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Fynjy0 [20]
3 years ago
11

The old car left on the side of the road showed signs of corrosion. We call this common chemical reaction rusting. The equation

for this reaction can be written as 4Fe + 3O2 → 2Fe2O3. What are the reactants in the rusting process? A) iron oxide B) water and iron C) iron and oxygen D) water and iron oxide
Chemistry
1 answer:
Ivanshal [37]3 years ago
4 0
<span>The old car left on the side of the road showed signs of corrosion. We call this common chemical reaction rusting. The equation for this reaction can be written as 4Fe + 3O2 → 2Fe2O3. What are the reactants in the rusting process? 
</span>
C. Iron and Oxygen

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For the reaction below, identify the type of reaction and determine the coefficient for Ag after balancing the reaction
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Answer:

single displacement reaction

Explanation:

coefficient if Ag is 2

after it is balanced

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0.01040 m as an integer
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Here,

0.01040 m as an integer= 1.04 × 10-2

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Compounds that have the exact same molecular formula but are arranged differently are<br> called?
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Answer:

isomers

Explanation:

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Weight is measured as the downward force of __________. a. gravity c. acceleration b. friction d. motion
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At 2000°C, the equilibrium constant for the reaction below is Kc = 4.10 ´ 10–4 . If 0.600 moles of NO is placed in a 1.0-L react
erastova [34]

Answer:

At equilibrium, the concentration of N_{2 (g)} is going to be 0.30M

Explanation:

We first need the reaction.

With the information given we can assume that is:

N_{2 (g)} + O_{2 (g)} ⇄ 2NO_{(g)}

If there is placed 0.600 moles of NO in a 1.0-L vessel, we have a initial concentration of 0.60 M NO; and no N_{2 (g)} nor  O_{2 (g)} present. Immediately, N_{2 (g)} andO_{2 (g)} are going to be produced until equilibrium is reached.

By the ICE (initial, change, equilibrium) analysis:

I: [N_{2 (g)}]=0   ;     [O_{2 (g)} ]= 0    ; [NO_{(g)}]=0.60M

C: [N_{2 (g)}]=+x   ;     [O_{2 (g)} ]= +x    ; [NO_{(g)}]=-2x

E: [N_{2 (g)}]=0+x   ;     [O_{2 (g)} ]= 0+x   ; [NO_{(g)}]=0.60-2x

Now we can use the constant information:

K_{c}=\frac{[products]^{stoichiometric coefficient} }{[reactants]^{stoichiometric coefficient} }

4.10* 10^{-4} =\frac{(0.60-2x)^{2}}{(x)*(x)}

4.10* 10^{-4}= \frac{(0.60-2x)^{2}}{x^{2} }

4.10* 10^{-4} * x^{2}= (0.60-2x)^{2}}

\sqrt{4.10* 10^{-4} * x^{2}}= \sqrt{(0.60-2x)^{2}}}

0.0202 x =0.60 - 2x

2x+0.0202x=0.60

x=\frac{0.60}{2.0202}= 0.30

At equilibrium, the concentration of N_{2 (g)} is going to be 0.30M

3 0
3 years ago
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