Answer:
6 moles of electrons
Explanation:
Let us consider the species NO3− and ClO−. The NO3− is oxidized to NO the oxidation number of nitrogen is decreased from +5 to +2.
The oxidation number of chlorine is increased from +1 to +3. This implies that six electrons were transferred in the balanced reaction equation shown in the question. Hence the answer.
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:
1.7 ppm
Explanation:
Original amount N' = 2.6 ppm
time to testing t = 24 hr
final amount N = 2.1 ppm
Using exponential inhibited decay, we have
N = N'e^(-kt)
Where
N is the new reading
N' is the original reading
t is the decay time
k is the decay constant
Substituting, we have
2.1 = 2.6 x e^(-k x 24)
2.1 = 2.6 x e^(-24k)
0.808 = e^(-24k)
We take the natural log of both sides of the equation
Ln 0.808 = Ln (e^(-24k))
-0.213 = - 24k
K = 0.213/24 = 0.00886
After 48 hrs, the reading of free chlorine will be
N = 2.6 x e^(-0.00886 x 48)
N = 2.6 x e^(-0.425)
N = 2.6 x 0.654
N = 1.7 ppm
<span>The mixture that is most likely to form a suspension is flour and liquid water mixed together, as in a mixture like gravy. A suspension mixture is a mixture that has large solid particles, particles that are large enough for sedimentation.</span>
<span>In a mole of anything, there are 6.023 x 10^23 units. So, in 3.9 moles of sulfur, there are 3.9 * 6.023 x 10^23 = 23 x 10^23 = 2.3 x 10^24 atoms (keeping only 2 sig figs). Hope I help!!
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