When dT = Kf * molality * i
= Kf*m*i
and when molality = (no of moles of solute) / Kg of solvent
= 2.5g /250g x 1 mol /85 g x1000g/kg
=0.1176 molal
and Kf for water = - 1.86 and dT = -0.255
by substitution
0.255 = 1.86* 0.1176 * i
∴ i = 1.166
when the degree of dissociation formula is: when n=2 and i = 1.166
a= i-1/n-1 = (1.166-1)/(2-1) = 0.359 by substitution by a and c(molality) in K formula
∴K = Ca^2/(1-a)
= (0.1176 * 0.359)^2 / (1-0.359)
= 2.8x10^-3
A) 2H₂(g) + O₂(g) → 2H₂O(l) + 285.83 kJ
Exothermic
B) 2Mg + O₂ → 2MgO + 1200kJ
Exothermic
Answer:
12.8 g of 
must be withdrawn from tank
Explanation:
Let's assume gas inside tank behaves ideally.
According to ideal gas equation- 
where P is pressure of , V is volume of , n is number of moles of , R is gas constant and T is temperature in kelvin scale.
We can also write, 
Here V, T and R are constants.
So, 
ratio will also be constant before and after removal of from tank
Hence, 
Here, 
and 
So, 
So, moles of must be withdrawn = (0.66 - 0.26) mol = 0.40 mol
Molar mass of = 32 g/mol
So, mass of must be withdrawn = 
Answer:
the specific shape of letters, e.g. their roundness or sharpness.
regular or irregular spacing between letters.
the slope of the letters.
the rhythmic repetition of the elements or arrhythmia.
the pressure to the paper.
the average size of letters.
the thickness of letters.
When you boil water, you aren't changing the elements. You're just making water vapor. However, when you burn paper, it becomes carbon (mostly). So physical changes will not change the substance, only chemical changes will.
