Glucose taken up by liver cells is rapidly phosphorylated to glucose 6-phosphate by ATP in the initial step of metabolism of the sugar. Two enzymes, which may be considered isozymes, are capable of catalyzing this reaction in the liver cell. Hexokinase has a low Km of ~0.05 mM for glucose, whereas glucokinase exhibits a Km of ~5 mM. After a meal, which of the following scenarios best describes the relative activity levels for the two enzymes? Hexokinase Glucokinase
A) not active not active
B) Vo = Vmax/2 not active
C) Vo = Vimax Vo =Vmax
D) Vo = Vmax/2 Vo = Vmax/2
E) Vo = Vmax Vo = Vmax/2

confused

confused             hewdejdnjnjsnd

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

hey there!

mass = 6.75 g

volume =   9 cm³

therefore:

d = mass / volume

d   = 6.75 / 9

d = 0.7 g/cm³

answer b