One mole of NaOH (Sodium hydroxide) is equal to 39.997 grams of NaOH.
Answer is: <span>he boiling point of a 1.5 m aqueous solution of fructose is </span>100.7725°C.
The boiling point
elevation is directly proportional to the molality of the solution
according to the equation: ΔTb = Kb · b.<span>
ΔTb - the boiling point
elevation.
Kb - the ebullioscopic
constant. of water.
b - molality of the solution.
Kb = 0.515</span>°C/m.
b = 1.5 m.
ΔTb = 0.515°C/m · 1.5 m.
ΔTb = 0.7725°C.
Tb(solution) = Tb(water) + ΔTb.
Tb(solution) = 100°C + 0.7725°C = 100.7725°C.
Answer:
Explanation:
q= mc theta
where,
Q = heat gained
m = mass of the substance = 670g
c = heat capacity of water= 4.1 J/g°C
theta =Change in temperature=(
66-25.7)
Now put all the given values in the above formula, we get the amount of heat needed.
q= mctheta
q=670*4.1*(66-25.7)
=670*4.1*40.3
=110704.1
First, we need to get the number of moles:
from the reaction equation when Y4+ takes 4 electrons and became Y, X loses 4 electrons and became X4+
∴ the number of moles n = 4
we are going to use this formula:
㏑K = n *F *E/RT
when K is the equilibrium constant = 4.98 x 10^-5
and F is Faraday's constant = 96500
and the constant R = 8.314
and T is the temperature in Kelvin = 298 K
and n is number of moles of electrons = 4
so, by substitution:
㏑4.98 x 10^-5 = 4*96500*E / 8.314*298
∴E = -0.064 V