Unless you're able to provide a diagram representing the problem, there's not much I can do so solve this problem.
<u>Answer:</u> No crystals of potassium sulfate will be seen at 0°C for the given amount.
<u>Explanation:</u>
We are given:
Mass of potassium nitrate = 47.6 g
Mass of potassium sulfate = 8.4 g
Mass of water = 130. g
Solubility of potassium sulfate in water at 0°C = 7.4 g/100 g
This means that 7.4 grams of potassium sulfate is soluble in 100 grams of water
Applying unitary method:
In 100 grams of water, the amount of potassium sulfate dissolved is 7.4 grams
So, in 130 grams of water, the amount of potassium sulfate dissolved will be 
As, the soluble amount is greater than the given amount of potassium sulfate
This means that, all of potassium sulfate will be dissolved.
Hence, no crystals of potassium sulfate will be seen at 0°C for the given amount.
Answer:
<h2>1.9</h2>
Explanation:
The pH of a solution can be found by using the formula
![pH = - log [ {H}^{+} ]](https://tex.z-dn.net/?f=pH%20%3D%20-%20log%20%5B%20%7BH%7D%5E%7B%2B%7D%20%5D)
From the question we have

We have the final answer as
<h3>1.9</h3>
Hope this helps you
Answer:
The reaction is not spontaneous in the forward direction, but in the reverse direction.
Explanation:
<u>Step 1: </u>Data given
H2(g) + I2(g) ⇌ 2HI(g) ΔG° = 2.60 kJ/mol
Temperature = 25°C = 25+273 = 298 Kelvin
The initial pressures are:
pH2 = 3.10 atm
pI2 = 1.5 atm
pHI 1.75 atm
<u>Step 2</u>: Calculate ΔG
ΔG = ΔG° + RTln Q
with ΔG° = 2.60 kJ/mol
with R = 8.3145 J/K*mol
with T = 298 Kelvin
Q = the reaction quotient → has the same expression as equilibrium constant → in this case Kp = [p(HI)]²/ [p(H2)] [p(I2)]
with pH2 = 3.10 atm
pI2 = 1.5 atm
pHI 1.75 atm
Q = (3.10²)/(1.5*1.75)
Q = 3.661
ΔG = ΔG° + RTln Q
ΔG = 2600 J/mol + 8.3145 J/K*mol * 298 K * ln(3.661)
ΔG =5815.43 J/mol = 5.815 kJ/mol
To be spontaneous, ΔG should be <0.
ΔG >>0 so the reaction is not spontaneous in the forward direction, but in the reverse direction.
Answer:
979 atm
Explanation:
To calculate the osmotic pressure, you need to use the following equation:
π = <em>i </em>MRT
In this equation,
-----> π = osmotic pressure (atm)
-----><em> i</em> = van't Hoff's factor (number of dissolved ions)
-----> M = Molarity (M)
-----> R = Ideal Gas constant (0.08206 L*atm/mol*K)
-----> T = temperature (K)
When LiCl dissolves, it dissociates into two ions (Li⁺ and Cl⁻). Therefore, van't Hoff's factor is 2. Before plugging the given values into the equation, you need to convert Celsius to Kelvin.
<em>i </em>= 2 R = 0.08206 L*atm/mol*K
M = 20 M T = 25°C + 273.15 = 298.15 K
π = <em>i </em>MRT
π = (2)(20 M)(0.08206 L*atm/mol*K)(298.15 K)
π = 979 atm