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3 years ago

How many atoms of Oxygen (O) in the reactants of the equation below:

Chemistry

2 answers:

Naya [18.7K]3 years ago

8 0

Answer: 18 atoms

Explanation: the reactant contains 6O2 and 6O and this gives a total of 18

olga2289 [7]3 years ago

8 0

Answer:

The correct option is 18

Explanation:

There are two reactant specie in the equation; carbon dioxide (CO₂) and water (H₂O). From the equation, there are 6 molecules of CO₂, hence the number of oxygen atoms from CO₂ is <em>6 × 2 = 12</em> and there are also 6 molecules of H₂O, hence the number of oxygen atoms from H₂O is <em>6 × 1 = 6</em>.

The total number of oxygen atoms in the reactants is therefore 12 + 6 = 18

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3 years ago

For the reaction:

Rom4ik [11]

Answer:

Half-life at 310 K is 6.54 × 10³ s.

Explanation:

Let's consider the following reaction:

2 N₂O₅(g) → 4 NO₂(g) + O₂(g)

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(Δ[O₂]/Δt) = k . [N₂O₅]

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where,

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$k$ is the rate constant

At 300 K,

$(t_{1/2})_{1}=\frac{ln2}{k_{1}}\\k_{1}=\frac{ln2}{(t_{1/2})_{1}} =\frac{ln2}{2.50 \times 10^{4} s} =2.77 \times 10^{-5} s^{-1}$

We can use two-point Arrhenius equation to solve for k₂ at 310 K

$ln\frac{k_{2}}{k_{1}} =\frac{-Ea}{R} (\frac{1}{T_{2}} -\frac{1}{T_{1}} )\\ln\frac{k_{2}}{k_{1}} =\frac{-103.3kJ/mol}{8.314 \times 10^{-3} kJ/mol.K} .(\frac{1}{310K}-\frac{1}{300K} )\\ln\frac{k_{2}}{k_{1}}=1.34\\k_{2}=1.06 \times 10^{-4} s^{-1}$

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