Explanation:
A volatile substance is defined as the substance which can easily evaporate into the atmosphere due to weak intermolecular forces present within its molecules.
Whereas a flammable substance is defined as a substance which is able to catch fire easily when it comes in contact with flame.
Hence, when we heat a flammable or volatile solvent for a recrystallization then it should be kept in mind that should heat the solvent in a stoppered flask to keep vapor away from any open flames so that it won't catch fire.
And, you should ensure that no one else is using an open flame near your experiment.
Thus, we can conclude that following statements are correct:
- You should heat the solvent in a stoppered flask to keep vapor away from any open flames.
- You should ensure that no one else is using an open flame near your experiment.
Answer is: the boiling point of the resulting solution of sucrose is 100.42°C.
m(H₂<span>O) = 15.2 g ÷ 1000 g/kg = 0.0152 kg.
</span>m(C₁₂H₂₂O₁₁<span>) = 4.27 g.
n</span>(C₁₂H₂₂O₁₁) = m(C₁₂H₂₂O₁₁) ÷ M(C₁₂H₂₂O₁₁).
n(C₁₂H₂₂O₁₁) = 4.27 g ÷ 342.3 g/mol.
n(C₁₂H₂₂O₁₁) = 0.0125 mol.
b(solution) = n(C₁₂H₂₂O₁₁) ÷ m(H₂O).
b(solution) = 0.0125 mol ÷ 0.0152 kg.
b(solution) = 0.82 m.
ΔT = b(solution) · Kb(H₂O).
ΔT = 0.82 m · 0.512°C/m.
ΔT = 0.42°C.
Tb = 100°C + 0.42°C = 100.42°C.
Answer:
0.12693 mg/L
Explanation:
First we <u>calculate the concentration of compound X in the standard prior to dilution</u>:
- 10.751 mg / 100 mL = 0.10751 mg/mL
Then we <u>calculate the concentration of compound X in the standard after dilution</u>:
- 0.10751 mg/mL * 5 mL / 25 mL = 0.021502 mg/L
Now we calculate the<u> concentration of compound X in the sample</u>, using the <em>known concentration of standard and the given areas</em>:
- 2582 * 0.021502 mg/L ÷ 4374 = 0.012693 mg/L
Finally we <u>calculate the concentration of X in the sample prior to dilution</u>:
- 0.012693 mg/L * 50 mL / 5 mL = 0.12693 mg/L
Answer:
The standard heat of formation of Compound X at 25°C is -3095.75 kJ/mol.
Explanation:
Mass of compound X = 7.00 g
Moles of compound X = 
Mass of water in calorimeter ,m= 35.00 kg = 35000 g
Change in temperature of the water in calorimeter = ΔT
ΔT = 2.113°C
Specific heat capacity of water ,c= 4.186 J/g °C
Q = m × c × ΔT
Heat gained by 35 kg of water is equal to the heat released on burning of 0.100 moles of compound X.
Heat of formation of Compound X at 25°C:
= -3095.75 kJ/mol
Answer:
Water moves into the cell
Explanation:
As shown in the question above, the cell is high in glucose and placed in a glass filled with water. This cell has a semi permeable membrane that allows only water to pass through, as the concentration of water within the cell is low, the cell will attempt to strike a balance with the medium it is inserted into. For this reason, what is likely to happen is the passage of water from the most concentrated to the least concentrated medium, that is, the water will pass from the cup to the cell.
water moves into the cell through osmosis.during osmosis water moves from a region of low concentration of solute to a region of high concentration of solute.the glucose introduced into the cell makes it more concentrated.
In this case the cell is hypertonic and water would enter into the cell through the semi permeable membrane.this membrane allows water to pass through but not glucose.this movement of water into the cell causes the cell to become turgid.
