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musickatia [10]
2 years ago
9

How to make Sodium thiosulfate

Chemistry
1 answer:
Arada [10]2 years ago
3 0

Explanation:

Method of prepration of sodium thiosulphate - definition

In the laboratory, this salt can be prepared by heating an aqueous solution of sodium sulphite with sulphur or by boiling aqueous NaOH and sulfur according to this equation:

6NaOH + 4 S _{2} Na _{2} S ->Na _{2} S_{2}O_{3}+ 3H _{2}O.

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What was the control in Pasteur's experiment?
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Answer:

a straight neck flask to allow air to get in

Explanation:

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3 years ago
Sodium metal reacts with water to produce hydrogen gas according to the following equation: 2Na(s) + 2H2O(l)2NaOH(aq) + H2(g) Th
Maru [420]

Answer:

Number of moles of sodium reacted = 0.707 moles

Explanation:

P(H₂) = P(T) – P(H₂O)

P(H₂) = 754 – 17.5 = 736.5 mm Hg

Use the ideal gas equation which

PV= nRT, where P is the pressure V is the volume, n is the number of moles R is the Gas Constant and T is temperature

<u>Re- arrange to calculate the number of moles and using the data provided</u>

n = P x V/R x T

n =736.5 x 8.77/62.36367 x (mmHg/mol K) x (20 + 273)

n = 0.35348668

n = 0.353 moles H₂

<u>from the equation we know that</u>

0.353 mole H₂ x 2mole Na/1mole H₂, So

0.353 x 2 = 0.707 mole Na

The number of moles of Sodium metal reacted were 0.707 moles.

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3 years ago
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Order the elements from smallest to largest atomic number
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the largest is francium, Z=87 , or caesium, Z=55 ; and the smallest is definitely helium, Z=2 .

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3 years ago
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
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