Identify all of the conditions that would facilitate the following conversion. > CI AICI3 1. , AICI 3 2. NaBH4, MeOH 1. CI AICI 3 2. NaBH, MeOH 1. Cl , AICI 3 2. Zn(Hg). HCI 1. , AICI 3 2. Zn(Hg), HCI only product Give detailed Solution with explanation needed with reaction. don't give Handwritten answer
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- Solve the problem correctly. I rate you helpful. Promise.Give a clear explanation handwritten answer!!Calcelate mokality of 52 (음)/ 아 to Solalicn. A) 0.65 p) 0 구2 () o.81 D) 0.94 Calulete pt of 0-001 M oK H Poa solation A) ) c) I.5 D) 2.5 NGOH 2 lit Hy soq mixed with Calulete pH of solution N of Nof slit of then A) 8.25 8) S.75 C) 7. 55 D) 9.45
- 3. Cal culate Nattz POy:(Ka, kaz and tas of HsPO4 6.32x10-8 and 4.5x10-13 he ott of a o.0400M Shutin of ane 7.11 メL65 respectively). 4. Con Sider the titration of 1w.00ml of a 0.200M Solution of he hypothetical acid tzA uith o.loc Soclium hydroxide. Enourhg hat Kay = (. 00 k1o-8, and kag = 1.00 xLo 12 Calculate ka after he addition of a) 1o. oml Na OH =l:00x16", E1:00x104 9 20.oml Na OH ) 30. omL Na ott d) 40.0ml Na OHMolarity of the standard solution NaOH is 0.0994MH1 gramof Sampleof an impure mixture af NARCOZ and Natteo, containing impurties is dissalved and tityated with SN Hel wit p colourless after addition5mD otacid-Methyt ge is then added and 22ontof, acid reguired orange tochange the of Naz coz benotphthalein as indicator, the sotudion Curns this indicator what is the percoulye and Na Heoz in the sample ?
- 0.2 g of carbonate sample dissolved in 40ml of 0.IM HCI. Back titrated required 20 ml of 0.1 M NaOH, calculate the molecular weight of the carbonate salt. O a. 100g/mole O b. 345g/mole O c. 200g/mole O d. 300g/mole Next page touch2- Ahmed has done four types of titrations of 1M and 50 ml for both acid and base (Conductometric Titration of Strong Acid with Strong Base , Potentiometric Titration of Weak Acid with Strong Base , Potentiometric Titration of Strong Acid with Strong Base and Conductometric Titration of Weak Acid with Strong Base ). He plotted the results in curves as shown below. Match between the type of titration and his results 3- Ahmed standardized NaOH (10 ml) with KHP( mass=1 g , MW= 204 g/mol ). Calculate the concentration of standardized NaOH? 4- Calculate pka for a weak acid titrated with strong base using the curve shown below? 14 12- 10- 8 6- 4- 2- 0- 10 20 30 40 50 5- If the solute moved 3 cm in paper chromatography and R, was 0.5. How long in cm the solvent has moved?Distribution Coefficient - Please help write conclusions based on results provided. Thank You! Results Equiv. wt. of acetic acid - 60.052 g/mol a. Volume of base required for first titration 10.10 mL- 0.0101 L b. Mass of acetic acid in the 10 mL aliquot 0.3033 g c. Volume of base required for second titration 6.53 ml. 0.00653 L d. Mass of acid left over in water layer 0.1961 g e. Mass of acid in ethyl acetate after a single 10 mL extraction 0.1072 g 6.29 ml.- 0.00629 L f. Volume of base required for third titration 0.1889 g g. Mass of acid in water layer 0.1144 g h. Mass of acid in ethyl acetate after two 5 mL extractions Kp (single extraction) 0.55 versus Kp (double extraction) 0.61
- Please answer this one, thank u101 Chem101 b My Questions | bartleby А аpp.101edu.co E Apps e Resource Library -. Dashboard 101 Chem101 E Reading list Word wco Writing Centers at. 11-0218 NCI Theor.. Question 5 of 11 Submit A solution of tartaric acid (H:C&H.Os) with a known concentration of 0.155 M H2C«H«Os is titrated with a 0.425 M NaOH solution. How many mL of NaOH are required to reach the second equivalence point with a starting volume of 70.0 mL H.C«H«Os , according to the following balanced chemical equation: H:CH.Os + 2 NAOH - Na:CaH.Os + 2 H:O STARTING AMOUNT ADD FACTOR ANSWER RESET *( ) 1000 0.425 0.001 0.0217 12.8 25.5 0.155 1 0.0511 51.1 5.11 x 104 70.0 g NaOH L H2CsH.Os mol H:CAH.O6 M H2CH«Os mol NaOH mL NaOH M NaOH mL H2CaH«Os L NAOH g H2C4H.O6 1:35 PM P Type here to search 46°F 2/27/2022Choose the answers from the box Please answer 4,5,6,7,8