The pH a 0.25 m solution of C₆H₅NH₂ is equal to 3.13.
<h3>How do we calculate pH of weak base?</h3>
pH of the weak base will be calculate by using the Henderson Hasselbalch equation as:
pH = pKb + log([HB⁺]/[B])
pKb = -log(1.8×10⁻⁶) = 5.7
Chemical reaction for C₆H₅NH₂ is:
C₆H₅NH₂ + H₂O → C₆H₅NH₃⁺ + OH⁻
Initial: 0.25 0 0
Change: -x x x
Equilibrium: 0.25-x x x
Base dissociation constant will be calculated as:
Kb = [C₆H₅NH₃⁺][OH⁻] / [C₆H₅NH₂]
Kb = x² / 0.25 - x
x is very small as compared to 0.25, so we neglect x from that term and by putting value of Kb, then the equation becomes:
1.8×10⁻⁶ = x² / 0.25
x² = (1.8×10⁻⁶)(0.25)
x = 0.67×10⁻³ M = [C₆H₅NH₃⁺]
On putting all these values on the above equation of pH, we get
pH = 5.7 + log(0.67×10⁻³/0.25)
pH = 3.13
Hence pH of the solution is 3.13.
To know more about Henderson Hasselbalch equation, visit the below link:
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Answer:False
Explanation:co2 or carbon dioxide is a compound not an element, you can even check this on the periodic table
<h3>
Answer: D) all of the above</h3>
Explanation:
The lungs pump oxygen in and carbon dioxide out, which goes through the blood stream to help with the cell's energy needs.
Nutrients pass through the blood stream as well. The nutrients start with the digestive system (mouth, esophagus, stomach, small intestine) before going into the blood stream. The nutrients are building blocks to help make new cells when old ones die off.
When those cells die off, the body sheds them like dead skin, but internal dead cells are passed off as waste. This waste and other byproducts the body doesn't need passes through the blood stream as well.
In short, the blood stream is basically the highway to help get desired materials (eg: oxygen and nutrients) and get rid of waste (eg: carbon dioxide and other unwanted byproducts or dead cell material)
So that's why the answer includes A, B and C.
Answer:
2Mg + O₂ ⟶ 2MgO
Explanation:
Step 1. Start with the most complicated-looking formula (O₂?).
Put a 1 in front of it.
Mg + 1O₂ ⟶ MgO
Step 2. Balance O.
We have fixed 2 O on the left. We need 2O on the right. Put a 2 in front of MgO.
Mg + 1O₂ ⟶ 2MgO
Step 3. Balance Mg.
We have fixed 2 Mg on the right-hand side. We need 2 Mg atoms on the left. Put a 2 in front of Mg.
2Mg + 1O₂ ⟶ 2MgO
Every formula now has a coefficient. The equation should be balanced. Let’s check.
<u>Atom</u> <u>On the left</u> <u>On the righ</u>t
Mg 2 2
O 2 2
All atoms are balanced.
The balanced equation is
2Mg + O₂ ⟶ 2MgO
