When writing an ionic compound formula, a "molecular" form is used. The formula is made with allowance for ion charges.
For example,
Ca²⁺ and NO₃⁻ ⇒ Ca(NO₃)₂
Al³⁺ and SO₄²⁻ ⇒ Al₂(SO₄)₃
Answer:
<u>During Metaphase</u>
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Explanation:
During Metaphase the cell chromosome align themselves in the middle of the cell.This occur due to a cellular process called "Tug of War".
The chromosome which have been replicated and joined at the central point called centromere are called sister chromatids
Prior to metaphase , Kinetochore type of protein are formed around the centromere. Long protein filament called kinetochore are extended from poles to other end of the cell attached to kinetochore.
Therse is important checkpoint in the middle of mitosis called<u> metaphase.</u>
At this point cell ensure that , it is ready to divide or not.
Once the cell ensure that everything is ready to divide. Only after then , yhe cell enters the fourth phase called <u>anaphase.</u>
Answer: from the Zn anode to the Cu cathode
Justification:
1) The reaction given is: Zn(s) + Cu₂⁺ (aq) -> Zn²⁺ (aq) +Cu(s)
2) From that, you can see the Zn(s) is losing electrons, since it is being oxidized (from 0 to 2⁺), while Cu²⁺, is gaining electrons, since it is being reduced (from 2⁺ to 0).
3) Then, you can already tell that electrons go from Zn to Cu.
4) The plate where oxidation occurs is called anode, and the plate where reduction occus is called cathode.
So you get that the electrons flow from the anode (Zn) to the cathode (Cu).
Always oxidation occurs at the anode, and reduction occurs at the cathode.
The complete balanced chemical reaction is written as:
AgNO3 + KCl ---> AgCl
+ KNO3
where AgCl is our
precipitate
So calculating for moles
of AgCl produced: MM AgCl = 143.5 g/mol
moles AgCl = 0.326 g /
(143.5 g/mol) = 2.27 x 10^-3 mol
we see that there is 1
mole of Ag per 1 mole of AgCl so:
moles Ag = 2.27 x 10^-3
mol
The molarity is simply
the ratio of number of moles over volume in Liters, therefore:
Molarity = 2.27 x 10^-3
mol / 0.0977 L
<span>Molarity = 0.0233 M</span>
