Answer:
Pentafluorobenzene: 11,92 min
Benzene: 12,14 min
Explanation:
<em>Retention time of pentafluorobenzene is 12,98 min and 13,20 min of benzene.</em>
The adjusted retention time is the time an analyte spends in the column not the stationary phase. As time of unretained solute is 1,06 min the adjusted retention time for an analyte is:
tr' = tr - 1,06min
For pentafluorobenzene:
tr' = 12,98min - 1,06min = <em>11,92 min</em>
For benzene:
tr' = 13,20 - 1,06min = <em>12,14 min</em>
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Answer:
1mpm(1 meters per minute) Explanation: Speed:Distance/time 25/25=1 plz mark as brainliest
The volume of the NaOH used is calculated as 14 mL.
<h3>What is stoichiometry?</h3>
The term stoichiometry has to do with the calculation of the amount of substance in a reaction using mass - mole or mass - volume relationship.
Here;
Number of moles of CaCO3 = 0.205 g/100.1 = 0.00205 moles
Number of moles of HCl = 2.00 M * 7/1000 L = 0.014 moles
2 moles of HCl reacts with 1 mole of CaCO3
x moles of HCl reacts with 0.00205 moles of CaCO3
x = 0.00205 moles * 2/1 = 0.0041 moles
Hence HCl is the excess reactant
Amount of excess HCl = 0.014 moles - 0.0041 moles = 0.0099 moles
Concentration of excess HCl reacted = 0.0099 moles/125 * 10^-3 = 0.0792 M
Using;
CAVA/CBVB = NA/NB
CAVANB = CBVBNA
VB = CAVANB/CBNA
VB = 0.0792 M * 10 mL * 1/ 0.058 M
VB = 14 mL
Missing parts;
A 0.205 g sample of caco3 (mr = 100.1 g/mol) is added to a flask along with 7.50 ml of 2.00 m hcl. caco3(aq) + 2hcl(aq) → cacl2(aq) + h2o(l) + co2(g) enough water is then added to make a 125.0 ml solution. a 10.00 ml aliquot of this solution is taken and titrated with 0.058 m naoh. naoh(aq) + hcl(aq) → h2o(l) + nacl(aq) how many ml of naoh are used?
Learn more about stoichiometry: brainly.com/question/9743981
No, their specific heats are different. For water, and ice
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