Discussion topic: Contributed Answer/Explanation to Q. 1
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| Home: alokrajiv | Original Message | Reply | | Subject: Contributed Answer/Explanation to Q. 73 |
<p>
<span style="white-space: nowrap" class="chemf">Fe + 2HCl
→ FeCl<sub>2</sub> + H<sub>2</sub></span>
</p>
<p>
With HCl, metals with variable valency forms always lower chloride.
</p>
Posted at: Mon Mar 8 17:59:17 2010 (GMT) |
| Page 1 of 1 | | From: alokrajiv | Reply 1 of 2 | Reply | | | Subject: Contributed Answer/Explanation to Q. 73 |
<p>
<span style="white-space: nowrap" class="chemf">Fe + 2HCl
→ FeCl<sub>2</sub> + H<sub>2</sub></span>
</p>
<p>
With HCl, metals with variable valency forms always lower chlorides.
</p>
Posted at: Mon Mar 8 18:03:26 2010 (GMT)
|
| From: alokrajiv | Reply 2 of 2 | Reply | | | Subject: Contributed Answer/Explanation to Q. 73 |
<p>
<span style="white-space: nowrap" class="chemf">Fe + 2HCl
→ FeCl<sub>2</sub> + H<sub>2</sub></span>
</p>
<p>
With HCl, metals with variable valency forms always lower chlorides.
</p>
Posted at: Mon Mar 8 18:04:18 2010 (GMT)
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