Answer:
The purpose of cryolite during electrolysis is that the mixture of cryolite and aluminium oxide has a lower melting point than pure aluminium oxide. Molten cryolite serves as a solvent for the molten aluminium oxide and increases the conductivity of the solution.
Answer:
2-hydroxy
Explanation:
The 4-hydroxy has a reduced dipole because its two electronegative substituents give it a dipole in opposite directions. On the other hand the 2-hydroxy has dipoles that add together.
<h2>Steps:</h2>
- Remember that Density = mass/volume, or D = m/v
So firstly, we have to find the volume of the rock. To do this, we need to subtract the volume of water A from the volume of the water B. In this case:
- Water A = 30 mL
- Water B = 40 mL
- 40 mL - 30 mL = 10 mL
<u>The volume of the rock is 10 mL.</u>
Now that we have the volume, we can plug that and the density of the rock into the density equation to solve for the mass.
For this, multiply both sides by 10:
<h2>Answer:</h2>
<u>Rounding to the tenths place, the mass of the rock is 36.8 g, or C.</u>
The answer of the given question above would be option C. Based on the example of a Hertzsprung-Russell diagram that is shown, what can we say about white dwarfs is that, they are <span>relatively dim or faint, but very hot. Hope this answers your question. Have a great day ahead!</span>
Answer:
Boiling point: 63.3°C
Freezing point: -66.2°C.
Explanation:
The boiling point of a solution increases regard to boiling point of the pure solvent. In the same way, freezing point decreases regard to pure solvent. The equations are:
<em>Boiling point increasing:</em>
ΔT = kb*m*i
<em>Freezing point depression:</em>
ΔT = kf*m*i
ΔT are the °C that change boiling or freezing point.
m is molality of the solution (moles / kg)
And i is Van't Hoff factor (1 for I₂ in chloroform)
Molality of 50.3g of I₂ in 350g of chloroform is:
50.3g * (1mol / 253.8g) = 0.198 moles in 350g = 0.350kg:
0.198 moles / 0.350kg = 0.566m
Replacing:
<em>Boiling point:</em>
ΔT = kb*m*i
ΔT = 3.63°C/m*0.566m*1
ΔT = 2.1°C
As boiling point of pure substance is 61.2°C, boiling point of the solution is:
61.2°C + 2.1°C = 63.3°C
<em>Freezing point:</em>
ΔT = kf*m*i
ΔT = 4.70°C/m*0.566m*1
ΔT = 2.7°C
As freezing point is -63.5°C, the freezing point of the solution is:
-63.5°C - 2.7°C = -66.2°C
