The mass (g) of the original sample after decomposition is 8.3983 g.
A decomposition reaction can be described as a chemical reaction wherein one reactant breaks down into or extra merchandise.
explanation:
Reaction 2KClO₃ ⇒ 2KCl + 3O₂
moles 2 2 3
molar mass 122.55 74.55 32
Given, Mass of O₂ = 3.29g ⇒ moles of O₂
= (3.29/32) = 0.1028
3 moles of O₂ produced by 2 moles of KClO₃
Therefore, 0.1028 moles of O₂ produced by (2*0.1028/3) = 0.06853 moles of Kclo₃
Mass of KClo₃ in original sample is = moles * molar mass
= 0.06853 * 122.55
= 8.3983 g
A decomposition response occurs whilst one reactant breaks down into or extra merchandise. this may be represented through the general equation: XY → X+ Y. Examples of decomposition reactions consist of the breakdown of hydrogen peroxide to water and oxygen, and the breakdown of water to hydrogen and oxygen.
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Answer:
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Explanation:
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the calculated value is Ea is 18.2 KJ and A is 12.27.
According to the exponential part in the Arrhenius equation, a reaction's rate constant rises exponentially as the activation energy falls. The rate also grows exponentially because the rate of a reaction is precisely proportional to its rate constant.
At 500K, K=0.02s−1
At 700K, k=0.07s −1
The Arrhenius equation can be used to calculate Ea and A.
RT=k=Ae Ea
lnk=lnA+(RT−Ea)
At 500 K,
ln0.02=lnA+500R−Ea
500R Ea (1) At 700K lnA=ln (0.02) + 500R
lnA = ln (0.07) + 700REa (2)
Adding (1) to (2)
700REa100R1[5Ea-7Ea] = 0.02) +500REa=0.07) +700REa.
=ln [0.02/0 .07]
Ea= 2/35×100×8.314×1.2528
Ea =18227.6J
Ea =18.2KJ
Changing the value of E an in (1),
lnA=0.02) + 500×8.314/18227.6
= (−3.9120) +4.3848
lnA=0.4728
logA=1.0889
A=antilog (1.0889)
A=12.27
Consequently, Ea is 18.2 KJ and A is 12.27.
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Answer:
Scientists debated the theories with each other and compared the theories with scientific observations
