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Tema [17]
3 years ago
8

Barium and lead both produce yellow precipitate with the chromate ion as part of their confirmation tests. if you had a sample t

hat possibly contained lead, barium, or both ions, outline a procedure for correctly identifying the cations present.
Chemistry
2 answers:
ehidna [41]3 years ago
8 0

Explanation:

Lead belongs to group 1 cations and barium belongs to the fifth group cations.

If the sample, contains lead then it can be precipitated by using dilute HCl as lead chloride. Barium, if present will not precipitate.

Confirmatory test for lead:

Dissolve the precipitate in hot water and centrifugate and divide the filtrate in 3 parts as lead chloride comes in the solution because it is soluble in hot water.

  • To the filtrate 1, add dilute sulfuric acid. The appearance of white precipitate confirms lead.
  • To the filtrate 2, add 2-3 mL of potassium chromate solution. The appearance of yellow precipitate confirms lead.
  • To the filtrate 3, add 2-3 mL of potassium iodide solution. The appearance of yellow precipitate confirms lead.

If barium is also present, it can be precipitated by using ammonium carbonate and can be confirmed by the addition of sulphurica cid which gives white precipitate and also addition of the potassium chromate solution. The appearance of yellow precipitate confirms barium. Also, barium imparts green color to the flame.

UkoKoshka [18]3 years ago
5 0
Barium is an element that is primarily used in fireworks because of its distinct green colour while on the other hand, lead is used in electrical industries due to its unique properties. The ions of these elements can simply be identified when its insoluble salts get precipitated by any process.
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D

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A substance that has high reactivity
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Analysis of an athletes urine found the presence of a compound with a molar mass of 312 g/mol. How many moles of this compound a
rewona [7]
<h3>Answer:</h3>

= 5.79 × 10^19 molecules

<h3>Explanation:</h3>

The molar mass of the compound is 312 g/mol

Mass of the compound is 30.0 mg equivalent to 0.030 g (1 g = 1000 mg)

We are required to calculate the number of molecules present

We will use the following steps;

<h3>Step 1: Calculate the number of moles of the compound </h3>

Moles=\frac{mass}{molar mass}

Therefore;

Moles of the compound will be;

=\frac{0.030}{312g/mol}

      = 9.615 × 10⁻5 mole

<h3>Step 2: Calculate the number of molecules present </h3>

Using the Avogadro's constant, 6.022 × 10^23

1 mole of a compound contains 6.022 × 10^23  molecules

Therefore;

9.615 × 10⁻5 moles of the compound will have ;

= 9.615 × 10⁻5 moles × 6.022 × 10^23  molecules

= 5.79 × 10^19 molecules

Therefore the compound contains 5.79 × 10^19 molecules

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3 years ago
The masses of carbon and hydrogen in samples of four pure hydrocarbons are given above. The hydrocarbon in which sample has the
enot [183]

Answer:

Sample B

Explanation:

In this case, we need to determine the empirical formula for each sample. The one that match the formula of the propene would be the sample.

Let's do Sample A:

C: 60 g;       H: 12 g

1. Calculate moles:

We need the atomic weights of carbon (12 g/mol) and hydrogen (1 g/mol):

C: 60 / 12 = 5

H: 12 / 1 = 12

2. Determine number of atoms in the formula

In this case, we just divide the lowest moles obtained in the previous part, by all the moles:

C: 5 / 5 = 1

H: 12 / 5 = 2.4    or rounded to two

3. Write the empirical formula:

Now, the prior results, represent the number of atoms in the empirical formula for each element, so, we put them with the symbol and the atoms as subscripted:

C₁H₂ = CH₂

Therefore, sample A is not the same as propene.

Sample B:

C: 72 g    H: 12 g

Following the same steps, let's determine the empirical formula for this sample

C: 72 / 12 = 6 ---> 6 / 6 = 1

H: 12 / 1 = 12 ----> 12 / 6 = 2

EF: CH₂

Sample C:

C: 84 g    H: 10 g

C: 84 / 12 = 7 ----> 7 / 7 = 1

H: 10 / 1 = 10    ----> 10 / 7 = 1.4 or just 1

EF: CH

Sample D

C: 90 g      H: 10 g

C: 90 / 12 = 7.5     -----> 7.5 / 7.5 = 1

H: 10 / 1 = 10  -------> 10 / 7.5 = 1.33 or just 1

EF: CH

Neither compound has the same empirical formula as C3H6, but C3H6 is a molecular formula, so, if we just simplify the formula we have:

C3H6  -----> CH₂

Therefore, sample B is the one that match completely. Sample B would be the one.

Hope this helps

8 0
2 years ago
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