False beacause that was one of the first things they thought
Answer:
1.12g/mol
Explanation:
The freezing point depression of a solvent for the addition of a solute follows the equation:
ΔT = Kf*m*i
<em>Where ΔT is change in temperature (Benzonitrile freezing point: -12.82°C; Freezing point solution: 13.4°C)</em>
<em>ΔT = 13.4°C - (-12.82) = 26.22°C</em>
<em>m is molality of the solution</em>
<em>Kf is freezing point depression constant of benzonitrile (5.35°Ckgmol⁻¹)</em>
<em>And i is Van't Hoff factor (1 for all solutes in benzonitrile)</em>
Replacing:
26.22°C = 5.35°Ckgmol⁻¹*m*1
4.90mol/kg = molality of the compound X
As the mass of the solvent is 100g = 0.100kg:
4.9mol/kg * 0.100kg = 0.490moles
There are 0.490 moles of X in 551mg = 0.551g, the molar mass (Ratio of grams and moles) is:
0.551g / 0.490mol
= 1.12g/mol
<em>This result has no sense but is the result by using the freezing point of the solution = 13.4°C. Has more sense a value of -13.4°C.</em>
Answer: Metals form cations.
The alkali metals (the IA elements) lose a single electron to form a cation with a 1+ charge.
The alkaline earth metals (IIA elements) lose two electrons to form a 2+ cation.
Aluminum, a member of the IIIA family, loses three electrons to form a 3+ cation.
Therefore, metals in the s and p block of the periodic table have 1, 2 or 3 electrons in their outermost orbit (or valence shell). Now to gain a stable octet metals lose either 1, 2 or 3 electrons from the valence shell thus forming cation with +1, +2 or +3 charge.
Answer:
Cost to supply enough vanillin is 
Explanation:
Threshold limit of vanillin in air is 
per litre means there should be of vanillin in 1L of air to detect aroma of vanillin.
So, 
So amount of vanillin should be present to detect = 
As cost of 50 g vanillin is 
therefore cost of vanillin = 
Answer:
78 moles of the solute
Explanation:
From the question;
- Molarity of the solution is 6.50 M
- Volume of the solution is 12.0 L
We want to determine the number of moles needed
We need to know that;
Molarity = Number of moles ÷ Volume
Therefore;
Number of moles = Molarity × Volume
Hence;
Number of moles = 6.50 M × 12.0 L
= 78 moles
Thus, the moles of the solute needed is 78 moles
