Answer:
0.1 m C6H12O6 < 0.1 m NaCl < 0.1 m Cu(NO3)2 < 0.1 m Fe(NO3)3
Explanation:
Step 1: Data given
ΔTb = Kb*m*i
⇒with Kb = 0.512°C/m
⇒with m = molality
⇒with i = Van't hoff factor = the number of particles into which the solute dissociates
a. 0.1 m NaCl
⇒ molality = 0.1 molal
⇒ i(NaCl) = 2 ( dissociates in Na+ and Cl-)
0.1 * 2 = 0.2
b. 0.1 m Cu(NO3)2
⇒ molality = 0.1 molal
⇒ i(Cu(NO3)2) = 3 ( dissociates in Cu^2+ and 2NO3-)
0.1 * 3 = 0.3
c. 0.1 m C6H12O6
⇒ molality = 0.1 molal
⇒ i(C6H12O6) = 1 ( doesn't dissociate in water)
0.1 * 1 = 0.1
d. 0.1 m Fe(NO3)3
⇒ molality = 0.1 molal
⇒ i(Fe(NO3)3) = 4 ( dissociates in Fe^3+ and 3NO3-)
0.1 * 4 = 0.4
0.1 m C6H12O6 < 0.1 m NaCl < 0.1 m Cu(NO3)2 < 0.1 m Fe(NO3)3
Answer 3
Ag is displaced By the more reactive al
Answer:
HI& HNO3= strong acid
H2CO3 &HF= weak acid
LiOH & Ba(OH)2= strong base
CH3NH2= weak base
NH4Cl = weak acid solution
Na3PO4 = weak base solution
Explanation:
HI& HNO3= strong acid because they can dissociate completely producing free Oxonium ions
H2CO3 &HF= weak acid because there dissociate partly in Solution
LiOH & Ba(OH)2= strong base because LiOH is a group one alkali which dissociate completely and Ba(OH)2 a group 2 alkali which dissociate completely as well
CH3NH2= weak base because in solution it dissociates partly forming NH4+ ion
NH4Cl = weak acid solution because it is a salt of strong acid and weak base
Na3PO4 = weak base solution
Because it's a salt of strong base and weak acid
