7-b. a solution of naoh was standardized by gravimetric titration of a known quantity of the primary standard, potassium hydrogen phthalate: ao — potassium hydrogen phthalate c8 h5 o4 k, fm 204.22 ß ß co2k co3h 2 ß ß co2k co2na b the naoh was then used to fi nd the concentration of an unknown solution of h2so4: h2so4 1 2naoh s na2so4 1 2h2o (a) verify from the structure of potassium hydrogen phthalate that its formula is c8h5o4k. (b) titration of 0.824 g of potassium hydrogen phthalate required 38.314 g of naoh solution to reach the end point detected by phenolphthalein indicator. find the concentration of naoh (mol naoh/kg solution). (c) a 10.00-ml aliquot of h2so4 solution required 57.911 g of naoh solution to reach the phenolphthalein end point. find the molarity of h2so4.

it is to be noted that an alpha particle is an helium nucleus. so it contains 2 protons 2 neutrons and has a mass number of 4. when alpha decay particles are ejected from the nucleus it gets transformed into an atom which has atomic number less than 2 and a mass number less than 4.

an equation that could represent alpha decay is radium 226 decay of alpha particle forms radon 222.

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while cooking a piece of 1/2 pound beef steak, a cast iron pan (2.4kg in weight) was heated 260 celsius.

as given the equation is

2al + 6 hcl > 2 alcl3 + 3 h2(g)

as per equation two moles of aluminum reacts with six moles of hcl to give two moles of alcl3 and three moles of hydrogen gas

given: three moles of al are present and 7 moles of hcl

for three moles of aluminum nine moles of hcl will be required

however as 7 moles of hcl are present so the limiting reagent is hcl

hcl is going to decide the number of moles of product formed

when six moles of hcl are present = 3 moles of h2 are formed

when one mole of hcl is present == 3/6 mole sof h2 will be formed

when 7 moles of hcl are present = 3 x 7/ 6 moles of h2 will be formed = 3.5 moles of h2

answerambulatory;