The volume of SO2 produced at 325k is calculated as below
calculate the moles of SO2 produced which is calculated as follows
write the reacting equation
K2SO3 +2 HCl = 2KCl +H2O+ SO2
find the moles of HCl used
=mass/molar mass = 15g/ 36.5 g/mol =0.411 moles
by use of mole ratio between HCl to SO2 which is 2:1 the moles of SO2 is therefore = 0.411 /2 =0.206 moles of SO2
use the idea gas equation to calculate the volume SO2
that is V=nRT/P
where n=0.206 moles
R(gas constant) = 0.082 L.atm/ mol.k
T=325 K
P=1.35 atm
V=(0.206 moles x 0.082 L.atm/mol.k x325 k)/1.35 atm = 4.07 L of SO2
Answer:
1000g
Explanation:because the higher mass takes up heat.
Answer:
0.20 m glucose < 0.40 m NaCl < 0.30 m BaCl2 < 0.50 m Na2SO4.
Explanation:
Step 1: Data given
ΔT = i*Kb*m
⇒ΔT = the boiling point elevation = Shows how much the boiling point increases
⇒i = the van't Hoff factor: Says in how many particles the compound will dissociate
⇒ Since all are aqueous solutions Kb for all solutions is the same (0.512 °C/m)
⇒m = the molality
Step 2:
0.20 m glucose
ΔT = i*Kb*m
⇒ΔT = the boiling point elevation = TO BE DETERMINED
⇒i = the van't Hoff factor for glucose = 1
⇒ Kb = 0.512 °C/m
⇒m = 0.20 m
ΔT = 1*0.512 * 0.20
<u>ΔT = 0.1024 °C</u>
0.30 m BaCl2
ΔT = i*Kb*m
⇒ΔT = the boiling point elevation = TO BE DETERMINED
⇒i = the van't Hoff factor for BaCl2 = Ba^2+ + 2Cl- : i = 3
⇒ Kb = 0.512 °C/m
⇒m = 0.30 m
ΔT = 3*0.512 * 0.30
<u>ΔT = 0.4608 °C</u>
0.40 m NaCl
ΔT = i*Kb*m
⇒ΔT = the boiling point elevation = TO BE DETERMINED
⇒i = the van't Hoff factor for NaCl = Na+ + Cl- : i = 2
⇒ Kb = 0.512 °C/m
⇒m = 0.40 m
ΔT = 2*0.512 * 0.40
<u>ΔT = 0.4096 °C</u>
0.50 m Na2SO4.
ΔT = i*Kb*m
⇒ΔT = the boiling point elevation = TO BE DETERMINED
⇒i = the van't Hoff factor for Na2SO4 = 2Na+ + SO4^2- : i =3
⇒ Kb = 0.512 °C/m
⇒m = 0.50 m
ΔT = 3*0.512 * 0.50
<u>ΔT = 0.768 °C</u>
0.20 m glucose < 0.40 m NaCl < 0.30 m BaCl2 < 0.50 m Na2SO4.
