Ammonia gas occupies a volume of 2,725ml at a pressure of 701 kPa. What volume would it occupy at 101 kPa?
1 answer:
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Depression in freezing point (Δ
) = 
×m×i,
where, = cryoscopic constant = 
,
m= molality of solution = 0.0085 m
i = van't Hoff factor = 2 (For
)
Thus, (Δ) = 1.86 X 0.0085 X 2 = 
Now, (Δ) = 
- T
Here, T = freezing point of solution
= freezing point of solvent = 
Thus, T = - (Δ) = -
where,
m= molality of solution = 0.0085 m
i = van't Hoff factor = 2 (For
Thus, (Δ
Now, (Δ
Here, T = freezing point of solution
Thus, T =
Answer:
Volume of lithium atom is found to be 1.47 X 10⁻²⁹ m³
Explanation:
Let us consider the volume of atom as a sphere (but it is little complex than that). This volume is mathematically expressed as,
----------------------------------------------------------------------------------------(Eq. 1)
Here, R is the radius of lithium atom. The radius is given in picometers, so firstly let us convert it into meters
placing this value in Eq.1 the required result is achieved
V= 1.47 X 10⁻²⁹ m³
Answer : The correct option is, the negative log of the hydroxide ion concentration.
Explanation :
pOH : It is defined as the negative logarithm of hydroxide ion concentration. It is a measure of the alkalinity of the solution.
Formula used :
is the concentration of
ions.
When pOH is less than 7, the solution is alkaline.
When pOH is more than 7, the solution is acidic.
When pOH is equal to 7, the solution is neutral.