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In-s [12.5K]
3 years ago
14

The overall reaction of ozone reacting to form oxygen has been proposed to occur in a reaction mechanism of: What is the role of

O(g) and what is the overall balanced equation? Question 14 options: 1) O(g) is an intermediate; 2O3(g) + O(g)→3O2(g) + O(g) 2) O(g) is a catalyst; 2O3(g) + O(g)→3O2(g) + O(g) 3) O(g) is an intermediate; 2O3(g)→3O2(g) 4) O(g) is a catalyst; 2O3(g)→3O2(g)
Chemistry
1 answer:
LekaFEV [45]3 years ago
8 0

Answer:

3) O(g) is an intermediate; 2O3(g)→3O2(g)

Explanation:

The decomposition of ozone to yield oxygen occurs in a sequence of steps. The various non-elementary reactions involved constitute the reaction mechanism. In the sequence of reaction steps O(g) serves as an intermediate.

The overall reaction involves the conversion of two moles of ozone to three moles of oxygen as shown in the answer. Thus the O(g) is merely a reaction intermediate.

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(part 1 of 3) Copper reacts with silver nitrate through a single replacement. If 1.29 g of silver are produced from the reaction
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Answer:

See explanation.

Explanation:

Hello there!

In this case, according to the described chemical reaction, we first write the corresponding equation to obtain:

Cu+2AgNO_3\rightarrow 2Ag+Cu(NO_3)_2

Thus, we proceed as follows:

Part 1 of 3: here, since the molar mass of silver and copper (II) nitrate are 107.87 and 187.55 g/mol respectively, and the mole ratio of the former to the latter is 2:1, we can set up the following stoichiometric expression:

m_{Cu(NO_3)_2}=1.29gAg*\frac{1molAg}{107.87gAg}*\frac{1molCu(NO_3)_2}{2molAg}*\frac{187.55gCu(NO_3)_2}{1molCu(NO_3)_2}   \\\\m_{Cu(NO_3)_2}=1.12gCu(NO_3)_2

Part 2 of 3: here, the molar mass of copper is 63.55 g/mol and the mole ratio of silver to copper is 2:1, the mass of the former that was used to start the reaction was:

m_{Cu}=1.29gAg*\frac{1molAg}{107.87gAg}*\frac{1molCu}{2molAg}*\frac{63.55gCu)_2}{1molCu}   \\\\m_{Cu}=0.380gCu

Part 3 of 3: here, the molar mass of silver nitrate is 169.87 g/mol and their mole ratio 2:2, thus, the mass of initial silver nitrate is:

m_{AgNO_3}=1.29gAg*\frac{1molAg}{107.87gAg}*\frac{2molAgNO_3}{2molAg}*\frac{169.87gAgNO_3}{1molAgNO_3}   \\\\m_{AgNO_3}=2.03gAgNO_3

Best regards!

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