Answer:
1) No shift
2) No shift
3) Leftward shift
4)Rightward sifht
Explanation:
1) 2) Adding N or Removing N in the equilibrium will produce No shift, because of its solid state, the N is not contemplated in the equilibrium equation:
![K=\frac{[L]^2}{[M]^3}](https://tex.z-dn.net/?f=K%3D%5Cfrac%7B%5BL%5D%5E2%7D%7B%5BM%5D%5E3%7D)
3) Increasing the volume produces a decrase in the preassure due to the expansion of the gases. This will cause a leftward shift, because the system will try to increase the moles of gas and in consecuence of this, also increase the preassure.
4) Decreasing the volume has the opposite effect of the item 3): the preassure will increase and the system will consume moles of gas to decrease it, producing a rightward shift.
It should be noted that when a reaction is occurring in a test tube, heat transmitted to the surroundings when Molecules collide with the glass, and the glass molecules then transmit that energy to the outside.
<h3>What is heat?</h3>
Heat can be regarded as a form of energy which is energy that is been transferred as a result of difference in temperature.
In the case above, Molecules collide with the glass, and the glass molecules then transmit that energy to the outside which is an exothermic reaction.
Therefore, option B is correct.
Learn more about heat at:
brainly.com/question/12072129
The percent yield of the reaction : 89.14%
<h3>Further explanation</h3>
Reaction of Ammonia and Oxygen in a lab :
<em>4 NH₃ (g) + 5 O₂ (g) ⇒ 4 NO(g)+ 6 H₂O(g)</em>
mass NH₃ = 80 g
mol NH₃ (MW=17 g/mol):
mass O₂ = 120 g
mol O₂(MW=32 g/mol) :
Mol ratio of reactants(to find limiting reatants) :
mol of H₂O based on O₂ as limiting reactants :
mol H₂O :
mass H₂O :
4.5 x 18 g/mol = 81 g
The percent yield :
It’s soft which makes It low energy
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.
