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ra1l [238]
3 years ago
9

Colorful fireworks often involve the decomposition of barium nitrate and potassium chlorate and the reaction of the metals magne

sium ,alminium and iron with oxygen.
(a)write the formulas of barium nitrate and potassium chlorate
(b)The decomposition of solid potassium chlorateleads to the formaion of solid potassium chloride and diatomic oxygen gas write an equation for the reaction
(c) The decomposition of the solid barium nitrate leads to the formation of solid barium oxide, nitrogen gas and diatomic oxygen gas
(d)write separate equations for the reactions of the solid Metals in magnesium aluminum with a diatomic oxygen gas to yield the corresponding metal oxide (Assume iron oxide contains Fe+ ions)
Chemistry
1 answer:
Fynjy0 [20]3 years ago
5 0

Fireworks owe their colors to reactions of combustion of the metals present. When Mg and Al burn, they emit a white bright light, whereas iron emits a gold light. Besides metals, oxygen is necesary for the combustion. The decomposition reactions of barium nitrate and potassium chlorate provide this element. At the same time, barium can burn emitting a green light.

(a) Barium nitrate is a <em>salt</em> formed by the <em>cation</em> barium Ba²⁺ and the <em>anion</em> nitrate NO₃⁻. Its formula is Ba(NO₃)₂. Potassium chlorate is a <em>salt</em> formed by the <em>cation</em> potassium K⁺ and the <em>anion</em> chlorate ClO₃⁻. Its formula is KClO₃.

(b) The balanced equation for the decomposition of potassium chloride is:

2KClO₃(s) ⇄ 2KCl(s) + 3O₂(g)

(c)  The balanced equation for the decomposition of barium nitrate is:

Ba(NO₃)₂(s) ⇄ BaO(s) + N₂(g) + 3O₂(g)

(d) The balanced equations of metals with oxygen to form metal oxides are:

  • 2 Mg(s) + O₂(g) ⇄ 2 MgO(s)
  • 4 Al(s) + 3 O₂(g) ⇄ 2 Al₂O₃(s)
  • 4 Fe(s) + 3 O₂(g) ⇄ 2 Fe₂O₃(s)

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NeX [460]
Compared to the charge and mass of a proton an electron has......

A proton has approximately the same mass as..........
5 0
3 years ago
How many grams of glucose are needed to prepare 144.3 mL of a 1.4%(m/v) glucose solution?
Montano1993 [528]

Answer:

2.0202 grams

Explanation:

1.4% (m/v) glucose solution means: 1.4g glucose/100mL solution.

so ?g glucose = 144.3 mL soln

Now apply the conversion factor, and you have:

?g glucose = 144.3mL soln x (1.4g glucose/100mL soln).

so you have (144.3x1.4/100) g glucose= 2.0202 grams

6 0
3 years ago
Read 2 more answers
Round the following numbers to the number of significant figures indicated:
Kruka [31]
  1. 3.5
  2. 9
  3. 24.7
  4. 8.01
  5. 0.006
  6. 13(1.25×1012=12.65)

Explanation:

Above we have simply looked the condition that where we put decimal point that gives your needed answer after round off.

For example:

we have to make 2 s.f from 345 so we kept decimal after 3 and we rounded off to make 3.5 from 3.45

Hope you got it

7 0
3 years ago
When 40.5 g of Al and 212.7 g of Cl2 combine in the reaction:
slamgirl [31]

Answer:

1.5 mole

Explanation:

Step 1:

The balanced equation for the reaction. This is illustrated below:

2Al(s) + 3Cl2(g) --> 2AlCl3(s)

Step 2:

Determination of the masses of Al and Cl2 that reacted from the balanced equation. This is illustrated below:

Molar Mass of Al = 27g/mol

Mass of Al from the balanced equation = 2 x 27 = 54g

Molar Mass of Cl2 = 2 x 35.5 = 71g/mol

Mass of Cl2 from the balanced equation = 3 x 71 = 213g

From the balanced equation,

54g of Al reacted.

213g of Cl2 reacted

Step 3:

Determination of the limiting reactant.

This is illustrated below:

From the balanced equation above,

54g of Al reacted with 213g of Cl2.

Therefore, 40.5g of Al will react with = (40.5 x 213)/54 = 159.75g of Cl2.

From the calculations made above, there are leftover of Cl2 as 159.75g reacted out of 212.7g. Therefore, Cl2 is the excess reactant and Al is the limiting reactant.

Step 4:

Determination of the number of mole in 40.5g of Al. This is illustrated below:

Molar Mass of Al = 27g/mol

Mass of Al = 40.5g

Number of mole of Al =?

Number of mole = Mass/Molar Mass

Number of mole of Al = 40.5/27

Number of mole of Al = 1.5 mole

Step 5:

Determination of the number of mole of AlCl3 produced When 40.5 g of Al and 212.7 g of Cl2 combine together. This is illustrated below:

2Al(s) + 3Cl2(g) --> 2AlCl3(s)

From the balanced equation above,

2 moles of Al produced 2 moles of AlCl3.

Therefore, 1.5 mole of Al will also produce 1.5 mole of AlCl3.

From the calculations made above, 1.5 mole of AlCl3 is produced When 40.5 g of Al and 212.7 g of Cl2 combine together.

8 0
3 years ago
Read 2 more answers
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Irina18 [472]
B. Na2O



Explanation:
8 0
3 years ago
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