Answer:
She can add 380 g of salt to 1 L of hot water (75 °C) and stir until all the salt dissolves. Then, she can carefully cool the solution to room temperature.
Explanation:
A supersaturated solution contains more salt than it can normally hold at a given temperature.
A saturated solution at 25 °C contains 360 g of salt per litre, and water at 70 °C can hold more salt.
Yasmin can dissolve 380 g of salt in 1 L of water at 70 °C. Then she can carefully cool the solution to 25 °C, and she will have a supersaturated solution.
B and D are wrong. The most salt that will dissolve at 25 °C is 360 g. She will have a saturated solution.
C is wrong. Only 356 g of salt will dissolve at 5 °C, so that's what Yasmin will have in her solution at 25 °C. She will have a dilute solution.
Answer: option A) True
Explanation:
Joseph Proust is the founder of the Law of Definite proportion. This law states that two elements will always combine together to form a chemical compound and this will be in the same proportion by mass.
For Example: 2 moles of Hydrogen (H2) ALWAYS combine with 1 atom of Oxygen (O) to yield water (H2O)
So, the answer is True
Yes, free electrons appear in balanced redox reaction equations. However, this is only true for half-reactions. This is because redox reactions primarily involve the transfer of electrons, which are better visualized if explicitly shown in the balanced reactions. In reduction reactions, electrons are placed on the left side of the equation. Oxidation reactions show electrons on the right side of the equation.
Explanation:
A half reaction is either the chemical reaction or reduction reaction part of an oxidoreduction reaction. A half reaction is obtained by considering the amendment in chemical reaction states of individual substances concerned within the oxidoreduction reaction. Half-reactions are usually used as a way of leveling oxidoreduction reactions.The half-reaction on the anode, wherever chemical reaction happens, is Zn(s) = Zn2+ (aq) + (2e-).
The metal loses 2 electrons to create Zn2+. The half-reaction on the cathode wherever reduction happens is Cu2+ (aq) + 2e- = Cu(s).
Here, the copper ions gain electrons and become solid copper.
