If a 0.1 m solution of glucose 1-phosphate is incubated with a catalytic amount of phospho-glucomutase, the glucose 1-phosphate is transformed to glucose 6-phosphate until equilibrium is reached. at equilibrium, the concentration of glucose 1-phosphate is 4.5 x 10–3 m and that of glucose 6-phosphate is 8.6 × 10–2 m. set up the expressions for the calculation of keq' and g'° for this reaction (in the direction of glucose 6-phosphate formation). (r = 8.315 j/mol·k; t = 298 k)

Explanation:

The given data us as follows.

At equilibrium, the concentration of glucose-1-phosphate = M

At equilibrium, the concentration of glucose-6-phosphate = M

As,  

                  =

                  = 19.11

As it is known that

Hence, putting the given values into the above formula as follows.

                          =

                          = 7309.3620 J/mol

or,                       = 7.309 kJ/mol         (as 1 kJ = 1000 J)

Thus, we can conclude that value of is equal to 7.309 kJ/mol.

the density of gold = 19.3 g/cm3

the relation between density and mass and volume is

density = mass / volume

given :

mass of gold = 13 g

density = 13g / volume = 19.3

volume = 13 /19.3 = 0.67 cm^3

thus volume of gold nugget = 0.67 cm^3

your answer is c) plasma

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