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egoroff_w [7]
3 years ago
11

when copper sulfide is partially roasted in air (reaction with o2), copper sulfite is formed first. subsequently, upon heating,

the copper sulfite thermally decomposes to copper oxide and sulfur dioxide. Write balanced chemical equations for these two reactions
Chemistry
1 answer:
Lyrx [107]3 years ago
5 0

The question is incomplete, complete question is:

When copper(I) sulfide is partially roasted in air (reaction with oxygen), copper(I) sulfite is formed first. subsequently, upon heating, the copper sulfite thermally decomposes to copper(I) oxide and sulfur dioxide. Write balanced chemical equations for these two reactions.

Answer:

The balanced chemical equations for these two reactions:

2Cu_2S(s)+3O_2(g)\rightarrow 2Cu_2SO_3(s)

Cu_2SO_3(s)\overset{heat}\rightarrow Cu_2O(s)+SO_2(g)

Explanation:

On  partial roasting of copper sulfide in an air. The balanced chemical reaction is given as:

2Cu_2S(s)+3O_2(g)\rightarrow 2Cu_2SO_3(s)

On further heating of copper(I) sulfite it get decomposes into copper oxide and sulfur dioxide. The balanced chemical reaction is given as:

Cu_2SO_3(s)\overset{heat}\rightarrow Cu_2O(s)+SO_2(g)

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Answer:

1429.32 mmHg

Explanation:

Initial Pressure P1 = 3000.0mmHg

Initial Temperature T1 = 500.0°C + 273 = 573 K ( Converting to kelvin temperature)

Final Temperature T2 = 0.00°C + 273 = 273 K ( Converting to kelvin temperature)

Final Pressure P2 = ?

The pressure of a given amount of gas is directly proportional to the absolute temperature provided volume remains constant.

This is given by the mathematical expression;

P1 / T1  = P2 / T2

Inserting the values;

3000 / 573 = P2 / 273

P2 = 273 * 3000 / 573

P2 = 1429.32 mmHg

5 0
3 years ago
Aluminum metal reacts with aqueous cobalt(II) nitrate to form aqueous aluminum nitrate and cobalt metal. What is the stoichiomet
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Answer:

the stoichiometric coefficient for cobalt is 3

Explanation:

the unbalanced reaction would be

Co(NO₃)₂+ Al → Al(NO₃)₃ + Co

One way to solve is to build a system of linear equations for each element (or group as NO₃) , knowing that the number of atoms of each element is conserved.

For smaller reactions a quick way to solve it can be:

- First the Co as product and as reactant needs to have the same stoichiometric coefficient

- Then the Al as product and as reactant needs to have the same stoichiometric coefficient

- After that we look at the nitrates . There are 2 as reactants and 3 as products . Since the common multiple is 6 then multiply the reactant by 3 and the product by 2.

Finally the balanced equation will be

3 Co(NO₃)₂+ 2 Al → 2 Al(NO₃)₃ + 3 Co

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7 0
4 years ago
Electric lights will not come on unless their electrical circuit is a?
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Answer:

closed circuit

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Hope this helps- Good luck! ^w

6 0
3 years ago
A compound made of two elements, iridium (Ir) and oxygen (O), was produced in a lab by heating iridium while exposed to air. The
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1) Compund Ir (x) O(y)

2) Mass of iridium = mass of crucible and iridium - mass of crucible = 39.52 g - 38.26 g = 1.26 g

3) Mass of iridium oxide = mass of crucible and iridium oxide - mass of crucible = 39.73g - 38.26g = 1.47g

4) Mass of oxygen = mass of iridum oxide - mass of iridium = 1.47g - 1.26g = 0.21g

5) Convert grams to moles

moles of iridium = mass of iridium / molar mass of iridium = 1.26 g / 192.17 g/mol = 0.00656 moles

moles of oxygen = mass of oxygen / molar mass of oxygen = 0.21 g / 15.999 g/mol = 0.0131

6) Find the proportion of moles

Divide by the least of the number of moles, i.e. 0.00656

Ir: 0.00656 / 0.00656 = 1

O: 0.0131 / 0.00656 = 2

=> Empirical formula = Ir O2 (where 2 is the superscript for O)

Answer: Ir O2
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