The theoretical yield is 204.4 g while the percent yield is 2.57%.
<h3>What is theoretical yield?</h3>
Theoretical yield is the amount of product obtained based on the stoichiometry of the reaction.
S8(s) + 8 Na2SO3(aq) + 40 H2O(l) --->8 Na2S2O3·5 H2O(s)
Number of moles of sulfur = 3.25 g /8(32) = 0.013 moles
Number of moles of sodium sulfite = 13.1 g/126 g/mol = 0.103 moles
Since 1 moles of sulfur reacts with 8 moles of sodium sulfite
0.013 moles reacts with 0.013 moles × 8 moles /1 mole = 0.104 moles
There is not enough sodium sulfite hence it is the limiting reactant.
1 mole of sodium sulfite yields 8 moles of product
0.103 moles of sodium sulfite yields 0.103 moles × 8 moles /1 mole = 0.824 moles
Mass of product = 0.824 moles × 248 g/mol = 204.4 g
percent yield = 5.26 g /204.4 g × 100/1
= 2.57%
Learn more about percent yield: brainly.com/question/2506978
Heated mater rises and cold mater sinks
Answer:
pH = 3.3
Explanation:
Buffer solutions minimize changes in pH when quantities of acid or base are added into the mix. The typical buffer composition is a weak electrolyte (wk acid or weak base) plus the salt of the weak electrolyte. On addition of acid or base to the buffer solution, the solution chemistry functions to remove the acid or base by reacting with the components of the buffer to shift the equilibrium of the weak electrolyte left or right to remove the excess hydronium ions or hydroxide ions is a way that results in very little change in pH of the system. One should note that buffer solutions do not prevent changes in pH but minimize changes in pH. If enough acid or base is added the buffer chemistry can be destroyed.
In this problem, the weak electrolyte is HNO₂(aq) and the salt is KNO₂(aq). In equation, the buffer solution is 0.55M HNO₂ ⇄ H⁺ + 0.75M KNO₂⁻ . The potassium ion is a spectator ion and does not enter into determination of the pH of the solution. The object is to determine the hydronium ion concentration (H⁺) and apply to the expression pH = -log[H⁺].
Solution using the I.C.E. table:
HNO₂ ⇄ H⁺ + KNO₂⁻
C(i) 0.55M 0M 0.75M
ΔC -x +x +x
C(eq) 0.55M - x x 0.75M + x b/c [HNO₂] / Ka > 100, the x can be
dropped giving ...
≅0.55M x ≅0.75M
Ka = [H⁺][NO₂⁻]/[HNO₂] => [H⁺] = Ka · [HNO₂]/[NO₂⁻]
=> [H⁺] = 6.80x010⁻⁴(0.55) / (0.75) = 4.99 x 10⁻⁴M
pH = -log[H⁺] = -log(4.99 x 10⁻⁴) -(-3.3) = 3.3
Solution using the Henderson-Hasselbalch Equation:
pH = pKa + log[Base]/[Acid] = -log(Ka) + log[Base]/[Acid]
= -log(6.8 x 10⁻⁴) + log[(0.75M)/(0.55M)]
= -(-3.17) + 0.14 = 3.17 + 0.14 = 3.31 ≅ 3.3
<u>Answer:</u>
Nitrogen gas be a mineral only, if it is in organic forms.
<u>Explanation:</u>
Most of the forms of organic nitrogen is not be taken by plants, with the exception in the form of small organic molecules. Also plants can promptly take the nitrogen when it is in other forms like ammonia and nitrate.
The microorganisms in the soil converts the organic forms of nitrogen to mineral form when they decompose organic matters and also fresh plant residues. This type of process is called mineralisation.
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