Answer:
Moles of silver iodide produced = 1.4 mol
Explanation:
Given data:
Mass of calcium iodide = 205 g
Moles of silver iodide produced = ?
Solution:
Chemical equation:
CaI₂ + 2AgNO₃ → 2AgI + Ca(NO₃)₂
Number of moles calcium iodide:
Number of moles = mass/ molar mass
Number of moles = 205 g/ 293.887 g/mol
Number of moles = 0.7 mol
Now we will compare the moles of calcium iodide with silver iodide.
CaI₂ : AgI
1 : 2
0.7 : 2×0.7 = 1.4
Thus 1.4 moles of silver iodide will be formed from 205 g of calcium iodide.
Enzymes in our body increases the speed of chemical digestion.
Answer:
Option A. KCl (aq)
Option D. Mg(OH)₂(s
Explanation:
We'll begin by writing the balanced equation for the reaction. This is illustrated below:
MgCl₂(aq) + KOH(aq) —>
In solution, MgCl₂(aq) and KOH(aq) will dissociate as follow:
MgCl₂(aq) —> Mg²⁺(aq) + 2Cl¯(aq)
KOH(aq) —> K⁺(aq) + OH¯(aq)
MgCl₂(aq) + KOH(aq) —>
Mg²⁺(aq) + 2Cl¯(aq) + 2K⁺(aq) + OH¯(aq) —> 2K⁺(aq) + 2Cl¯(aq) + Mg(OH)₂ (s)
MgCl₂(aq) + KOH(aq) —> 2KCl (aq) + Mg(OH)₂(s)
Thus, the products of the above reaction are: KCl(aq) and Mg(OH)₂(s)
Thus, option A and D gives the correct answer to the question.
