You might have meant hemiacetal, not hemicetal.
Acetals contain two –OR groups, one –R group and a –H atom. In hemiacetals, one of the –OR groups in acetals is replaced by a –OH group<span>.
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The correct answer is D.
Increasing the pressure shifts the position of equilibrium towards the side with fewer gas molecules.
Answer:
The rate of disappearance of C₂H₆O = 2.46 mol/min
Explanation:
The equation of the reaction is given below:
2 K₂Cr₂O₇ + 8 H₂SO₄ + 3 C₂H₆O → 2 Cr₂(SO₄)₃ + 2 K₂SO₄ + 11 H₂O
From the equation of the reaction, 3 moles of C₂H₆O is used when 2 moles of Cr₂(SO₄)₃ are produced, therefore, the mole ratio of C₂H₆O to Cr₂(SO₄)₃ is 3:2.
The rate of appearance of Cr₂(SO₄)₃ in that particular moment is given 1.64 mol/min. This would than means that C₂H₆O must be used up at a rate which is approximately equal to their mole ratios. Thus, the rate of of the disappearance of C₂H₆O can be calculated from the mole ratio of Cr₂(SO₄)₃ and C₂H₆O.
Rate of disappearance of C₂H₆O = 1.64 mol/min of Cr₂(SO₄)₃ * 3 moles of C₂H₆O / 2 moles of Cr₂(SO₄)₃
Rate of disappearance of C₂H₆O = 2.46 mol/min of C₂H₆O
Therefore, the rate of disappearance of C₂H₆O = 2.46 mol/min
The new pressure would be = 4.46 atm
<h3>Further explanation</h3>
Given
V₁=6.7 L(at STP, 1 atm 273 K)
V₂=1.5 L
Required
The new pressure
Solution
Boyle's Law
At a constant temperature, the gas volume is inversely proportional to the pressure applied
P₂ = (P₁V₁)/V₂
P₂ = (1 atm x 6.7 L)/1.5 L
P₂ = 4.46 atm
