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Three colorless solutions in test tubes, with no labels, are in a test tube rack on the laboratory bench. Lying beside the tests tubes are three labels : 0.10 M Na2CO3, 0.10 M HCL, and 0.10 M KOH. You are to place the labels on the test tubes using only the three solutions present. Here are your tests:
A few drops of the solutions from test tube 1 added to a similar volume of the solution in test tube 2 produces no visible reaction but the solution becomes warm.
A few drops of the solution from test tube 1 added to a similar volume of the solution in test tube 3 produces carbon dioxide gas.
Identify the labels for test tubes 1, 2, and 3

Answer :

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Answer:

Test tube 1  0.10 M HCL

Test tube 2  0.10 M KOH

Test tube 3 0.10 M Na2CO3

Explanation:

From the question we are told that

    A few drops of the solutions from test tube 1 added to a similar volume of the solution in test tube 2 produces no visible reaction but the solution becomes warm

Generally this warmth is as a result of a reaction between an acid and a base and the acid is 0.10 M HCL and the base is  0.10 M KOH , the heat generated is know as the heat of neutralization,

The reaction is  

       [tex]HCl_{(aq)} + KOH_{(aq)} \rightarrow KCl_{(aq)} + H_2O_{(l)} + \Delta H[/tex]

We are also told from the reaction that

A few drops of the solution from test tube 1 added to a similar volume of the solution in test tube 3 produces carbon dioxide gas.

Generally  carbon dioxide gas  is produced is as a result of  a reaction between the acid HCl  and Na2CO3.

The reaction is

        [tex]2HCl -{(aq)} + Na_2 CO_3_{(aq)} \rightarrow 2 NaCl _{(aq)} + CO_2_{(g)} + H_2O_{(l)}[/tex]

Hence from this explanation above we see that the solution in test tube 1 is  0.10 M HCL while solution in test tube 2 is 0.10 M KOH and then solution in test tube three is  0.10 M Na2CO3

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