16 H + + 2Cr2O72- + C2H5OH → 4 Cr3 + +11H2O +2CO2
The reducing agent is C2H5OH
Explanation
reducing agent is a substance that loses or donate electrons in a chemical reaction. C2H5OH is the one which donate electrons in the above chemical equation.
27/208 = normality
12 x 10^-2 approx = normality
nw Ka = 14.3 x 10^-3
pKa = 3 - log 14
now, after getting the pKa put it in formula :
pH = pKa + log concn of ion/concn of salt and you'll get it
hope this helps
Q. How many molecules of H2O can be produced from reactants in container below?
A. 3 molecule of molecules H2O will be produced from reactants in container.
<em><u>Explanation</u><u>:</u></em>
There are seven molecules of H2 and three molecule of O2 are there in the container Q, 6 molecules of H2 will react with 3 molecules of O2 to produce 3 molecules of H2O. One molecule of Hydrogen will not take part in reaction and will be present in container Q after then reaction, and the mass in overall reaction is conserved!
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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
