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Ivanshal [37]
2 years ago
14

A student wants to carry out flame tests on these three chemicals. Describe how to carry out a flame test.

Chemistry
1 answer:
neonofarm [45]2 years ago
3 0

Answer:

Flame test is one of the chemical tests that can be used to identify a metal or metaloid

in an ionic compound.

Explanation:

Flame test is guided by a principle that when an unknown metal is heated, heat of the flame converts the metal ions into atoms which become excited and emit visible light.

To carry out flame test in a laboratory the following instruments are required:

--> a wire loop

--> Bunsen burner

--> the chemical compound under observation

--> protective wears.

The procedure involves the following steps:

--> turn on the Bunsen burner and adjust the barrel of the burner roll the flame is blue.

--> using a clean wire loop deep into a beaker containing hydrochloric acid.

--> to ensure that the loop is clean, it will give the same blue colour of the flame when heated.

--> deep the loop in the acid again and then use it to pick few grains of the metal.

--> place the wire loop in the side of the flame and observe the colour of the flame.

Result of flame tests of some metallic ion will show the following:

--> sodium ion will give yellow colour flame

--> potassium ion will give lilac colour flame

--> calcium ion will give orange-red colour flame

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Answer:

Answer: 3.01 * 10^35

Explanation:

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Explain why nitrogen diffuses faster than clorine
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Please help me this is my fourth attempt.
Vaselesa [24]

Explanation:

CH4 + 4S ---> CS2 + 2H2S

4) 0.75 mol S × (1 mol CS2/4 mol S) = 0.19 mol CS2

5) 3 mol H2S × (1 mol CH4/2 mol H2S) = 1.5 mol CH4

Fe2O3 + 2Al ---> 2Fe + Al2O3

6) 25 g FeO3 × (1 mol Fe2O3/159.69 g Fe2O3) = 0.16 mol Fe2O3

0.16 mol Fe2O3 × (2 mol Al/1 mol Fe2O3) = 0.32 mol Al

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7) Given:

45 g Al × (1 mol Al/26.98 g Al) = 1.6 mol Al

85 g Fe2O3 ×(1 mol Fe2O3/159.69 g Fe2O3)

= 0.53 mol Fe2O3

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1.6 mol Al × (2 mol Fe/2 mol Al) = 1.6 mol Fe

0.53 mol Fe2O3 × (2 mol Fe/1 mol Fe2O3) = 1.1 mol Fe

Note that the given amount of Fe2O3 will give us fewer Fe. Therefore, Fe2O3 is the limiting reactant.

8) Al will produce 1.6 mol Fe × (55.845 g Fe/1 mol Fe)

= 89 g Fe

Fe2O3 will produce 1.1 mol Fe × (55.845 Fr/1 mol Fe)

= 61 g Fe

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%yield = (25 g Fe/61 g Fe) × 100% = 41%

10) If we only got 25 g Fe, then the amount of Al actually used in the reaction is

25 g Fe × (1 mol Fe/55.845 g Fe) = 0.45 mol Fe

0.45 mol Fe × (2 mol Al/2 mol Fe) = 0.45 mol Al

0.45 mol Al × (26.98 g Al/1 mol Al) = 12 g

Therefore, the leftover amount of Al is

25 g Al - 12 g Al = 13 g Al

8 0
2 years ago
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amm1812

Answer:

8moles

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We are given the mass of ammonia to be 136g

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so;

   Number of moles  = \frac{136}{17}   = 8moles

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