiene menos partículas, lo que hace que las partículas se muevan en un movimiento aleatorio. El agua pura tiene una baja concentración de iones y, por lo tanto, no conduce electricidad.

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4 years ago

Using your value of Ksp, and starting with an equilibrium system consisting of a saturated solution of calcium hydroxide, predic

Ugo [173]

Answer:

(A) $HNO_{3}$ Solubility will increase.

(B) $NaOH$ Solubility will decrease.

(C) $Ca(NO_{3} )_{2}$ Solubility will decrease.

Explanation:

a) According to Le Chatelier's Principle when a change is introduced in a reaction system at equilibrium, the system responds by the reaction shifting in the direction to counter the change introduced.

When the change introduced is addition of acid, some $OH^{-}$ are consumed due to it and the system would respond by reaction shifting towards creation of more $OH^{-}$ in the system by increased dissolution. Thus solubility increase.

(b) When $OH^{-}$ are added, the reaction shifts to counter the change introduced of increased $OH^{-}$ by shifting to side that decreases the increased $OH^{-}$ . That happens by reaction shifting towards some $Ca (OH)_{2}$ precipitating back. Hence solubility decreases.

(c) The change introduced is addition of $Ca^{2+}$ which will be countered by reaction shifting to side that decreases increased $Ca ^{2+}$ , which happens by some $Ca(OH)_{2}$ precipitating back. Hence solubility decreases.