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lozanna [386]
3 years ago
6

E. solid copper sulfide and silver nitrate react to form copper (ii) nitrate and solid silver sulfide. write a balanced chemical

equation that describes the reaction. identify the oxidation number of each element in the reaction. (you do not need to include the total contribution of charge.) is this reaction a redox reaction or a non-redox reaction? explain your answer.
Chemistry
2 answers:
Rudiy273 years ago
5 0
It is reaction of ion exchange. Substances react in the melt. As a result we receive alloy of four salts.

CuS + 2AgNO₃ ⇄ Cu(NO₃)₂ + Ag₂S
andreyandreev [35.5K]3 years ago
4 0

Answer:

The given reaction is not a redox reaction.

Explanation:

Redox reaction is defined as chemical reaction in oxidation and reduction reactions occurs simultaneously.

Oxidation reaction is defined as the reaction in which an atom looses its electrons. Here, oxidation state of the atom increases.

Reduction reaction is defined as the reaction in which an atom gains electrons. Here, oxidation state of the atom decreases.

Solid copper sulfide when immersed in silver nitrate solution it reacts to form silver sulfide precipitate and copper(II) nitrate solution.

CuS(s)+2AgNO_3(aq)\rightarrow Cu(NO_3)_2(aq)+Ag_2S(s)

Oxidation states of elements on reactant side:

Oxidation state of copper  = +2

Oxidation state of sulfur = -2

Oxidation number of silver  +1

Oxidation number of nitrogen = +5

Oxidation number of oxygen = -2

Oxidation states of elements on product side:

Oxidation state of copper  = +2

Oxidation state of sulfur = -2

Oxidation number of silver  +1

Oxidation number of nitrogen = +5

Oxidation number of oxygen = -2

No change in oxidation states of elements is observed which means that the given reaction is not a redox reaction.

You might be interested in
If 5.85 grams of cobalt metal react with 15.8 grams of silver nitrate, how many grams of silver metal can be formed and how many
vladimir2022 [97]
Answers:
<span>Answer 1: 10.03 g of siver metal can be formed.</span>
Answer 2: 3.11 g of Co are left over.

Work:

1) Unbalanced chemical equation (given):

<span>Co + AgNO3 → Co(NO3)2 + Ag

2) Balanced chemical equation
</span>
<span>Co + 2AgNO3 → Co(NO3)2 + 2Ag

3) mole ratios

1 mol Co : 2 mole AgNO3 : 1 mol Co(NO3)2 : 2 mol Ag

4) Convert the masses in grams of the reactants into number of moles

4.1) 5.85 grams of Co

# moles = mass in grams / atomic mass

atomic mass of Co = 58.933 g/mol

# moles Co = 5.85 g / 58.933 g/mol = 0.0993 mol

4.2) 15.8 grams of Ag(NO3)

# moles Ag(NO3) = mass in grams / molar mass

molar mass AgNO3 = 169.87 g/mol

# moles Ag(NO3) = 15.8 g / 169.87 g/mol = 0.0930 mol

5) Limiting reactant

Given the mole ratio 1 mol Co : 2 mol Ag(NO3) you can conclude that there is not enough Ag(NO3) to make all the Co react.

That means that Ag(NO3) is the limiting reactant, which means that it will be consumed completely, whilce Co is the excess reactant.

6) Product formed.

Use this proportion:

2 mol Ag(NO3)           0.0930mol Ag(NO3)    
--------------------- =      ---------------------------
    2 mol Ag                              x

=> x = 0.0930 mol

Convert 0.0930 mol Ag to grams:

mass Ag = # moles * atomic mass = 0.0930 mol * 107.868 g/mol = 10.03 g

Answer 1: 10.03 g of siver metal can be formed.

6) Excess reactant left over

    1 mol Co                             x
----------------------- =  ----------------------------
2 mole Ag(NO3)       0.0930 mol Ag(NO3)

=> x = 0.0930 / 2 mol Co = 0.0465 mol Co reacted

Excess = 0.0993 mol - 0.0465 mol = 0.0528 mol

Convert to grams:

0.0528 mol * 58.933 g/mol = 3.11 g

Answer 2: 3.11 g of Co are left over.
</span>


8 0
3 years ago
with strong heating calcium carbonate undergoes thermal decomposition how many mole of CaCO3 are there in 50g of calcium carbona
aev [14]
Ca=40
C=12
O=16
1 mole of CaCO3 has 100 grams
So 50 grams is 0.5 mole
6 0
2 years ago
Read 2 more answers
Ammonia, NH3 is a common base with Kb of 1.8 X 10-5. For a solution of 0.150 M NH3:
Vesnalui [34]

The concentrations : 0.15 M

pH=11.21

<h3>Further explanation</h3>

The ionization of ammonia in water :

NH₃+H₂O⇒NH₄OH

NH₃+H₂O⇒NH₄⁺ + OH⁻

The concentrations of all species present in the solution = 0.15 M

Kb=1.8 x 10⁻⁵

M=0.15

\tt [OH^-]=\sqrt{Kb.M}\\\\(OH^-]=\sqrt{1.8\times 10^{-5}\times 0.15}\\\\(OH^-]=\sqrt{2.7\times 10^{-6}}=1.64\times 10^{-3}

\tt pOH=-log[OH^-]\\\\pOH=3-log~1.64=2.79\\\\pH=14-2.79=11.21

8 0
3 years ago
Ill give the brainliest answer to whoever helps me with this equation
vampirchik [111]

Answer: The percent yield for the NaBr is, 86.7 %

Explanation : Given,

Moles of FeBr_3 = 2.36 mol

Moles of NaBr = 6.14 mol

First we have to calculate the moles of NaBr

The balanced chemical equation is:

2FeBr_3+3Na_2S\rightarrow Fe_2S_3+6NaBr

From the reaction, we conclude that

As, 2 moles of FeBr_3 react to give 6 moles of NaBr

So, 2.36 moles of FeBr_3 react to give \frac{6}{2}\times 2.36=7.08 mole of NaBr

Now we have to calculate the percent yield for the NaBr.

\text{Percent yield}=\frac{\text{Experimental yield}}{\text{Theoretical yield}}\times 100

Experimental yield = 6.14 moles

Theoretical yield = 7.08 moles

Now put all the given values in this formula, we get:

\text{Percent yield}=\frac{6.14mol}{7.08mol}\times 100=86.7\%

Therefore, the percent yield for the NaBr is, 86.7 %

6 0
3 years ago
Why do South Pole and North Pole and South Pole attract?
vfiekz [6]
They have a magnetic attraction. 
6 0
3 years ago
Read 2 more answers
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