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- Lead hydroxide, Pb(0H)2(s) | Pb(0H)2(s)Pb+2(aq)+20H-(aq) | Ksp=1.4x10-20 H2O(aq)H*(aq)+OH-(aq) Pb(OH)2 Water Kw=1.00x10-14 Molar masses 241.21 Pb 207.20 OH- 17.01 If Pb+2(aq) is controlled by the solubility of Pb(OH)2, what Pb+2(aq) concentration is expected if pH=p0H=7.00? b. What pH would be required to lower Pb+2(aq) to 2.07 ppb? a.College of Teacher Bachelor of Secondary Education *promoting pedagogical excellence elor of Sec omoting pedage RIN ht: (a) (b) Br. C. CH3 b1. HNO3, H2SO4 2. Fe, H30*r H2/Pd s / oD Br NO2 (c) KMN04 7 (d) Cl CH3CH2CH2CI AICI3 H20 7. OCH3 4. Predict the major product(s) of the following reactions: OM (a) CI (b) CH CH CH3CH2C! AICI3 CH3CH2COCI AICI3 (c) CO2H HNO3 (d) N(CH2CH3)2 H2SO4 SO, H2SO4 ASITYOral rehydration salts are stated to contain the following components: Sodium Chloride 3.5g Potassium Chloride 1.5g Sodium Citrate 2.9g Anhydrous Glucose 20.0g 8.342 g of oral rehydration salts are dissolved in 500 ml of water. 5 ml of the solution is diluted to 100 ml and then 5 ml is taken from the diluted sample and is diluted to 100 ml. The sodium content of the sample is then determined by flame photometry. The sodium salts used to prepare the mixture were: Trisodium citrate hydrate (C6H5Na3O7, 2H2O) MW 294.1 and sodium chloride (NaCl) NW 58.5. Atomic weight of Na = 23. The content of Na in the diluted sample was determined to be 0.3210 mg/100 ml. Determine the % of stated content of Na in the sample.
- Oral rehydration salts are stated to contain the following components: Sodium Chloride 3.5g Potassium Chloride 1.5g Sodium Citrate 2.9g Anhydrous Glucose 20.0g 8.342 g of oral rehydration salts are dissolved in 500 ml of water. 5 ml of the solution is diluted to 100 ml and then 5 ml is taken from the diluted sample and is diluted to 100 ml. The sodium content of the sample is then determined by flame photometry. The sodium salts used to prepare the mixture were: Trisodium citrate hydrate (C6H5Na3O7, 2H2O) MW 294.1 and sodium chloride (NaCl) NW 58.5. Atomic weight of Na = 23. The content of Na in the diluted sample was determined to be 0.3210 mg/100 ml. Determine the % of stated content of Na in the sample. The stated should be 104.5, how??A solution of HCl was titrated against sodium carbonate. What is the average normality of acid in the given data? (MW Na₂CO3 = 106) T2 0.3479 0.3562 0.3042 Weight (g) Initial V (ml) 0.80 1.60 35.20 36.70 39.80 Final V (mL) Vol HCl used (mL) 35.10 39.40 N of HCl (eq/L) 0.1954 0.1870 Average N of HCl (eq/L)Q1) what is The concentration of Hg in PPm in the solution in equilibrium with Hg2(CN)2 ? PKSP ( Hg2(CN)2 )=39.3 Pka(HCN) = 9.2 Mw(Hg) =200.6.
- 2b) What is the primary consideration?1)Please express the normal salt (NaCl) concentration in body fluid into molarity (mM).2) Commercial fuming Sulphuric acid (Oleum-H2S2O6) is 99.9%. solution. Please convert it into molarity.3) Find out the Volume (dm3) of product (gas) at RTP when 0.58 M, 150 mL NaOH (aq.) reacts with 350 mL, 0.25 NH4Cl.4) The above reaction has the product Ammonia, which when dissolved in 650 mL ethanol makes an alkaline ethanolic solution. Find its molarity (M) 5) Calculate the adult dose as per the BW of the baby. (Child dose-50 mg and the BW of the baby is 48 lb (British pound) (1lb=0.453 Kg)In the standardization of HCl using pure anhydrous sodium carbonate as the primarystandard for methyl orange as an indicator, 1.0 mL HCl was found to be equivalent to 0.05gof sodium carbonate (MW =106). The normality of HCl is:
- Qq.1. Subject :- Account(d) A sample contains a mixture of ammonium carbonate (NH4)2CO3, sodium carbonate Na2CO3, and sodium chloride NaCl. Thermogravimetric analysis indicates that a 0.0965-g portion of this sample loses 0.0574 g over a temperature range of 50-75 oC and then loses another 0.0124 g at 800 oC.(i) What causes the weight loss at 50-75 oC? (ii) What species will remain after being heated to 800 oC? (iii) What is the composition of the original sample? Given: Atomic masses: H = 1.008; C = 12.011; N = 14.007; Na = 22.99; O = 15.999;Cl = 35.45.(7) KMNO4, NaOH