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elixir [45]
3 years ago
7

Which element reacts spontaneously with 1.0M HCl (aq) at room temperature?

Chemistry
2 answers:
Ad libitum [116K]3 years ago
8 0
<span>(4) zinc.................</span>
Kaylis [27]3 years ago
4 0
<em><u>(4) Zinc is correct answer. Others don't react at room temperature with HCl because their reactivity (capability to react) is quite low.</u></em>
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Answer:

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P=\frac{(mol)(\frac{L.atm}{mol.K})(K)}{L-(mol)(\frac{L}{mol})}-\frac{(mol^{2})(\frac{atm.L^{2}}{mol^{2}})}{L^{2}}

Then you should multiply and eliminate the same units which they are dividing each other (Please see the photo below), so you have:

P=\frac{L.atm}{L-L}-atm

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P=atm

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(V-nb)=\frac{nRT}{(P+\frac{n^{2}a}{V^{2}})}

nb=V-\frac{nRT}{(P+\frac{n^{2}a}{V^{2}})}

b=\frac{V-\frac{nRT}{(P+\frac{n^{2}a}{V^{2}})}}{n}

Replacing units:

b=\frac{L-\frac{(mol).(\frac{L.atm}{mol.K}).K}{(atm+\frac{mol^{2}.\frac{atm.L^{2}}{mol^{2}}}{L^{2}})}}{mol}

Multiplying and dividing units,(please see the second photo below), we have:

b=\frac{L-\frac{L.atm}{atm}}{mol}

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b=\frac{L}{mol}

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