Explanation:
Depression in Freezing point
= Kf × i × m
where m is molality , i is Van't Hoff factor, m = molality
Since molality and Kf remain the same
depression in freezing point is proportional to i
i= 2 for CuSO4 ( CuSO4----------> Cu+2 + SO4-2
i=1 for C2h6O
i= 3 for MgCl2 ( MgCl2--------> Mg+2+ 2Cl-)
So the freezing point depression is highest for MgCl2 and lowest for C2H6O
so freezing point of the solution = freezing point of pure solvent- freezing point depression
since MgCl2 has got highest freezing point depression it will have loweest freezing point and C2H6O will have highest freezing point
Simply put, Potential energy is the "build up". If I had a ball on the top a 5ft slide, it would have potential energy, as long as it hasn't slid down yet.
If I had another ball on a 10ft slide, it would have twice the potential energy the first ball had.
What comes next is kinetic energy, which is the energy used when the object is moving, like the ball as it goes down the slide. The faster it moves, the more kinetic energy.
Basically, <em>Potential</em> is the "build up" but it does not, I repeat does not move.
<em>Kinetic</em> energy is the use of the "build up" through movement.
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a. KCl = strong electrolyte
b. CCl4 = non-electrolyte
c. LiCl = strong electrolyte
d. Na2SO4 = strong electrolyte
The answer is 19.9 grams cadmium.
Assuming there was no heat leaked from the system, the heat q lost by cadmium would be equal to the heat gained by the water:
heat lost by cadmium = heat gained by the water
-qcadmium = qwater
Since q is equal to mcΔT, we can now calculate for the mass m of the cadmium sample:
-qcadmium = qwater
-(mcadmium)(0.850J/g°C)(38.6°C-98.0°C)) = 150.0g(4.18J/g°C)(38.6°C-37.0°C)
mcadmium = 19.9 grams
Answer:
1.840 x 10⁻³ mol HBrO₃
Explanation:
To find the moles of bromic acid (HBrO₃), you should (1) convert milligrams to grams (by dividing by 1,000) and then (2) convert moles to grams (via molar mass from periodic table).
Molar Mass (HBrO₃): 1.008 g/mol + 79.904 g/mol + 3(15.998 g/mol)
Molar Mass (HBrO₃): 128.906 g/mol
2.372 x 10² mg HBrO₃ 1 g 1 mole
---------------------------------x----------------x------------------ = 1.840 x 10⁻³ mol HBrO₃
1,000 mg 128.906 g
