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SIZIF [17.4K]
3 years ago
5

Atoms in 3.57 g carbon?

Chemistry
1 answer:
N76 [4]3 years ago
8 0

Answer:

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Explanation:

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Explain the effects of nh3 and hcl on the cuso4 solution in terms of le chatelier's principle
Fittoniya [83]

The Principle of Le Chatelier states that if a system in equilibrium is subjected to a disturbance, the system will react in such a way that it will diminish the effect of that disturbance. Thus, when the concentration of one of the substances in an equilibrium system is changed, the equilibrium varies in such a way that it can compensate for this change.

For example, if the concentration of one of the reactants is increased, the equilibrium shifts to the right or to the side of the products. Also, if you add more reagents, the reaction will move even more to the right until the balance is re-established again, increasing the quantity of products.

In this way, adding HCl to a solution of CuSO4 will produce the following reaction:

CuSO4 (aq) + 2HCl (aq) ⇔ CuCl2 (aq) + H2SO4 (aq)

Initially the solution of CuSO4 in water will be blue, but when adding HCl the solution will change color to green, since the aqueous solutions of CuCl2 are green. By adding more HCl this color will intensify as the balance shifts to the right, producing more CuCl2 and H2SO4.

On the other hand, adding NH3 to a solution of CuSO4 will produce the following reaction

CuSO4 (aq) + 4NH3 (aq) ⇔ [Cu(NH3)4] SO4 (s)

Thus, by adding NH3 to the CuSO4 solution we will observe the formation of a precipitate corresponding to [Cu(NH3) 4] SO4. <u>When adding more NH3, the formation of more precipitate will be observed as the equilibrium moves to the right, producing a greater quantity of [Cu (NH3) 4] SO4.</u>

6 0
4 years ago
Separate this redox reaction into its balanced component half-reactions. What is the oxidation and reduction half reactions
erma4kov [3.2K]
<span>Separate this redox reaction into its component half-reactions. 
Cl2 + 2Na ----> 2NaCl 

reduction: Cl2 + 2 e- ----> 2Cl-1 
oxidation: 2Na ----> 2Na+ & 2 e- 


2) Write a balanced overall reaction from these unbalanced half-reactions: 

oxidation: Sn ----> Sn^2+ & 2 e- 
reduction: 2Ag^+ & 2e- ----> 2Ag 

giving us 
2Ag^+ & Sn ----> Sn^2+ & 2Ag </span>Steve O <span>· 5 years ago </span><span>
</span>
3 0
4 years ago
Read 2 more answers
Two moles of MgsO4•7H2O contain ___ grams of MgSO4•7H2O
gogolik [260]

Answer:

492.6g

Explanation:

Given parameters:

Number of moles of MgSO₄.7H₂O = 2moles

Unknown:

Mass = ?

Solution:

To find the mass of this compound;

       Mass  = number of moles x molar mass

Molar mass of MgSO₄.7H₂O = 24.3 + 32 + 4(16) + 7[2(1) + 16]

                                                = 246.3g/mol

Mass  = 2 x 246.3 = 492.6g

5 0
3 years ago
Will mark the BEST answer as BRANLIEST!
BigorU [14]

Answer:

1. 2Fe + 3CuSO₄ →  Fe₂(SO₄)₃ + 3Cu.

2. Pb(NO₃)₂+ 2Kl →  PbI₂ + 2KNO₃.

3. Mg + 2HCl →  MgCl₂ + H₂.

4. H₂O →  H₂ + 1/2O₂.

5. 2Mg + O₂ → 2MgO.

<h3 /><h2>explanation: 1. Combine iron and copper (II) sulfate solution. (Hint: Iron will form the iron (III) ion)</h2>

2Fe + 3CuSO₄ →  Fe₂(SO₄)₃ + 3Cu.

It is a redox reaction including replacing Cu with Fe and changing their oxidation states.

That Fe replaces Cu and resulting in ferric sulfate.

2. Combine lead (II) nitrate and potassium iodide solutions.

Pb(NO₃)₂+ 2Kl →  PbI₂ + 2KNO₃.

It is a double replacement reaction that lead nitrate reacts with potassium iodide resulting in lead iodide and potassium nitrate.

3. Combine magnesium metal and hydrochloric acid solution.

Mg + 2HCl →  MgCl₂ + H₂.

It is a dissolution reaction that HCl dissolve Mg and resulting in Magnesium chloride and hydrogen gas is evolved.

4. Electrolysis (splitting) of water.  

H₂O →  H₂ + 1/2O₂.

Water electrolysis resulting in splitting of water to produce hydrogen and water.

5. Burning magnesium.

2Mg + O₂ → 2MgO.

It is a combustion reaction that Mg is burned with oxygen to produce magnesium oxide.

Explanation:1. Combine iron and copper (II) sulfate solution. (Hint: Iron will form the iron (III) ion)

2Fe + 3CuSO₄ →  Fe₂(SO₄)₃ + 3Cu.

It is a redox reaction including replacing Cu with Fe and changing their oxidation states.

That Fe replaces Cu and resulting in ferric sulfate.

2. Combine lead (II) nitrate and potassium iodide solutions.

Pb(NO₃)₂+ 2Kl →  PbI₂ + 2KNO₃.

It is a double replacement reaction that lead nitrate reacts with potassium iodide resulting in lead iodide and potassium nitrate.

3. Combine magnesium metal and hydrochloric acid solution.

Mg + 2HCl →  MgCl₂ + H₂.

It is a dissolution reaction that HCl dissolve Mg and resulting in Magnesium chloride and hydrogen gas is evolved.

4. Electrolysis (splitting) of water.  

H₂O →  H₂ + 1/2O₂.

Water electrolysis resulting in splitting of water to produce hydrogen and water.

5. Burning magnesium.

2Mg + O₂ → 2MgO.

It is a combustion reaction that Mg is burned with oxygen to produce magnesium oxide.

3 0
4 years ago
The effort is always greater than the load in a third class lever.
Lubov Fominskaja [6]
The third class lever s<span>have </span>the effort<span> placed amongst  </span>load<span> and the fulcrum.</span>
7 0
4 years ago
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