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:
Mass of chemical = 50 gram
Explanation:
Given:
Volume of chemical = 5 ml
Density of chemical = 10 g/mL
Find:
Mass of chemical
Computation:
Mass of chemical = Density of chemical × Volume of chemical
Mass of chemical = 10 × 5
Mass of chemical = 50 gram
In nitrogen-14, there are 7 protons, 7 neutrons, and 7 electrons. The protons and neutrons are in the nucleus, and the electrons are in the electron shells. The atomic number is the number of protons, the mass number is the number of protons AND neutrons, and the atomic mass is the average of the masses of all isotopes.
So what you do is you write out each element separately and how many of that element on either side of the arrow S-8|S-1
O-2|O-2
on one side theres 8 (S) and 2 (O) and 1(S) and 2(O) if you put a number in front to balance it you are multiplying it the elements number like
SO2 if I put a 2 in front now I will have 2(S) and 4(O)
Answer is 1,8,8
Answer: The thermocline begins at 100 meters of depth.
Step-to-step explanation:-
Thermocline is a transition oceanic water layer between deep and surface water in which water temperature decreases rapidly with increasing depth.
From the given graph it cam be seen that at 100 meters the thermoline begins such that the temperature drops from 
to 
.
Hence, the thermocline begins at 100 meters of depth.
