Solution :
Comparing the solubility of silver chromate for the solutions :
----- Less soluble than in pure water.
----- Less soluble than in pure water.
----- Similar solubility as in the pure water
----- Similar solubility as in the pure water
The silver chromate dissociates to form :
![$AgCrO_4 (s) \rightleftharpoons 2Ag^+ (aq) +CrO_4^{2-}(aq)$](https://tex.z-dn.net/?f=%24AgCrO_4%20%28s%29%20%5Crightleftharpoons%202Ag%5E%2B%20%28aq%29%20%2BCrO_4%5E%7B2-%7D%28aq%29%24)
When 0.1 M of
is added, the equilibrium shifts towards the reverse direction due to the common ion effect of
, so the solubility of
decreases.
Both
and
are neutral mediums, so they do not affect the solubility.
Given the volume of HCl solution = 30.00 mL
Molarity of HCl solution = 0.1000 M
Molarity, moles and volume are related by the equation:
Molarity = ![\frac{Moles of solute}{Volume of solution (L)}](https://tex.z-dn.net/?f=%5Cfrac%7BMoles%20of%20solute%7D%7BVolume%20of%20solution%20%28L%29%7D)
Converting volume of HCl from mL to L:
![30.00 mL * \frac{1 L}{1000mL}=0.030000 L](https://tex.z-dn.net/?f=30.00%20mL%20%2A%20%5Cfrac%7B1%20L%7D%7B1000mL%7D%3D0.030000%20L)
Calculating moles of HCl from volume in L and molarity:
![0.03000 L * \frac{0.1000mol}{L}= 0.003000 mol HCl](https://tex.z-dn.net/?f=0.03000%20L%20%2A%20%5Cfrac%7B0.1000mol%7D%7BL%7D%3D%200.003000%20mol%20HCl)
The final moles would be reported to 4 sig figs. So the correct answer will be 0.03000 mol HCl
Correct option: C. 0.03000mol
Answer:
The answer is Frost Point.
Explanation:
The temperature to which the air must be cooled, with constant pressure, to reach saturation (in relation to liquid water), is called the dew point. The dew point gives a measure of the water vapor content in the air. The higher, the greater the concentration of water vapor in the air. However, when cooling produces saturation at a temperature of 0 ° C or less, the temperature is called a frost point. The water vapor is deposited as frost on a surface whose temperature is below the dew point.
Answer:
2.7 × 10⁻⁴ bar
Explanation:
Let's consider the following reaction at equilibrium.
SbCl₅(g) ⇄ SbCl₃(g) + Cl₂(g)
The pressure equilibrium constant (Kp) is 3.5 × 10⁻⁴. We can use these data and the partial pressures at equilibrium of SbCl₅ and SbCl₃, to find the partial pressure at equilibrium of Cl₂.
Kp = pSbCl₃ × pCl₂ / pSbCl₅
pCl₂ = Kp × pSbCl₅ / pSbCl₃
pCl₂ = 3.5 × 10⁻⁴ × 0.17 / 0.22
pCl₂ = 2.7 × 10⁻⁴ bar