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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

SOVA2 [1]

Answer:

1,812 wt%

Explanation:

The reactions for this titration are:

2Ce⁴⁺ + 3I⁻ → 2Ce³⁺ + I₃⁻

I₃⁻ + 2S₂O₃⁻ → 3I⁻ + S₄O₆²⁻

The moles in the end point of S₂O₃⁻ are:

0,01302L×0,03428M Na₂S₂O₃ = 4,463x10⁻⁴ moles of S₂O₃⁻. As 2 moles of S₂O₃⁻ react with 1 mole of I₃⁻, the moles of I₃⁻ are:

4,463x10⁻⁴ moles of S₂O₃⁻× $\frac{1molI_{3}^-}{2molS_{2}O_{3}^-}$ = 2,2315x10⁻⁴ moles of I₃⁻

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

2,2315x10⁻⁴ moles of I₃⁻× $\frac{2molCe^{4+}}{1molI_{3}^-}$ = 4,463x10⁻⁴ moles of Ce(IV). These moles are:

4,463x10⁻⁴ moles of Ce(IV)× $\frac{140,116g}{1mol}$ = <em>0,0625 g of Ce(IV)</em>

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

0,0625g of Ce(IV) / 3,452g × 100 = <em>1,812 wt%</em>

I hope it helps!

5 0

3 years ago

How many moles are in 1.2x10^3 grams of ammonia(NH3) ?

miskamm [114]

Answer : The number of moles present in ammonia is, 70.459 moles.

Solution : Given,

Mass of ammonia = $1.2\times 10^3g$

Molar mass of ammonia = 17.031 g/mole

Formula used :

$\text{Moles of }NH_3=\frac{\text{ given mass of }NH_3}{\text{ molar mass of }NH_3}$

$\text{Moles of }NH_3=\frac{1.2\times 10^3g}{17.031g/mole}=70.459moles$

Therefore, the number of moles present in ammonia is, 70.459 moles.

5 0

3 years ago

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A sample of phosphonitrilic bromide, PNBr2, contains 2.01 mol of the compound. Determine the amount (in mol) of each element pre

nordsb [41]

Answer:

2.01 moles of P → 1.21×10²⁴ atoms

2.01 moles of N → 1.21×10²⁴ atoms

4.02 moles of Br → 2.42×10²⁴ atoms

Explanation:

We begin from this relation:

1 mol of PNBr₂ has 1 mol of P, 1 mol of N and 2 moles of Br

Then 2.01 moles of PNBr₂ will have:

2.01 moles of P

2.01 moles of N

4.02 moles of Br

To determine the number of atoms, we use the relation:

1 mol has NA (6.02×10²³) atoms

Then: 2.01 moles of P will have (2.01 . NA) = 1.21×10²⁴ atoms

2.01 moles of N (2.01 . NA) = 1.21×10²⁴ atoms

4.02 moles of Br (4.02 . NA) = 2.42×10²⁴ atoms

6 0

3 years ago

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Dalton's law of partial pressures states that the total pressure exerted by a mixture of gases is the sum of the pressures exert

Damm [24]

Answer: The given statement is true.

Explanation:

According to the Dalton's law, total pressure of a mixture of gases that do not react with each other is equal to the partial pressure exerted by each gas.

The relationship is as follows.

$p_{total} = \sum_{i=1}^{n} p_{i}$

or, $p_{total} = p_{1} + p_{2} + p_{3} + p_{4} + ......... + p_{n}$

where, $p_{1}, p_{2}, p_{3}$ ....... = partial pressure of individual gases present in the mixture

Also, relation between partial pressure and mole fraction is as follows.

$p_{i} = p_{total} \times x_{i}$

where, $x_{i}$ = mole fraction

Thus, we can conclude that the statement Dalton's law of partial pressures states that the total pressure exerted by a mixture of gases is the sum of the pressures exerted independently by each gas in the mixture, is true.

5 0

3 years ago

Calculate the ratio of the mass ratio of SS to OO in SOSO to the mass ratio of SS to OO in SO2SO2. Consider a sample of SOSO in

Veronika [31]

Answer:

The ratio of the mass ratio of S to O; in SO, to the mass ratio of S to O; in SO₂, is 2:1

Explanation:

According to the consideration, let us first find the ratio of S and O in both the compounds

For SO:

$\frac{m_{S} }{m_{O} }= \frac{32}{16}\\\\ \frac{m_{S} }{m_{O} }= 2$