Answer:
1.457*10^-8 grams
Explanation:
First we want to find the molar concentration of MgF2. We can do this by dividing 0.016 (the solubility in grams of MgCl2 in a litre of water) by its molar mass (approx. 62.3 grams). Thus, the molar solubility of MgF2 is 2.57*10^-4 M.
Next, we must calculate the Ksp of MgF2. The equilibrium expression is:
MgF2⇄Mg+2F
Thus
This means that, in equilibrium, there are 2.57*10^-4 M of and 5.136*10^-4 M of
Plugging in the above information, our Ksp for MgF2 is approximately 6.78*10^-11
Next we will need to use the RICE table. Since there is already 0.29M of NaF dissolved, there is initially 0.29M of .
R: MgF2 ⇄ +2
I: N/A 0 0.29M
C: N/A +x +x
--------------------------------------------
E: N/A x 0.29+x
To make calculations easier, we will assume that 0.29+x≈0.29
This means that Ksp=0.29x=6.78*10^-11
Therefore, x≈2.338*10^-10M
Multiply that by 62.3 and we get around 1.457*10^-8 grams.
Answer:
49.2°C is the temperature
Explanation:
Based on Charles's law, the absolute temperature of a gas is directly proportional to the volume under constant pressure. The equation is:
V1 / T1 = V2 / T2
<em>Where V is volume and T is absolute temperature of 1, initial state and 2, final state of the gas.</em>
<em />
Computing the values of the problem:
V1 = 0.432L
T1 = -20.0°C + 273.15K = 253.15K
V2 = 0.550L
T2 = ?
0.432L / 253.15K = 0.550L / T2
322.3K = T2
322.3K - 273.15 =
<h3>49.2°C is the temperature</h3>
Could be from liquid (most common refrigerant) to gas phase.