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Alenkinab [10]
3 years ago
10

Consider the following equilibrium. 2 SO2 (g) + O2 (g) 2 SO3 (g) The equilibrium cannot be established when ________ is/are plac

ed in a 1.0-L container.
Chemistry
1 answer:
Dmitry_Shevchenko [17]3 years ago
7 0

Answer:

Question is incomplete, the complete question is pasted below.

Consider the following equilibrium. 2SO2(g) + O2(g) ⇌2SO3(g) The equilibrium cannot be established when ________ is/are placed in a 1.0-L container?

A) 0.50 mol O2 (g) and 0.50 mol SO3 (g)

B) 1.0 mol SO3 (g)

C) 0.25 mol SO2 (g) and 0.25 mol O2 (g)

D) 0.75 mol SO2 (g)

E) 0.25 mol of SO2 (g) and 0.25 mol of SO3 (g)

Answer:

The correct answer is : D) 0.75 mol SO2 (g)

Explanation:

Option A is not true because both SO3 can be converted into SO2 and O2 and equilibrium can be achieved.

Option B is not true because SO3 will be decomposed into SO2 and O2 and can achieve equilibrium.

Option C is not true because both SO2 and O2 will react and make SO3 and can achieve equilibrium.

Option D is correct because SO2 requires O2 to react and form SO3. Since O2 is not present, thus reaction will not occur and equilibrium will not be achieved.

Option E is not true because SO3 can decompose into O2 and SO2 and can achieve equilibrium.

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3 years ago
How many ml of 12.0 M H2S04 are needed to make 500.0 ml of a 1.00 M solution? What happens to the freezing point and the boiling
lora16 [44]

Answer:

41.66 mL of 12.0 M sulfuric acid are needed.

Explanation:

Concentration of sulfuric acid solution taken =M_1=12.0M

Volume of the 12.0 M Solution = V_1

Concentration of required solution = M_2=1.00M

Volume of required 1.00 M solution = V_2=500.0 mL

M_1V_1=M_2V_2 (Dilution)

V_1=\frac{1.00 M\times 500.0 mL}{12.0 M}=41.66 mL

41.66 mL of 12.0 M sulfuric acid are needed.

The freezing point and the boiling point of a solvent when a non-volatile solute is dissolved in it decrease and increase respectively.

3 0
4 years ago
How many electrons are in the π system of the ozone molecule, o3?
algol13
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7 0
3 years ago
The atomic symbol superscript 40 subscript 19 upper k. represents potassium-40 (k-40), a radioactive isotope that has 19 protons
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Answer:

40/18AR

Explanation:

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8 0
2 years ago
Complete and balance the molecular equation for the reaction of aqueous copper(II) chloride, CuCl2, and aqueous potassium phosph
aleksandrvk [35]

<u>Answer:</u> The net ionic equation is written below.

<u>Explanation:</u>

Net ionic equation of any reaction does not include any spectator ions.

Spectator ions are defined as the ions which do not get involved in the chemical equation. It is also defined as the ions which are found on both the sides of the chemical reaction when it is present in ionic form.

The chemical equation for the reaction of copper (II) chloride and potassium phosphate is given as:

3CuCl_2(aq.)+2K_3PO_4(aq.)\rightarrow Cu_3(PO_4)_2(s)+6KCl(aq.)

Ionic form of the above equation follows:

3Cu^{2+}(aq.)+6Cl^-(aq.)+6K^+(aq.)+2PO_4^{3-}(aq.)\rightarrow Cu_3(PO_4)_2(s)+6K^+(aq.)+6Cl^-(aq.)

As, potassium and chloride ions are present on both the sides of the reaction, thus, they will not be present in the net ionic equation.

The net ionic equation for the above reaction follows:

3Cu^{2+}(aq.)+2PO_4^{3-}(aq.)\rightarrow Cu_3(PO_4)_2(s)

Hence, the net ionic equation is given above.

7 0
3 years ago
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