
<h3><u>Basic </u><u>Characteristic </u><u>of </u><u>acids </u></h3>
- Acids are sour in taste
- Acid turns blue litmus paper or solution into red litmus paper or solution
- Acids are good conductor of electricity because it dissociate into cation in aqueous solution
- Acids classified into edible acids and non edible acids. Non edible acids are very hazardous
- Generally, All acids are soluble in water
- Acids have PH smaller than 7
<u>Arrhenius </u><u>definition </u><u>of </u><u>acids </u><u>:</u><u>-</u>
According to Arrehinus,
- Acids are those substances which when dissolve in water given H positive ions . Then, this hydrogen ions combine with water to form H30 + ions
<u>Second </u><u>definition </u><u>of </u><u>Acids </u><u>was </u><u>given </u><u>by </u><u>Bonsted </u><u>Lowry </u><u>:</u><u>-</u>
According to Bonsted Lowry
- Acids are the proton donors that is when acids dissociate into water gives hydrogen ions that is H+ ions
<u>3rd </u><u>definition </u><u>was </u><u>given </u><u>by </u><u>Lewis </u>
According to Lewis
- Acids are those substances which have the ability to accept a pair of electrons .
Example of Acids
- HNO3 :- Nitric acid
- H2SO4 :- Sulfuric acid
- HCl :- Hydrochloric acid
<h3><u>Basic </u><u>characterists </u><u>of </u><u>bases </u></h3>
- Bases are bitter in taste
- Bases turns red litmus paper into blue litmus paper or solution
- Bases are also good conductor of electricity because on dissociation it produces anion in aqueous solution
- Bases are also good conductor of electricity
- When bases are soluble in water then they are known as alkaline base
- Bases have PH greater than 7
<u>Arrehinus definition of bases :-</u>
According to Arrehinus ,
- Bases are those substances which when dissolve in water produce OH negative ions that is hydroxide ions
<u>Bonsted Lowry definition </u>
According to Bonsted Lowry
- Bases are the proton donors as they produce OH negative ions in dissociation in aqueous solution
<u>Lewis </u><u>definition </u>
According to Lewis
- Acids are those substances which have the ability to lose electrons that is they are electron donors.
Example of bases
- Ca(OH)2 :- Calcium hydroxide
- NaOH :- Sodium hydroxide
- KOH :- Potassium hydroxide
[ Note :- There are so many Lewis acids and bases but they are not Arrhenius or Lowry acids or bases ]
<h3><u>Basic </u><u>characteristic </u><u>of </u><u>salt </u></h3>
- Salts are the ionic compounds which are composed of acids and bases that cation and anion
- Salts are generally found in oceans and seas in the forms of crystals
- As they are composed of acids and bases so they are neutral in nature but the salt of strong acid or weak base is acidic in nature or vice versa
- Salts are also good conductor of electricity as they form ionic bond
- Generally, All salts are soluble in water.
- The PH of common Salt is 7
Example of salts
- NH4Cl :- Ammonium chloride
- CuSO4 :- Copper sulphate
- NaCl :- Sodium chloride
Answer:
The solutions are ordered by this way (from lowest to highest freezing point): K₃PO₄ < CaCl₂ < NaI < glucose
Option d, b, a and c
Explanation:
Colligative property: Freezing point depression
The formula is: ΔT = Kf . m . i
ΔT = Freezing T° of pure solvent - Freezing T° of solution
We need to determine the i, which is the numbers of ions dissolved. It is also called the Van't Hoff factor.
Option d, which is glucose is non electrolyte so the i = 1
a. NaI → Na⁺ + I⁻ i =2
b. CaCl₂ → Ca²⁺ + 2Cl⁻ i =3
c. K₃PO₄ → 3K⁺ + PO₄⁻³ i=4
Potassium phosphate will have the lowest freezing point, then we have the calcium chloride, the sodium iodide and at the end, glucose.
Answer:
0.0468 g.
Explanation:
- The decay of radioactive elements obeys first-order kinetics.
- For a first-order reaction: k = ln2/(t1/2) = 0.693/(t1/2).
Where, k is the rate constant of the reaction.
t1/2 is the half-life time of the reaction (t1/2 = 1620 years).
∴ k = ln2/(t1/2) = 0.693/(1620 years) = 4.28 x 10⁻⁴ year⁻¹.
- For first-order reaction: <em>kt = lna/(a-x).</em>
where, k is the rate constant of the reaction (k = 4.28 x 10⁻⁴ year⁻¹).
t is the time of the reaction (t = t1/2 x 8 = 1620 years x 8 = 12960 year).
a is the initial concentration (a = 12.0 g).
(a-x) is the remaining concentration.
∴ kt = lna/(a-x)
(4.28 x 10⁻⁴ year⁻¹)(12960 year) = ln(12)/(a-x).
5.54688 = ln(12)/(a-x).
Taking e for the both sides:
256.34 = (12)/(a-x).
<em>∴ (a-x) = 12/256.34 = 0.0468 g.</em>
the answer is in the picture
Answer:
D
Explanation:
the production of an odor would indicate that the heat ignited a chemical reaction