Do you think placing an Alka-Seltzer in water causes a chemical reaction? Explain Why.
1 answer:
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Answer:
The Photosynthesis process
Explanation:
Plants, algae, and some other organisms can transform the sunlight energy into chemical energy. The photosynthesis process occur thanks to the chloroplasts. The chloroplast is an organelle found in all green plants. Inside of the chloroplast you can find the thylakoids which are arranged in stacks named grana, they have membranes with chloropyll a photosynthetic pigment, also you can find the photosystems, they are functional and structural units of protein complexes. The thylakoids capture the light and allow the reactions to transform CO2. The set of reactions that occurs in the chloroplasts are known as the Calvin cycle.
The general equation of photosynthesis is:
6 CO2 + 6 H2O + Energy -> C6H12O6 + 6 O2
Carbon Dioxide + water + Light -> Glucose (sugar) + Oxygen
After, this glucose is transformed into pyruvate, and it allowed the release of denosine triphosphate (ATP) by cellular respiration. The ATP is an organic chemical that is requires for the cell to perform any process (any kind or work).
Answer:
pH = 8.0
Explanation:
First, we have to calculate the moles of NaOH.
Let's consider the balanced equation.
C₂H₄O₃ + NaOH ⇒ C₂H₃O₃Na + H₂O
The molar ratio C₂H₄O₃: NaOH: C₂H₃O₃Na is 1: 1: 1. So, when 7.2 × 10⁻⁴ moles of NaOH react completely with 7.2 × 10⁻⁴ moles of C₂H₄O₃ they form 7.2 × 10⁻⁴ moles of C₂H₃O₃Na.
The concentration of C₂H₃O₃Na is:
C₂H₃O₃Na dissociates according to the following equation:
C₂H₃O₃Na(aq) ⇒ C₂H₃O₃⁻(aq) + Na⁺(aq)
C₂H₃O₃⁻ comes from a weak acid so it undergoes basic hydrolisis.
C₂H₃O₃⁻ + H₂O ⇄ C₂H₄O₃ + OH⁻
If we know that pKa for C₂H₄O₃ is 3.9, we can calculate pKb for C₂H₃O₃⁻ using the following expression:
pKa + pKb = 14
pKb = 14 -3.9 = 10.1
10.1 = -log Kb
Kb = 7.9 × 10⁻¹¹
We can calculate [OH⁻] using the following expression:
[OH⁻] = √(Kb.Cb) <em>where Cb is the initial concentration of the base</em>
[OH⁻] = √(7.9 × 10⁻¹¹ × 0.012M) = 9.7 × 10⁻⁷ M
Now, we can calculate pOH and pH.
pOH = -log [OH⁻] = -log (9.7 × 10⁻⁷) = 6.0
pH + pOH = 14
pH = 14 - pOH = 14 - 6.0 = 8.0