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noname [10]
3 years ago
8

How many grams of KCIO3 are needed to produce 5.00 of O2 at STP?

Chemistry
1 answer:
Arlecino [84]3 years ago
6 0
2KClO₃ → 2KCl + 3O₂

mole ratio of KClO₃ to O₂ is 2 : 3

∴ if moles of O₂ = 5 mol

then moles of KClO₃ = \frac{5 mol   *   2}{3}

                            = 3.33 mol


Mass of KClO₃ needed = mol of KClO₃ × molar mass of KClO₃

                                      = 3.33 mol × ((39 × 1) + (35.5 × 1) + (16 × 3) g/mol

                                      = 407.93 g
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If a tree dies and the trunk remains undisturbed for 13,750 years, what percentage of the original 14c is still present? (the ha
defon

Answer:

18.94%.

Explanation:

  • The decay of carbon-14 is a first order reaction.
  • The rate constant of the reaction (k) in a first order reaction = ln (2)/half-life = 0.693/(5730 year) = 1.21 x 10⁻⁴ year⁻¹.
  • The integration law of a first order reaction is:

<em>kt = ln [A₀]/[A]</em>

k is the rate constant = 1.21 x 10⁻⁴ year⁻¹.

t is the time = 13,750 years.

[A₀] is the initial percentage of carbon-14 = 100.0 %.

[A] is the remaining percentage of carbon-14 = ??? %.

∵ kt = ln [Ao]/[A]

∴ (1.21 x 10⁻⁴ year⁻¹)(13,750 years) = ln (100.0%)/[A]

1.664 =  ln (100.0%)/[A]

Taking exponential for both sides:

5.279 = (100.0%)/[A]

<em>∴ [A]</em> = (100.0%)/5.279 = <em>18.94%.</em>

8 0
3 years ago
The pyrolysis of ethane proceeds with an activation energy of about 300 kJ/mol. How much faster is the decomposition at 625°C th
Alona [7]

Answer:

The decomposition of ethane is 153.344 times much faster at 625°C than at 525°C.

Explanation:

According to the Arrhenius equation,

K=A\times e^{\frac{-Ea}{RT}}

\log (\frac{K_2}{K_1})=\frac{Ea}{2.303\times R}[\frac{1}{T_1}-\frac{1}{T_2}]

where,

K_2 = rate of reaction at T_2

K_1 = rate of reaction at T_1

Ea = activation energy of the reaction

R = gas constant = 8.314 J/K mol

E_a=300 kJ/mol=300,000 J/mol

T_2=625^oC=898.15 K,T_1=525^oC=798.15 K

\log (\frac{K_2}{K_1})=\frac{300,000 J/mol}{2.303\times 8.314 J/K mol}[\frac{1}{798.15 K}-\frac{1}{898.15 K}]

\log (\frac{K_2}{K_1})=2.185666

K_2=153.344\times K_1

The decomposition of ethane is 153.344 times much faster at 625°C than at 525°C.

4 0
3 years ago
Calculate the mass in grams of each sample.
OlgaM077 [116]

Answer: The answer is A. - 4.88x10^20 H2O2 molecules

Explanation: I hope this helps!

4 0
3 years ago
Ni-cad (nickel–cadmium) batteries have a slightly lower cell potential than the common alkaline value of 1.5 v normally used in
Lorico [155]
The half-reaction are:

Cd ---> Cd(OH)₂
The oxidation number of Cd changed from 0 to +2. So, the number of mol electron transferred here is 2.

NiO(OH) --> Ni(OH)₂
The oxidation number of Cd changed from +3 to +2. So, the number of mol electron transferred here is 1.

Now, the greatest common factor would be 2. So, we use n=2 for the formula for ΔG°. F is Faraday's constant equal to 96,485 J/mol e.

ΔG° = nFE° = (2)(96,485)(1.5) =<em> 289,455 J</em>


6 0
3 years ago
What is the equilibrium equation for the dissociation of formic acid in water? hcooh (aq) + h2o (l) ⇌ h3o+ (aq) + hcoo- (aq)?
Pani-rosa [81]
Formic acid when in water would dissociate into ions just like any acids. It would dissociate into the hydrogen ion and the formate ion. The equilibrium dissociation equation would be written as:

<span>HCOOH (aq) + H2O (l) ⇌ H+ (aq) + HCOO- (aq)

Formic acid is a weak acid which means that when in aqueous solution it does not completely dissociate into its corresponding ions. Only a certain amount that would be dissociated so in the solution there will be HCOOH, HCOO- and H+ molecules. It is also known as Methanoic acid and an important substance for the synthesis of a number of substances. It is naturally occurring in ants.</span>
7 0
3 years ago
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