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kari74 [83]
3 years ago
14

HCL is slowly added to aqueous Na2CO3 forming NaCl, H2O, and CO2. Follow the steps above to write a balanced equation for this r

eaction .
Chemistry
1 answer:
user100 [1]3 years ago
4 0

 The  balanced  equation is as  below

Na2CO3 +2 HCl → 2 NaCl  +H2O  + CO2

<u><em>Explanation</em></u>

 step 1 : write  the unbalanced chemical  equation

that is  Na2CO3  +  HCl→  NaCl  + H2O  +CO2

Step 2: change  the coefficient ( <em>number in front  of chemical  formula)  </em>to make  sure  the  number of  atoms  of  an  element  are the same in both side.


By adding 2 in front  of HCl  and  in front  of NaCl  balance the    equation.

Therefore the balanced equation is as  below

Na2CO3  +2HCl  → 2NaCl  +H2O +Co2

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ivann1987 [24]

Answer:

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3 years ago
If 2.0 ml of 6.0m hcl is used to make a 500.0-ml aqueous solution, what is the molarity of the dilute solution?
svp [43]
Make sure that you understand what they are asking you from this question, as it can be confusing, but the solution is quite simple. They are stating that they want you to calculate the final concentration of 6.0M HCl once a dilution has been made from 2.0 mL to 500.0 mL. They have given us three values, the initial concentration, initial volume and the final volume. So, we are able to employ the following equation:

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