Answer:
32.6%
Explanation:
Equation of reaction
2KClO₃ (s) → 2KCl (s) + 3O₂ (g)
Molar mass of 2KClO₃ = 245.2 g/mol ( 122.6 × 2)
Molar volume of Oxygen at s.t.p = 22.4L / mol
since the gas was collected over water,
total pressure = pressure of water vapor + pressure of oxygen gas
0.976 = 0.04184211 atm + pressure of oxygen gas at 30°C
pressure of oxygen = 0.976 - 0.04184211 = 0.9341579 atm = P1
P2 = 1 atm, V1 = 789ml, V2 = unknown, T1 = 303K, T2 = 273k at s.t.p
Using ideal gas equation
=
V2 =
V2 = 664.1052 ml
245.2 yielded 67.2 molar volume of oxygen
0.66411 will yield =
= 2.4232 g
percentage of potassium chlorate in the original mixture =
= 32.6%
Answer:
V₂ = 50.93 L
Explanation:
Initial volume, 
Initial temperature, 
Final temperature, 
We need to find the final volume of the gas. The relation between the volume and the temperature is given by :

So, the final volume of the gas is 50.93 L.
2 Li(s) +Cl₂→ 2 Li⁺ (aq) + 2Cl⁻ (aq)
The cell potential of the reaction above is +4.40V
<em><u>calculation</u></em>
Cell potential =∈° red - ∈° oxidation
in reaction above Li is oxidized from oxidation state 0 to +1 therefore the∈° oxid = -3.04
Cl is reduce from oxidation state 0 to -1 therefore the ∈°red = +1.36 V
cell potential is therefore = +1.36 v -- 3.04 = + 4.40 V
Answer:
T2 = 550K
Explanation:
From Charles law;
V1/T1 = V2/T2
Where;
V1 is initial volume
V2 is final volume
T1 is initial temperature
T2 is final temperature
We are given;
V1 = 20 mL
V2 = 55 mL
T1 = 200 K
Thus from V1/T1 = V2/T2, making T2 the subject;
T2 = (V2 × T1)/V1
T2 = (55 × 200)/20
T2 = 550K