Coe caco t 2k acid base neutralization for an acid to

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CoE Caco t 2K Acid Base neutralization For an acid to be neutral the ph is in the middle ph 0 is acid ph 14 is base HN Oz NaOH cap Na Nacht H Oh
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Molarity solute into mates molarity Yi IEm 500mL of 0.15 M Nau molarity 4.39 moles Nall 250mL of 1.55 M NoHo Solution 1,55 M 0.2504 0.3875 moles Naito 0.3875 moles Na Hoo tkH 32.55 g NoHo how much you need to make 1.55M Molarity is the concentration numerically Dilution M V Miz initial find 50.0mL IN Agnos dilutes in 475mL that is the M 1 6.05 0.425 M O It M
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3M Hasa 1kg Hz looosltafh.no I YmIf ff.I.sou 165L of 3MHz Soa to produce 1kg Ha Polar compounds are soluble acids sucrose and ethanol are soluble hydrocarbons gasoline oils and Cola are NOT Nat Nos NHI are soluble ALL Electrolytes are all soluble and ionic all acids and bases but are they ionic as well f Ar h V ti 2.35 si fomIm 8425n CYI Ippm utlT 5.55xio8cm 8425cm 1,000,000g solution 1.47 10 3cm ML 1.240mi O m 1.76 10 s 23.0 1,9 9.7 2 1.76 109 3.90 10 o Nat
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24.5g Mgs in 735mL 0.735L 24.5g My s y O 435M µ o.u3 nd 0.7352 7.46g Carlos in 0.1M volume 7.96gCano f Tsq 0.059 mot 0.1M 0.09mi V 0.59L 590mL 6 ON Hull 55mL XM 1750mL 0.44 XM 77mL 0.4M 0.1M x my 308 Xml 308mL 77mL 53mL I 70M 1268mL XM 0.34M A total solution 134mL 0.34M 267mL XM 0.171 M
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CmHzzQ 342.296,4mi 0.85M o sc 0.21254 342.296 1 72.74g 0.2125 Mol Mol d p6NQ 0.5396 0.250M 1 mold NHy I 1.07 X 4.28L 4280mL
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