
<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:
Plants need energy from the sun, water from the soil, and carbon from the air to grow. Air is mostly made of nitrogen, oxygen, and carbon dioxide. So how do plants get the carbon they need to grow? They absorb carbon dioxide from the air.
please make most brainlyest
Every science experiment should follow the basic principles of proper investigation so that the results presented at the end are seen as credible.
Observation and Hypothesis. ...
Prediction and Modeling. ...
Testing and Error Estimation. ...
Result Gathering and Presentation. ...
Conclusions. ...
Law Formation.
1) Chemical equation
Na2 SiO3 (s) + 8 HF (aq) ---> H2 Si F6 (aq) + 2 Na F (aq) + 3H2O (l)
It is balanced
2) Molar ratios
1 mol Na2 SiO3 : 8 mol HF.
3) Proportion
0.340 mol Na2 SiO3 * 8 mol HF / 1mol Na2SiO3 = 2.72 mol HF.
Answer: 2.72 mol HF
Answer:
7200 kPa
Explanation:
Applying,
PV/T = P'V'/T'................ Equation 1
Where P = Initial pressure of neon gas, V = Initial volume of neon gas, T = Initial temperature of neon gas, P' = Final pressure of neon gas, V' = Final volume of neon gas, T' = Final Temperature of neon gas
Make P' the subject of the equation
P' = PVT'/V'T.............. Equation 2
Given: P = 900 kPa, V = 8.0 L, T = 300 K, V' = 2.0 L, T' = 600 K
Substitute these values into equation 2
P' = (900×8×600)/(2×300)
P' = 7200 kPa