Answer:Sodium chloride solid is not a mixture. ... It cannot be physically separated into its components, Na+ and Cl−
Explanation:Sodium chloride is formed when sodium atoms interact with chlorine atoms. When this occurs, sodium will donate an electron (which is a negatively-charged particle) to chlorine. This makes sodium slightly positive and chlorine slightly negative.
Opposite charges attract, right? So then, sodium ions will attract chloride ions and form an ionic bond. By the way, chloride is the term used to designate the anion form of chlorine. The result is a crystallized salt that has properties that are different from the two parent elements (sodium and chlorine). The chemical formula for sodium chloride is NaCl, which means that for every sodium atom present, there is exactly one chloride atom.
The following are neutral salts: NANO3, BaBr2 AND Ba[OH]2.
Neutral salts are those salts that are formed when a strong base react with a strong acid. In those neutralization reactions, the base completely neutralize the acid to form a neutral salts.
Answer:
Hope this is helpful to you!
The independent variable is the one that is changed by the scientist. To insure a fair test, a good experiment has only ONE independent variable. As the scientist changes the independent variable, he or she records the data that they collect.
Answer: pH of HCl =5, HNO3 = 1,
NaOH = 9, KOH = 12
Explanation:
pH = -log [H+ ]
1. 1.0 x 10^-5 M HCl
pH = - log (1.0 x 10^-5)
= 5 - log 1 = 5
2. 0.1 M HNO3
pH = - log (1.0 x 10 ^ -1)
pH = 1 - log 1 = 1
3. 1.0 x 10^-5 NaOH
pOH = - log (1.0 x 10^-5)
pOH = 5 - log 1 = 5
pH + pOH = 14
Therefore , pH = 14 - 5 = 9
4. 0.01 M KOH
pOH = - log ( 1.0 x 10^ -2)
= 2 - log 1 = 2
pH + pOH = 14
Therefore, pH = 14 - 2 = 12
Answer:
0.32M
Explanation:
<u>Step 1:</u> Balance the reaction
K2CO3 + Ba(NO3)2 ⇔ KNO3 + BaCO3
We have a 20 mL 0.2 M K2CO3 and a 30mL 0.4M Ba(NO3)2 solution
SinceK2CO3 is the limiting reactant, there will remain Ba(NO3)2 after it's consumed and produced KNO3 + BaCO3
<u>Step 2: </u>Calculate concentration
To find the concentration of the barium cation we use the following equation:
Concentration = moles of the <u>solute</u> / volumen of the <u>solution</u>
<u />
<u>[Ba2+] </u> = (20 * 10^-3 * 0.2M + 30 * 10^-3 * 0.4M) / ( 20 + 30mL) *10^-3
[Ba2+] = 0.32 M
The concentration of Barium ion in solution is 0.32 M
