This is an application of Le Chatlier's principle: What happens when we add a reagent to one side of an equation? The reaction will shift to the other side. So heat is a reactant and we're adding more of it, the reaction must therefore, shift to the right ( or the products side).
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
Anaphase 1 is when centromeres divide
Answer:
pH= 2- log3
Explanation:
H2SO4 + H2O -> HSO4^(-) + H30^(+)
0.03M ___ ___
___ 0.03M 0.03M
H30^(+) : C = 0.03M
pH= - log( [H3O^(+)] ) => pH= - log {3× 10^(-2)} => pH = 2 - log3
