Answer:
Add Iodine-KI reagent to a solution or directly on a potato or other materials such as bread, crackers, or flour. A blue-black color results if starch is present. If starch amylose is not present, then the color will stay orange or yellow
Assuming an ebullioscopic constant of 0.512 °C/m for the water, If you add 30.0g of salt to 3.75kg of water, the boiling-point elevation will be 0.140 °C and the boiling-point of the solution will be 100.14 °C.
<h3>What is the boiling-point elevation?</h3>
Boiling-point elevation describes the phenomenon that the boiling point of a liquid will be higher when another compound is added, meaning that a solution has a higher boiling point than a pure solvent.
- Step 1: Calculate the molality of the solution.
We will use the definition of molality.
b = mass solute / molar mass solute × kg solvent
b = 30.0 g / (58.44 g/mol) × 3.75 kg = 0.137 m
- Step 2: Calculate the boiling-point elevation.
We will use the following expression.
ΔT = Kb × m × i
ΔT = 0.512 °C/m × 0.137 m × 2 = 0.140 °C
where
- ΔT is the boiling-point elevation
- Kb is the ebullioscopic constant.
- b is the molality.
- i is the Van't Hoff factor (i = 2 for NaCl).
The normal boiling-point for water is 100 °C. The boiling-point of the solution will be:
100 °C + 0.140 °C = 100.14 °C
Assuming an ebullioscopic constant of 0.512 °C/m for the water, If you add 30.0g of salt to 3.75kg of water, the boiling-point elevation will be 0.140 °C and the boiling-point of the solution will be 100.14 °C.
Learn more about boiling-point elevation here: brainly.com/question/4206205
Salutations!
Determine the length of the object shown below.
The length of the scale is 5.6 cm.
Hope I helped.
Answer:
Value of n in MnSO₄.nH₂O is one.
Explanation:
The n represents the number of moles of water attached to the formula unit manganese sulfate. These moles (n) can be determined by taking the ratio of the moles of anhydrous salt and the moles of water. The moles of water can be determined by taking the difference of final and initial mass of the salt. This difference is equal to the mass of the water, mathematically it can be represented as,
Mass of H₂O = initial mass of the salt (g) - final mass of the salt (g)
Mass of H₂O = 16.260 g - 14.527 g
Mass of H₂O = 1.733 g
moles of H₂O = (1.733 g) ÷ (18.015 g/mole)
moles of H₂O = 0.0962
For the moles of anhydrous salt:
moles of MnSO₄ = mass of MnSO₄ ÷ molar mass of MnSO₄
moles of MnSO₄ = 14.5277 ÷ 151.001
moles of MnSO₄= 0.0962
Now for n:
n = moles of water ÷ moles of MnSO₄
n = 0.0962 ÷ 0.0962
n = 1
The above calculations show that one mole of H₂O is attached to the one formula unit of MnSO₄
