what is the chemical reaction and mole ratio between oxalate and permanganate in the titration reaction

1 answer:

Tom [10]6 months ago

6 0

5C2O4^(2-)(aq) + 2MnO4^-(aq) + 16H+(aq) → 10CO2(g) + 2Mn2+(aq) + 8H2O(l)` is the chemical reaction and mole ratio between oxalate and permanganate in the titration reaction.

A chemical reaction is a procedure that causes one group of chemical components to change chemically into another. Chemical reactions, which can frequently be described by a chemical equation, traditionally include changes that only affect the locations of electrons in the formation and dissolution of chemical bonds between atoms, with no change to the nuclei (no change to the elements present). The study of chemical processes involving unstable and radioactive elements, where both electronic and nuclear changes may take place, is known as nuclear chemistry.

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There are 342 g of sucrose in 1.00 mol of sucrose. What is the molar concentration (molarity) of a solution containing 171 g suc

AVprozaik [17]

For the answer to the question above, use these formulas in solving your problem and as a guide.

MM = 342 (g/mol) 

171 (g) / 342(g/mol) = x mol of sucrose 
x moles of sucrose/ 1.25 L = Molarity of soultion

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7 0

2 years ago

Calculate the following quantity: molarity of a solution prepared by diluting 45.45 mL of 0.0404 M ammonium sulfate to 550.00 mL

dybincka [34]

Answer:

$M_2=3.34x10^{-3}M$

Explanation:

Hello!

In this case, since a dilution process implies that the moles of the solute remain the same before and after the addition of diluting water, we can write:

$M_1V_1=M_2V_2$

Thus, since we know the volume and concentration of the initial sample, we compute the resulting concentration as shown below:

$M_2=\frac{M_2V_2}{V_1} =\frac{45.45mL*0.0404M}{550.00mL}\\\\M_2=3.34x10^{-3}M$

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5 0

2 years ago

The mass percent of solute in a solution containing 3.73 g KBr dissolved in 131 g of H2O is:

Elina [12.6K]

$\huge \underbrace \mathfrak \red{Answer}$

28%

Explanation:

mass of solute(KBr) = 3.73g

mass of solvent(H2O) = 131g

mass of solution = mass of solute + mass of solvent

= 3.73 + 131

= 134.73g

$\sf \large {mass \: percentage = \frac{mass \: of \: solute}{mass \: of \: solvent} \times 100} \\ \\ \sf mass \: percentage = \frac{3.73}{134.73} \times 100 \\ \\ \sf mass \: percentage = 0.028 \times 100 \\ \\ \sf mass \: percentage = 28\%$