What is the more reactive metal CA or Ra

Lena [83]

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8 0

3 years ago

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A 3.452 g sample containing an unknown amount of a Ce(IV) salt is dissolved in 250.0 mL of 1 M H2SO4. A 25.00 mL aliquot is anal

Olenka [21]

Answer:

The weight percent in the sample is 17,16%

Explanation:

The dissolution of the Ce(IV) salt provides free Ce⁴⁺ that reacts, thus:

2Ce⁴⁺ + 3I⁻ → 2Ce³⁺ + I₃⁻

I₃⁻ + 2S₂O₃²⁻ → 3I⁻ + S₄O₆²⁻

The moles in the end point of S₂O₃⁻ are:

0,01302L×0,03247M Na₂S₂O₃ = 4,228x10⁻⁴ moles of S₂O₃²⁻.

As 2 moles of S₂O₃⁻ react with 1 mole of I₃⁻, the moles of I₃⁻ are:

4,228x10⁻⁴ moles of S₂O₃⁻× $\frac{1molI_{3}^-}{2molS_{2}O_{3}^{2-}}$ = <em>2,114x10⁻⁴ moles of I₃⁻</em>

As 2 moles of Ce⁴⁺ produce 1 mole of I₃⁻, the moles of Ce⁴⁺ are:

2,114x10⁻⁴ moles of I₃⁻× $\frac{2molCe^{4+}}{1molI_{3}^-}$ = <em>4,228x10⁻⁴ moles of Ce(IV)</em>.

These moles are:

4,228x10⁻⁴ moles of Ce(IV)× $\frac{140,116g}{1mol}$ = 0,05924 g of Ce(IV)

As was taken an aliquot of 25,00mL from the solution of 250,0mL:

0,05924 g of Ce(IV)× $\frac{250,0mL}{25,00mL}$ =0,5924g of Ce(IV) in the sample

As the sample has 3,452g, the weight percent is:

0,5924g of Ce(IV) / 3,452g × 100 = <em>17,16 wt%</em>

I hope it helps!

3 0

3 years ago

what separation procedure is most likely to happen? .a. freezing a compound will break it down into elements .b. filtering a mix

SOVA2 [1]

Explanation:

i think its B but i'm not fully sure (Correct)

8 0

3 years ago

12. A voltaic cell consists of a chromium electrode dipped in a 1.20 M chromium (III) nitrate

Lilit [14]

For a voltaic cell consisting of chromium, an electrode dipped in a 1.20 M chromium (III) nitrate solution and a tin electrode dipped in a 0.400 M tin (II) nitrate solution, the cell potential at 298 K is mathematically given as

Ecell = 0.577 V

<h3 /><h3>What is the cell potential at 298 K?</h3>

Generally, the equation for the Oxidation and Reduction is mathematically given as

Cr(s) ------------------ Cr+3(aq) + 3e- ] x 2 ...O

Sn+2(aq) + 2e- ------------ Sn(s) ] x 3 ...R

Reaction

2 Cr(s) + 3 Sn+2(aq) --------------- 2 Cr+3(aq) + 3 Sn(s)

Therefore

Eicell = - 0.14 - ( - 0.74)

Eicell = 0.60

In conclusion

$Ecell= E0cell - \frac{0.0591}{n} * \frac{log[Cr+3]^2}{ [ Sn+2]^3}$

$Ecell = 0.60 - \frac{0.0591 }{6} \frac{log( 1.20)^2}{ ( 0.200)^3}$

Ecell = 0.577 V