What fraction of the formula weight of each
of the ff compounds represents the equivalent
weight in a redox process in which the product
formed is as indicated:
(a) Ce(SO4)2.2(NH4)2SO4.2H2O (->Ce3+)
(b) As2O5 (-> As3+)
(c) KIO3 (-> ICl32-)
(d) Na2SeO4 (-> SeO32-)
(e) VOSO4 (-> VO3-)
(f) Mo2O3 (-> H2MoO4


KMnO4:KCl = 1:1

HCl:H2O = 2:1

KMnO4:Cl2 = 2:5

HCl:MnCl2 = 8:1


It's pretty straightforward; the ratio of each substance pair is derived from their coefficients in a balanced chemical equation. Just don't forget to make them into their lowest terms.


How many mL of a 12.5% (by volume) solution of ethanol are required to produce 175 mL of 0.664 M acetic acid? (density of pure ethanol = 0.789 g/mL)

C_2H_5OH(aq) + O_2(g) \Longrightarrow HC_2H_3O_2(aq) + H_2O

We have the following data:

% m1/m (percent by solute mass/solution mass) = 12.5

Molar Mass (MM) of ethanol (C2H5OH) ≈ 46.07 g/mol

M1 (molarity) = 0.664 M (or 0.664 mol/L)

V1 (initial volume) = 175 ml = 0.175 L

n (number of mol) = ?

m1 (solute mass - ethanol) = ?

m (solution mass) = ?

d (density of pure ethanol) = 0.789 g/ml

v (volume of pure ethanol) = ? (in ml)


[First Step] Let's find the number of moles or matter number, let's see:


n_1 = n_2

M_1*V_1 = M_2*V_2

M_1*V_1 = n_2

n_2 = M_1*V_1

n_2 = 0.664\:\dfrac{mol}{\diagup\!\!\!\!L} * 0.175\:\diagup\!\!\!\!L

\boxed{n_2 = 0.1162\:mol}

[Second Step]  We apply the data to the Molarity or Molar Concentration formula and find the mass of the solute (m1), let's see:

M_1 = \dfrac{m_1}{MM*V_1}

m_1 = M_1*V_1*MM

m_1 = n_2*MM

m_1 = 0.1162\:mol\!\!\!\!\!\!\!\!\!\!\!\dfrac{\hspace{0.6cm}}{~}*46.07\:\dfrac{g}{mol\!\!\!\!\!\!\!\!\!\!\!\dfrac{\hspace{0.6cm}}{~}}

m_1 = 5.353334\:g\to \boxed{m_1 \approx 5.36\:g}

[Third Step] We apply the data to the formula of percent solute mass per solution mass and let's find the solution mass, let's see:

\%\:m_1/m = \dfrac{m_1}{m}*100

12.5 = \dfrac{5.36}{m}*100

12.5*m = 5.36*100

12.5\:m = 536

m = \dfrac{536}{12.5}

\boxed{m = 42.88\:g}

[Fourth Step] We apply the data to the density formula and let's find the volume of ethanol, let's see:

d = \dfrac{m}{v}

v = \dfrac{m}{d}

v = \dfrac{42.88\:\diagup\!\!\!\!\!g}{0.789\:\dfrac{\diagup\!\!\!\!\!g}{ml} }

\boxed{\boxed{v \approx 54.3\:ml}}\:\:\:\:\:\:\bf\blue{\checkmark}

The volume is approximately 54.3 ml


\bf\red{I\:Hope\:this\:helps,\:greetings ...\:Dexteright02!}\:\:\ddot{\smile}


Bake it at 350°F (175°C) for 50 minutes, until golden brown. Remove the pie from the oven, let it cool for a hot minute and then remove the tin foil and weights. Put the pie back into the oven for 10 minutes. Bake all pies in a rack fitted to the middle of your oven and at 350°F (175°C), always.

The volume of unmixed sugar and water is less than the volume of mixture of sugar and water

Do you know the answer?

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