Answer:
Explanation:
phosphorus belongs to group 5 of the periodic table because it has 5 electron in its outermost shell the number of electron in the outermost shell of electron determine the group of the element in the periodic table
Answer:
Explanation:
Unit 10 - Acid/Base ... (a) Mg(OH. 2. ) (b) Mg(OH). 2. (c) Mg. 2. OH. (d) MgOH. 2. Standard: ... balanced equation for these neutralization reactions: 3. HCl + NaOH → ... H2CO3 + Ca(OH)2 → ... C5.7B Predict products of an acid-base neutralization. 8. 2 NH4OH + H2S ...An Arrhenius base is a compound that increases the OH − ion concentration in ... and a base is called a neutralization reaction and can be represented as follows: ... chemical equation for the neutralization reaction between HCl and Mg(OH) 2. ... acid, an Arrhenius base, or neither. a) NaOH. b) C 2H 5OH. c) H 3PO 4. 6
Answer:They are movement, nutrition, respiration, sensitivity, reproduction, excretion,and growth.
Explanation:
"Silver chloride is essentially insoluble in water" this statement is true for the equilibrium constant for the dissolution of silver chloride.
Option: b
<u>Explanation</u>:
As silver chloride is essentially insoluble in water but also show sparing solubility, its reason is explained through Fajan's rule. Therefore when AgCl added in water, equilibrium take place between undissolved and dissolved ions. While solubility product constant 
for silver chloride is determined by equilibrium concentrations of dissolved ions. But solubility may vary also at different temperatures. Complete solubility is possible in ammonia solution as it form stable complex as water is not good ligand for Ag+.
To calculate firstly molarity of ions are needed to be found with formula: 
Then at equilibrium cations and anions concentration is considered same hence:
![\left[\mathbf{A} \mathbf{g}^{+}\right]=[\mathbf{C} \mathbf{I}]=\text { molarity of ions }](https://tex.z-dn.net/?f=%5Cleft%5B%5Cmathbf%7BA%7D%20%5Cmathbf%7Bg%7D%5E%7B%2B%7D%5Cright%5D%3D%5B%5Cmathbf%7BC%7D%20%5Cmathbf%7BI%7D%5D%3D%5Ctext%20%7B%20molarity%20of%20ions%20%7D)
Hence from above data can be calculated by: = ![\left[\mathbf{A} \mathbf{g}^{+}\right] \cdot[\mathbf{C} \mathbf{I}]](https://tex.z-dn.net/?f=%5Cleft%5B%5Cmathbf%7BA%7D%20%5Cmathbf%7Bg%7D%5E%7B%2B%7D%5Cright%5D%20%5Ccdot%5B%5Cmathbf%7BC%7D%20%5Cmathbf%7BI%7D%5D)
