Answer:

Explanation:
Given that:

From equation (3) , multiplying (-1) with equation (3) and interchanging reactant with the product side; we have:

Multiplying (2) with equation (4) ; we have:

From equation (1) ; multiplying (-1) with equation (1); we have:

From equation (2); multiplying (3) with equation (2); we have:

Now; Adding up equation (5), (6) & (7) ; we get:



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(According to Hess Law)


Answer:
1 mole of ferric contains 2 moles of iron,and 12 moles of oxygen atoms, and three moles of sulphate ions
The answer is A: Areas where the geologic process occurred now have major petroleum reserves
Answer:
The suitable equation for this reaction is
2CO + O₂ -----> 2CO₂
Here, we are given that we have 2 grams of O₂
From the equation, we can see that 2 * Moles of O₂ = Moles of CO₂
Moles of O₂:
2/32 = 1/16 moles
Therefore, the number of moles of CO₂ is twice the moles of O₂
Moles of CO₂ = 2 * 1/16
Moles of CO₂ formed = 1/8 moles
Mass of CO₂ formed = Molar mass of CO₂ * Moles of CO₂
Mass of CO₂ formed = 44 * 1/8
Mass of CO₂ formed = 5.5 grams
Hence, option B is correct
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Answer:
5.0 x 10⁹ years.
Explanation:
- It is known that the decay of a radioactive isotope isotope obeys first order kinetics.
- Half-life time is the time needed for the reactants to be in its half concentration.
- If reactant has initial concentration [A₀], after half-life time its concentration will be ([A₀]/2).
- Also, it is clear that in first order decay the half-life time is independent of the initial concentration.
- The half-life of K-40 = 1.251 × 10⁹ years.
- For, first order reactions:
<em>k = ln(2)/(t1/2) = 0.693/(t1/2).</em>
Where, k is the rate constant of the reaction.
t1/2 is the half-life of the reaction.
∴ k =0.693/(t1/2) = 0.693/(1.251 × 10⁹ years) = 5.54 x 10⁻¹⁰ year⁻¹.
- Also, we have the integral law of first order reaction:
<em>kt = ln([A₀]/[A]),</em>
where, k is the rate constant of the reaction (k = 5.54 x 10⁻¹⁰ year⁻¹).
t is the time of the reaction (t = ??? year).
[A₀] is the initial concentration of (K-40) ([A₀] = 100%).
[A] is the remaining concentration of (K-40) ([A] = 6.25%).
∴ (5.54 x 10⁻¹⁰ year⁻¹)(t) = ln((100%)/( 6.25%))
∴ (5.54 x 10⁻¹⁰ year⁻¹)(t) = 2.77.
∴ t = 2.77/(5.54 x 10⁻¹⁰ year⁻¹) = 5.0 x 10⁹ years.