Protozoa can be broadly classified into four phyla based on their locomotion and morphology:
1. Rhizopoda: Also known as amoebas, these protozoa move by extending and retracting pseudopodia from their cell body. They lack a definite shape and can change their shape rapidly. Examples include Amoeba proteus and Entamoeba histolytica.
2. Zoomastigophora: These protozoa have flagella that they use for movement. They have a more defined shape compared to Rhizopoda. Examples include Trypanosoma and Giardia.
3. Ciliophora: These protozoa have numerous hair-like projections called cilia that cover their surface and are used for locomotion and feeding. Examples include Paramecium and Stentor.
4. Sporozoa: These protozoa are non-motile and usually have complex life cycles involving multiple hosts. They typically reproduce through spore formation. Examples include Plasmodium (the causative agent of malaria) and Toxoplasma gondii.
classification of protozoan
3 answers
Balance the following ionic equation
ClO- + CrO2- Cl - + CrO42- ( under basic solution)
ClO- + CrO2- Cl - + CrO42- ( under basic solution)
To balance the given ionic equation under basic solution, follow these steps:
Step 1: Write the unbalanced equation:
ClO⁻ + CrO₂⁻ ⟶ Cl⁻ + CrO₄²⁻
Step 2: Balance the atoms other than hydrogen and oxygen.
For this equation, we have Cl, Cr, and O.
a) Balance Cl by adding Cl⁻ to the left side:
Cl⁻ + ClO⁻ + CrO₂⁻ ⟶ Cl⁻ + CrO₄²⁻
b) Balance Cr by adding a coefficient of 2 in front of CrO₂⁻ on the left side:
Cl⁻ + ClO⁻ + 2CrO₂⁻ ⟶ Cl⁻ + CrO₄²⁻
c) Count the number of oxygen atoms. We have 5 on the left side and 7 on the right side.
Add water molecules (H₂O) to the left side to balance the oxygen atoms:
Cl⁻ + ClO⁻ + 2CrO₂⁻ ⟶ Cl⁻ + CrO₄²⁻ + H₂O
Step 3: Balance hydrogen atoms:
Add hydrogen ions (H⁺) to the right side to balance the hydrogen atoms:
Cl⁻ + ClO⁻ + 2CrO₂⁻ ⟶ Cl⁻ + CrO₄²⁻ + H₂O + H⁺
Step 4: Balance charge:
To balance the charge, add electrons (e⁻) to the left side to balance the charge on the right side:
Cl⁻ + ClO⁻ + 2CrO₂⁻ + 6e⁻ ⟶ Cl⁻ + CrO₄²⁻ + H₂O + H⁺
Step 5: Combine the half-reactions:
Combine the half-reactions by multiplying the first half-reaction by 2 and the second half-reaction by 6 to equalize the electron transfer:
2Cl⁻ + 2ClO⁻ + 4CrO₂⁻ + 12e⁻ ⟶ 2Cl⁻ + 6CrO₄²⁻ + 6H₂O + 6H⁺
Step 6: Simplify the equation:
Cancel out the Cl⁻ ions on both sides:
2ClO⁻ + 4CrO₂⁻ + 12e⁻ ⟶ 6CrO₄²⁻ + 6H₂O + 6H⁺
Step 7: Verify the balance:
Check if the number of each atom and the charge are balanced on both sides of the equation.
The balanced ionic equation for the given reaction under basic solution is:
2ClO⁻ + 4CrO₂⁻ + 12e⁻ ⟶ 6CrO₄²⁻ + 6H₂O + 6H⁺
Step 1: Write the unbalanced equation:
ClO⁻ + CrO₂⁻ ⟶ Cl⁻ + CrO₄²⁻
Step 2: Balance the atoms other than hydrogen and oxygen.
For this equation, we have Cl, Cr, and O.
a) Balance Cl by adding Cl⁻ to the left side:
Cl⁻ + ClO⁻ + CrO₂⁻ ⟶ Cl⁻ + CrO₄²⁻
b) Balance Cr by adding a coefficient of 2 in front of CrO₂⁻ on the left side:
Cl⁻ + ClO⁻ + 2CrO₂⁻ ⟶ Cl⁻ + CrO₄²⁻
c) Count the number of oxygen atoms. We have 5 on the left side and 7 on the right side.
Add water molecules (H₂O) to the left side to balance the oxygen atoms:
Cl⁻ + ClO⁻ + 2CrO₂⁻ ⟶ Cl⁻ + CrO₄²⁻ + H₂O
Step 3: Balance hydrogen atoms:
Add hydrogen ions (H⁺) to the right side to balance the hydrogen atoms:
Cl⁻ + ClO⁻ + 2CrO₂⁻ ⟶ Cl⁻ + CrO₄²⁻ + H₂O + H⁺
Step 4: Balance charge:
To balance the charge, add electrons (e⁻) to the left side to balance the charge on the right side:
Cl⁻ + ClO⁻ + 2CrO₂⁻ + 6e⁻ ⟶ Cl⁻ + CrO₄²⁻ + H₂O + H⁺
Step 5: Combine the half-reactions:
Combine the half-reactions by multiplying the first half-reaction by 2 and the second half-reaction by 6 to equalize the electron transfer:
2Cl⁻ + 2ClO⁻ + 4CrO₂⁻ + 12e⁻ ⟶ 2Cl⁻ + 6CrO₄²⁻ + 6H₂O + 6H⁺
Step 6: Simplify the equation:
Cancel out the Cl⁻ ions on both sides:
2ClO⁻ + 4CrO₂⁻ + 12e⁻ ⟶ 6CrO₄²⁻ + 6H₂O + 6H⁺
Step 7: Verify the balance:
Check if the number of each atom and the charge are balanced on both sides of the equation.
The balanced ionic equation for the given reaction under basic solution is:
2ClO⁻ + 4CrO₂⁻ + 12e⁻ ⟶ 6CrO₄²⁻ + 6H₂O + 6H⁺