Answer:
72.6 mL
Explanation:
A quick way to solve this titration problem when you have a monoprotic acid is to use the Dilution equation, M1V1=M2V2.
.589(x)=.821(52.1)
X=72.6 mL
Answer:
506.912 L
Explanation:
From the question given above, the following data were obtained:
Number of mole of O₂ = 22.63 moles
Volume of O₂ =?
Recall:
1 mole of a gas occupy 22.4 L at STP.
With the above information, we obtained the volume occupied by 22.63 moles of O₂ as follow:
1 mole of O₂ occupied 22.4 L at STP.
Therefore, 22.63 moles of O₂ will occupy = 22.63 × 22.4 = 506.912 L at STP.
Thus, 22.63 moles of O₂ is equivalent to 506.912 L.
Answer:
The α‑helix is held together by hydrogen bonds between the amide N−H and C=O groups.
Disulfide bonds stabilize secondary structure.
Explanation:
Proteins have primary, secondary, tertiary and quartinary structures.
The secondary structure of a protein is the regular, recurring sequence of amino acid in a polypeptide chain. Secondary structure of proteins give rise to the folding observed in the structure of a protein.
The major secondary structures of a protein are α-helices and β-structures.
Purely for crystalline structure, "twinning" or repetition of crystal forms can bring about a striated texture on the mineral. Crystal defects and chemical impurities can alter the physical and electrical properties of a mineral. Some minerals can exist in different crystal forms and exhibit "polymorphism." The range in crystal structure can change the mineral's hardness, strength, solubility, electrical properties, melting points, etc.
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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