Answer:
This question is incomplete, however, the unknown compound can be inferred to be "Lithium Bromide"
Explanation:
The unknown compound firstly is said to be an ionic compound. An ionic/electrovalent compound is a compound in which it's constituent ions transfer/receive electron(s). They are mostly made of group 1 and group 7 elements. Examples include NaCl, NaF, LiF and KCl.
Also, the ion (metallic ion) that produces a red flame test colour in a flame test is the <u>Lithium ion (Li⁺).</u> Also, when dissolved in water or hexane, the only halogen that produces a red/orange colour is bromine. Hence, the unknown ionic compound can be inferred to be Lithium Bromide.
Answer:
I. Changing the pressure:
Increasing the pressure: the amount of H₂S(g) will increase.
Decreasing the pressure: the amount of H₂S(g) will decrease.
II. Changing the temperature:
Increasing the temperature: the amount of H₂S(g) will decrease.
Decreasing the temperature: the amount of H₂S(g) will increase.
III. Changing the H₂ concentration:
Increasing the H₂ concentration: the amount of H₂S(g) will increase.
Decreasing the H₂ concentration: the amount of H₂S(g) will decrease.
Explanation:
Le Châtelier's principle states that when there is an dynamic equilibrium, and this equilibrium is disturbed by an external factor, the equilibrium will be shifted in the direction that can cancel the effect of the external factor to reattain the equilibrium.
I. Changing the pressure:
When there is an increase in pressure, the equilibrium will shift towards the side with fewer moles of gas of the reaction. And when there is a decrease in pressure, the equilibrium will shift towards the side with more moles of gas of the reaction.
For the reaction: CH₄(g) + 2H₂S(g) ⇄ CS₂(g) + 4H₂(g),
The reactants side (left) has 3.0 moles of gases and the products side (right) has 5.0 moles of gases.
Increasing the pressure: will shift the reaction to the side with lower moles of gas (left side), amount of H₂S(g) will increase.
Decreasing the pressure: will shift the reaction to the side with lower moles of gas (right side), amount of H₂S(g) will decrease.
II. Changing the temperature
The reaction is endothermic since the sign of ΔH is positive.
So the reaction can be represented as:
CH₄(g) + 2H₂S(g) + heat ⇄ CS₂(g) + 4H₂(g).
Increasing the temperature:
The T is a part of the reactants, increasing the T increases the amount of the reactants. So, the reaction will be shifted to the right to suppress the effect of increasing T and the amount of H₂S(g) will decrease.
Decreasing the temperature:
The T is a part of the reactants, increasing the T decreases the amount of the reactants. So, the reaction will be shifted to the left to suppress the effect of decreasing T and the amount of H₂S(g) will increase.
III. Changing the H₂ concentration:
H₂ is a part of the products.
Increasing the H₂ concentration:
H₂ is a part of the products, increasing H₂ increases the amount of the products. So, the reaction will be shifted to the left to suppress the effect of increasing H₂ and the amount of H₂S(g) will increase.
Decreasing the H₂ concentration:
H₂ is a part of the products, decreasing H₂ decreases the amount of the products. So, the reaction will be shifted to the right to suppress the effect of decreasing H₂ and the amount of H₂S(g) will decrease.
Answer:
correct option: A and D
Explanation:
generally element from d block can for ion with multiple charge and having incomplete outershell means element do not have fullfiled or half filed orbital because in these case it takes more energy to remove valence shell electron.
The white dwarf has the lowest density of the stars listed. Hope this helps and have a nice day!
Answer:
1) Increase
2) Decreases
3) increases
4) Increase
Explanation:
These questions can only be answered by considering the principle which states that, "When a constraint such as a change in concentration, pressure or volume is imposed on a reaction system in equilibrium. The system will readjust itself in order to annul the constraint."
Now, if more reactants are added, the equilibrium position will shift towards the right, If more products are added, the equilibrium position will shift to the left.
Similarly, the removal of H2S causes the O2 concentration to increase since the equilibrium position now shifts to the left.
Also, addition of O2 causes H2S to be removed as the equilibrium moves to the right.