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Exercise 18 - Chapter 2

edited June 2018 in Exercises

Here is an exercise for people familiar with reaction equations: check that conditions (a), (b), (c), and (d) of Definition 2.2 hold.

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  • 1.
    edited April 2018

    Let \( Mat \) be the set of materials, \( \rightarrow \) a reaction, and \( + \) a mixture or combination of materials. Let \( 0 \) signify nothing, e.g. no material. Then we have \( (Mat, \rightarrow, 0, +) \) and

    (a) \( 2{H}{2}O \rightarrow 2{H}{2}O \) and \( 2Na \rightarrow 2Na \), then \( 2{H}{2}O + 2Na \rightarrow 2NaOH + {H}{2} \)

    (b) \( 0 + {H}{2}O = {H}{2}O + 0 = {H}_{2}O \)

    (c) \( ({H}{2}O + {H}{2}O) + 2Na = {H}{2}O + ({H}{2}O + 2Na) \)

    (d) \( 2{H}{2}O + 2Na = 2Na + 2{H}{2}O \)

    But this doesn't express the "for all" quantification and I don't know how.

    Comment Source:Let \\( Mat \\) be the set of materials, \\( \rightarrow \\) a reaction, and \\( + \\) a mixture or combination of materials. Let \\( 0 \\) signify nothing, e.g. no material. Then we have \\( (Mat, \rightarrow, 0, +) \\) and (a) \\( 2{H}_{2}O \rightarrow 2{H}_{2}O \\) and \\( 2Na \rightarrow 2Na \\), then \\( 2{H}_{2}O + 2Na \rightarrow 2NaOH + {H}_{2} \\) (b) \\( 0 + {H}_{2}O = {H}_{2}O + 0 = {H}_{2}O \\) (c) \\( ({H}_{2}O + {H}_{2}O) + 2Na = {H}_{2}O + ({H}_{2}O + 2Na) \\) (d) \\( 2{H}_{2}O + 2Na = 2Na + 2{H}_{2}O \\) But this doesn't express the "for all" quantification and I don't know how.
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