Answer: A balloon is charged by a process of frictional charging and the object is getting charged by the process of induction.
Explanation:
When two bodies are rubbed against each other, charging by friction or rubbing occurs. The electropositive object loses electrons to electronegative object. Thus, when balloon is rubbed on a wall, it becomes charged.
The charged balloon is able to attract an uncharged object by inducing charge on it without the two objects touching each other. Electrostatic force acts between two charged objects. Charged balloon causes electrons to move at one end thereby inducing opposite charge in the object and thus, charged balloon is able to attract uncharged object.
Answer:
Correct answer is: "Cold Water"
Explanation:
EDGE 2020
Answer:
B
Explanation:
As you can see in these 4 examples, B- looks completely different from A, C, D! In B: The reactants and products are completely different in the Element Figures.
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.
D. A ll of the above. catalyst reduces the activation energy thereby making the reaction faster. an increase in temperature increases the kinetic energy and no of collisions making reaction faster for an endothermic reaction while decrease in temperature favors an exothermic reaction.
increase in concentration increases the molecules reacting so the reaction is faster and vice versa.
