I think is true not sure but it might be
The titrant for this exercise. suppose Ca(OH)₂ were used as the titrant, instead of NaOH. This will make the titrant twice as concentrated in hydroxide ion. the analyte will still be HC₂H₃O₂. the stoichiometry ratio of HC₂H₃O₂ to Ca(OH)₂ is 1 : 2.
The balanced reaction of the given condition as follow :
Ca(OH)₂ + 2HC₂H₃O₂ ------> Ca(C₂H₃O₂)₂ + 2H₂O
from the equation it is clear that stoichiometry of Ca(OH)₂ is 1 and the stoichiometry of HC₂H₃O₂ is 2. therefore the stoichiometry ratio of HC₂H₃O₂ to Ca(OH)₂ is 1 : 2.
Thus, The titrant for this exercise. suppose Ca(OH)₂ were used as the titrant, instead of NaOH. This will make the titrant twice as concentrated in hydroxide ion. the analyte will still be HC₂H₃O₂. the stoichiometry ratio of HC₂H₃O₂ to Ca(OH)₂ is 1 : 2.
To learn more about stoichiometry here
brainly.com/question/13145466
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<span>2 CH3OH + 3 O2 = 2 CO2 + 4 H2O
2 moles CH3OH ---------------- 3 moles O2
1.33 moles CH3OH ----------- ( moles O2)
moles O2 = 1.33 x 3 / 2
moles O2 = 3.99 / 2
= 1.995 moles of O2</span>
please mark my answer brainliest...
condition...pressure remains constant.
- for 720 ml temp is 15°...so for
- 960ml temp will be 15/720×960=20°...(.answer for 1st part...)
- for 900cmcube temp is 270°C...so for
- 300cmcube temp will be 270/900×300=90°....(answer for 2nd part)...
- I hope it helps the dear students...and if it is then let me know through ur comments...and please mark my answer as brainliest...plz...
