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Irina-Kira [14]
3 years ago
5

When a clean iron nail is placed in an aqueous solution of copper(ii) sulfate, the nail becomes coated with a brownish black mat

erial. (a) what is the name of the material coating the iron? (b) what are the oxidizing and reducing agents? (omit states-of-matter in your answer.) oxidizing agent: chempadhelp?
Chemistry
1 answer:
Tresset [83]3 years ago
5 0
<span>(a) What is the material coating the iron? Cu(s) 

(b) What is the oxidizing agent? The oxidizing agent is reduced (goes down in oxidation number) since Cu starts off as Cu^2+ and ends up as Cu^0, it went down in oxidation number, was reduced, and thus was the oxidizing agent. 

</span>
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Avocados and tomatoes were mainly eaten by the Aztecs and Maya.

<h3>How did Mayans cook their food?</h3>

"Mayans cooking method includes digging a shallow pit, lining it with stones or clay balls.

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Avocados and tomatoes were mainly eaten by the Aztecs and Maya.

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A/An _____ is described as a flow of charged particles.
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How do the following changes affect the value of the equilibrium constant for a gas-phase exothermic reaction:
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Answer:

A. increase equilibrium constant                    

B. decrease equilibrium constant

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3 years ago
A vessel of volume 22.4 dm3 contains 20 mol h2 and 1 mol n2 ad 273.15 k initially. All of the nitrogen reacted with sufficient h
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Nitrogen combine with hydrogen to produce ammonia \text{NH}_3 at a 1:3:2 ratio:

\text{N}_2 \; (g) + 3 \;  \text{H}_2 \; (g) \leftrightharpoons 2\; \text{NH}_3 \; (g)

Assuming that the reaction has indeed proceeded to completion- with all nitrogen used up as the question has indicated. 3 \; \text{mol} of hydrogen gas would have been consumed while 2 \; \text{mol} of ammonia would have been produced. The final mixture would therefore contain

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Apply the ideal gas law to find the total pressure inside the container and the respective partial pressure of hydrogen and ammonia:

  • \begin{array}{lll} P(\text{container}) &= & n \cdot R \cdot T / V \\ & = & (17 + 2) \; \text{mol} \times 8.314 \; \text{L} \cdot \text{kPa} \cdot \text{mol}^{-1} \cdot \text{K}^{-1} \\ & &\times 273.15 \; \text{K} / (22.4 \; \text{L}) \\ &=&  1.926 \times 10^{3} \; \text{kPa} \end{array}
  • \begin{array}{lll} P(\text{H}_2) &= & n \cdot R \cdot T / V \\ & = & (17) \; \text{mol} \times 8.314 \; \text{L} \cdot \text{kPa} \cdot \text{mol}^{-1} \cdot \text{K}^{-1} \\ & &\times 273.15 \; \text{K} / (22.4 \; \text{L}) \\ &=&  1.723 \times 10^{3} \; \text{kPa} \end{array}
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6 0
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