1) How much total sea salt is present in a 95 Kg person? (a normal person (70 kg BW) has approximately 43L body fluid). 2) Please express the normal blood glucose level into molarity (mM). 3) Commercial HCl is 36%. Please convert it into molarity. 4) Find out the Volume (dm3) of product at STP when 0.28 M, 200 mL NaOH (aq.) reacts with 250 mL 85% H3PO4 (Sp. G.-1.067). 5) The above reaction has the product Ammonia, which when dissolved in 300 mL water makes an alkaline solution. Find its molarity (M). 6) Please re-calculate the child dose as per the BW of the baby. (Adult dose-750 mg and the BW of the baby is 50 lb (British pound) (1 lb=0.453 Kg).
1) How much total sea salt is present in a 95 Kg person? (a normal person (70 kg BW) has approximately 43L body fluid). 2) Please express the normal blood glucose level into molarity (mM). 3) Commercial HCl is 36%. Please convert it into molarity. 4) Find out the Volume (dm3) of product at STP when 0.28 M, 200 mL NaOH (aq.) reacts with 250 mL 85% H3PO4 (Sp. G.-1.067). 5) The above reaction has the product Ammonia, which when dissolved in 300 mL water makes an alkaline solution. Find its molarity (M). 6) Please re-calculate the child dose as per the BW of the baby. (Adult dose-750 mg and the BW of the baby is 50 lb (British pound) (1 lb=0.453 Kg).
Chapter79: Solubility
Section: Chapter Questions
Problem 1P
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1) How much total sea salt is present in a 95 Kg person? (a normal person (70 kg BW) has approximately 43L body fluid).
2) Please express the normal blood glucose level into molarity (mM).
3) Commercial HCl is 36%. Please convert it into molarity.
4) Find out the Volume (dm3) of product at STP when 0.28 M, 200 mL NaOH (aq.) reacts with 250 mL 85% H3PO4 (Sp. G.-1.067).
5) The above reaction has the product Ammonia, which when dissolved in 300 mL water makes an alkaline solution. Find its molarity (M).
6) Please re-calculate the child dose as per the BW of the baby. (Adult dose-750 mg and the BW of the baby is 50 lb (British pound) (1 lb=0.453 Kg).
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