Atoms or molecule after gaining of electron possesses negative charge and is known as anion.
For the given sets:
The given elements are alkali metals and have tendency to lose electrons easily and form cations.
The given elements are non-metals and are electronegative. So, they gain electrons easily and form anion.
Carbon has tendency to form bond by sharing of electrons, Sulfur has tendency to gain electrons and form anion whereas Lead has tendency to lose electron.
Potassium and Iron has tendency to lose electron and form cation whereas Bromine has tendency to gain electron to form anion.
Hence, from the given sets, all elements of set: 
have tendency to form anions in binary ionic compounds.
<u>Answer:</u> The molar solubility of 
is 
<u>Explanation:</u>
Solubility is defined as the maximum amount of solute that can be dissolved in a solvent at equilibrium.
Solubility product is defined as the product of concentration of ions present in a solution each raised to the power its stoichiometric ratio.
The balanced equilibrium reaction for the ionization of calcium fluoride follows:
s 2s
The expression for solubility constant for this reaction will be:
![K_{sp}=[Pb^{2+}][I^-]^2](https://tex.z-dn.net/?f=K_%7Bsp%7D%3D%5BPb%5E%7B2%2B%7D%5D%5BI%5E-%5D%5E2)
We are given:
Putting values in above equation, we get:
Hence, the molar solubility of is
That is false.... I hope this is true or false
Answer:
Ei
Explanation:
De acordo com a teoria de Arrhenius, os ácidos são os compostos que se dissociam no meio aquoso para gerar os íons hidrogênio H + no meio aquoso.
Answer:
Mass of chemical = 1.5 mg
Explanation:
Step 1: First calculate the concentration of the stock solution required to make the final solution.
Using C1V1 = C2V2
C1 = concentration of the stock solution; V1 = volume of stock solution; C2 = concentration of final solution; V2 = volume of final solution
C1 = C2V2/V1
C1 = (6 * 25)/ 0.1
C1 = 1500 ng/μL = 1.5 μg/μL
Step 2: Mass of chemical added:
Mass of sample = concentration * volume
Concentration of stock = 1.5 μg/μL; volume of stock = 10 mL = 10^6 μL
Mass of stock = 1.5 μg/μL * 10^6 μL = 1.5 * 10^6 μg = 1.5 mg
Therefore, mass of sample = 1.5 mg
