Answer:
yes it is called the windchill factor.
The windchill factor is the temperature that a person feels because of the wind. For example, if a thermometer reads 35 degrees Fahrenheit outside and the wind is blowing at 25 miles per hour (mph), the windchill factor causes it to feel like it is 8 degrees F.
Answer:
The correct answer is B.
The 
is samller than 
of the reaction . So,the reaction will shift towards the left i.e. towards the reactant side.
Explanation:
Equilibrium constant is defined as the ratio of concentration of products to the concentration of reactants each raised to the power their stoichiometric ratios. It is expressed as
K is the constant of a certain reaction when it is in equilibrium, while Q is the quotient of activities of products and reactants at any stage other than equilibrium of a reaction.
For the given chemical reaction:
The expression for is written as:
![Q=\frac{[PCl_3][Cl_2]}{[[PCl_5]^1}](https://tex.z-dn.net/?f=Q%3D%5Cfrac%7B%5BPCl_3%5D%5BCl_2%5D%7D%7B%5B%5BPCl_5%5D%5E1%7D)
Given : = 0.0454
Thus as 
, the reaction will shift towards the left i.e. towards the reactant side.
According to this equation:
AgCl(s) ↔ Ag+(aq) + Cl-(aq)
so K1 = [Ag+][Cl-]
when [Ag+] = [Cl-] we can assume both = X
and when we have X the solubility = 1.33 x 10^-5 mol / L
by substitution:
∴ K1 = X^2
= (1.33 x 10^-5)^2
= 1.77 x 10^-10
by using vant's Hoff equation:
ln(K2/K1) = (ΔH/R)*(1/T2-1/T1)
when ΔH = 65700 J / mol
R = 8.314
T1 = 25+273 = 298 K
T2 = 47.7 +273 =320.7
by substitution:
∴㏑(K2/1.77 x 10^-10) = (65700/8.314) * ( 1/320.7 - 1/ 298)
by solving for K2
∴K2 = 2.7 x 10^-11
and when K2 = X^2
∴ the solubility X = √(2.7 x 10^-11)
= 5.2 x 10^-6 mol/L
Is this from connections academy?
