502g sample of dry cac03

  1. 502g sample of dry cac03 and cacl2 mixture was dissolved in 25ml of 0.925 M HCL solution. What was cacl2 percentage in original
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  2. 1.64g of a mixture of cac03 and mhco3 was dissolved in 50ml of 0.8M hcl. The excess of acid required 16ml of 0.25M naoh for
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  3. 1.64g of a mixture of cac03 and mgc03 was dissolved in 50ml of 0.8 m Hcl. The excess of acid required 16ml of 0.25< NAOH for
    1. answers icon 2 answers
    2. Fai asked by Fai
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  4. 1.64g of a mixture of cac03 and mgc03 was dissolved in 50ml of 0.8M hcl. The excess of acid required 16ml of 0.25M naoh for
    1. answers icon 0 answers
    2. Fai asked by Fai
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  5. 1.64g of a mixture of cac03 and mgc03 was dissolved in 50ml of 0.8M hcl. The excess of acid required 16ml of 0.25M naoh for
    1. answers icon 1 answer
    2. Fai asked by Fai
    3. views icon 441 views
  6. 80ml of hcl is added to 2.5gm of pure cac03 when the reaction is over, then 0.5gm cac03 is left. Find the normality of the acidT
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  7. 10L of hard water required 5.6g of line for removing hardness. Hence temporary hardness in ppm of cac03.The answer 1000 ppm=
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    2. ken asked by ken
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  8. 1.64g of a mixture of cac03 and mgc03 was dissolved in 50ml of 0.8M hcl. The excess of acid required 16ml of 0.25M naoh for
    1. answers icon 0 answers
    2. Fai asked by Fai
    3. views icon 370 views
  9. one litre of a sample hard water contain 4.44mg cacl2 and 1.9mg of mgcl2. What is the total hardness in terms ppm of cac03?
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    2. ken asked by ken
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  10. 10L of hard water required 5.6g of lime for removing hardness. Hence temporary hardness in ppm of cac03.The answer 1000 I think
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    2. ken asked by ken
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