Answer:
and 
Explanation:
The equation for the reaction is AgNO3(aq) + KCl(aq) ==> AgCl(s) + KNO3(aq)
With all the ions, it is
(aq) +
(aq) +
(aq) +
(aq) ==> AgCl(s) +
and
do not change, so they are the spectator ions and are removed
The ionic equation is:
(aq) +
(aq) ==> AgCl(s)
Answer:
0. 414
Explanation:
Octahedral interstitial lattice sites.
Octahedral interstitial lattice sites are in a plane parallel to the base plane between two compact planes and project to the center of an elementary triangle of the base plane.
The octahedral sites are located halfway between the two planes. They are vertical to the locations of the spheres of a possible plane. There are, therefore, as many octahedral sites as there are atoms in a compact network.
The Octahedral interstitial void ratio range is 0.414 to 0.732. Thus, the minimum cation-to-anion radius ratio for an octahedral interstitial lattice site is 0. 414.
Answer:
Explanation:
To solve this problem, we need to obtain the number of moles of the solute we desired to prepare;
Number of moles = molarity x volume
Parameters given;
volume of solution = 500mL = 0.5L
molarity of solution = 0.5M
Number of moles = 0.5 x 0.5 = 0.25moles
Now to know the volume stock to take;
Volume of stock =
molarity of stock = 4M
volume =
= 0.0625L or 62.5mL
It’s Tin (II) Phosphate!! (Hope this helps!)