Since the half-reaction is occurring in a basic solution, add 32OH− to each side of the equation to eliminate the H+ ions.
P₄ +16H₂O + 32OH⁻ ⟶ 4PO₃⁻⁴ + 32H⁺ +32OH⁻
Final reaction :
P₄ + 32OH⁻ ⟶ 4PO₃⁻⁴ + 16H₂O + 20e⁻
A half reaction is either the oxidation or reduction reaction component of a redox reaction. A half reaction is obtained by considering the change in oxidation states of individual substances involved in the redox reaction.
The concept of half-reactions is used to describe what occurs in an electrochemical cell, such as a Galvanic cell battery. Half-reactions can be written to describe both the metal undergoing oxidation (known as the anode) and the metal undergoing reduction (known as the cathode).
Half-reactions are often used as a method of balancing redox reactions. For oxidation-reduction reactions in acidic conditions, after balancing the atoms and oxidation numbers, one will need to add H+ ions to balance the hydrogen ions in the half reaction.
For oxidation-reduction reactions in basic conditions, after balancing the atoms and oxidation numbers, first treat it as an acidic solution and then add OH- ions to balance the H+ ions in the half reactions (which would give H2O).
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Equation for Half life :
A = a(0.5)^(t/h)
A is current amount, "a" is initial amount, h is halflife, t is time
5 = 40(0.5)^(t/1.3x10^9)
5/40 = (0.5)^(t/1.3x10^9)
take the log of both sides , power rule
Log(5/40) = (t/1.3x10^9) * Log(0.5)
(1.3x10^9) * Log(5/40) / Log(0.5) = t
3.9x10^9 years = t
And if you think about what a half life is, the time it take for the amount to reduce to half.
40/2 = 20
20/2 = 10
10/2 = 5
It went through 3 half-lifes
3 * 1.3x10^9 = 3.9x10^9 years
Its both A and C because both A and C have only one type each so it can only be those two :)
The percent yield of CO₂ is 93.3%.
<h3>What is the percent yield of CO₂?</h3>
The percent yield of a substance is given as follows:
- Percent yield = actual yield/theoretical yield * 100 %
The equation of the reaction is used to determine the theoretical yield.
- NaHCO₃ + CH₃COOH → CH₃COONa + H₂O + CO₂
Moe ratio of sodium bicarbonate and CO₂ is 1 : 1.
Given that sodium bicarbonate is the limiting reactant, the theoretical yield of CO₂ will be:
Moles of NaHCO₃ reacting = 2.01/84 = 0.0239 moles
Theoretical yield of CO₂ = 0.0239 moles * 22.4L/mol = 0.536 L
Actual yield = 0.50 L
Percent yield = 0.50/0.536 * 100%
Percent yield = 93.3%
In conclusion, the percent yield is the ratio of the actual yield and theoretical yield.
<em>Note that the complete question is given below:</em>
<em>Calculate your % yield of co2 in the reaction based on the grams of nahco3 being the limiting reagent in the reaction between 2.01 g of sodium bicarbonate and 24.6 mL of 1.5 M acetic acid? They produce 0.50 L of at s.t.p.</em>
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