Answer:
O: 3.41 / 3.41 = 1.00.
Explanation:
To find the simplest whole number ratio, divide each number by the smallest number of moles
Answer: GROUP 2 is the only group to include only metals. Group one includes the most metalic metals but hydrogen is usually put into group one and obviously hydrogen is not a metal. Group 2 includes Br, Mg, Ca, Sr, Ba,Ra.
Explanation:
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
D) warm and wet.
Water can be a catalyst, and heat speeds up reactions.
AgF consists of Ag+ and F- ions, which are fully dissociated in aqueous solution. When solving electrolysis problems, it is important to remember that water itself may also be a subject to electrolysis. Therefore, determining which species is oxidized and which species is reduced depends on selecting the processes that are the most energetically favorable. The most preferred reduction reaction will be Ag+ + e- = Ag (Emf=0.7996 V) which will occur at the cathode, on the other hand, the most favorable oxidation reaction will be
2H2O = O2 +4H+ + 4e- (Emf = -1.3 V) that will occur at the anode. Thus, the product at the anode is oxygen gas and at the cathode electrode is silver metal.