Answer:
1.7 * 10^-5
Explanation:
1- get the number of moles of PbCl2:
number of moles = mass / molar mass
number of moles = 0.45 / 278.1 = 1.618 * 10^-3 moles
2- get the concentration of Pb2+:
molarity = number of moles of solute / volume of solution in liters
molarity = (1.618 * 10^-3) / (0.1) = 0.0162 M
3- getting concentration of Cl-:
<span>PbCl2(s) <==> Pb2+(aq) + 2Cl-(aq)
</span>We can note that:
For a certain amount of Pb2+ formed, twice this amount of Cl- is formed.
This means that:
for 0.0162 M of Pb2+, 2*0.0168 = 0.0324 M of Cl- is formed
4- getting Ksp:
Ksp = [Pb2+][Cl-]²
Ksp = (0.0162)*(0.0324)²
Ksp = 1.7 * 10^-5
Hope this helps :)
Answer: all elements in the periodic table is classified as elements
Explanation:
The structure of the table shows periodic trends. The seven rows of the table, called periods, generally have metals on the left and nonmetals on the right. The columns, called groups, contain elements with similar chemical behaviours. Six groups have accepted names as well as assigned numbers: for example, group 17 elements are the halogens; and group 18 are the noble gases
Answer: alkaline earth metals (group-IIA)
Explanation:
The element which donates the electron is known as electropositive element and forms a positively charged ion called as cation. The element which accepts the electrons is known as electronegative element and forms a negatively charged ion called as anion.
Alkaline earth metals donate 2 valence electrons to acquire noble gas configuration.
For example: Berrylium is the first alkaline earth metal with atomic number of 4 and thus has 4 electrons
Electronic configuration of berrylium:
![[Be]:4:1s^22s^2](https://tex.z-dn.net/?f=%5BBe%5D%3A4%3A1s%5E22s%5E2)
Berrylium atom will loose two electrons to gain noble gas configuration and form berrylium cation with +2 charge.
![[Be^{2+}]:2:1s^2](https://tex.z-dn.net/?f=%5BBe%5E%7B2%2B%7D%5D%3A2%3A1s%5E2)
Thus Elements donate 2 electron to produce a cation with a 2+ charge are alkaline earth metals.
Answer:

Explanation:
Given a reaction mechanism, we will typically have catalysts and intermediates which will not be observed in the final balanced overall reaction. In order to obtain a net reaction, we need to add all the separate steps of the given reaction.
First of all, the reactants are combined followed by a combination of products. The repeating species on both sides of the final equation are then canceled out. Let's sum everything we have in our steps:

Repeating species that can be canceled out are:

This leaves a net reaction of:
