Answer:
m H2(g) = 2.241 g H2(g)
Explanation:
- 2Al(s) + 3H2SO4(aq) → Al2(SO4)3(aq) + 3 H2(g)
limit reagent:
∴ Mw Al = 26.982 g/mol
∴ Mw H2SO4 = 98.0785 g/mol
⇒ n Al = (20 g Al)×(mol/26.982 g) = 0.7412 mol Al
⇒ n H2SO4 = ( 115 g H2SO4 )×(mol/98.0785 g) = 1.173 mol H2SO4
⇒ n H2 = (0.7412 mol Al)×(3 mol H2/ 2 mol Al) = 1.112 mol H2
∴ Mw H2 = 2.016 g/mol
⇒ g H2 = (1.112 mol H2)×(2.016 g/mol) = 2.241 g H2
Answer:
a. True
b. True
c. False
d. True
Explanation:
a). A a very low substrate concentration , 
. Thus according to the Machaelis-Menten equation becomes
![$V_0 = \frac{V_{max} \times [S]}{Km}$](https://tex.z-dn.net/?f=%24V_0%20%3D%20%5Cfrac%7BV_%7Bmax%7D%20%5Ctimes%20%5BS%5D%7D%7BKm%7D%24)
Here since the 
varies directly to the substrate concentration [S], the initial velocity is lower than the maximal velocity. Thus option (a) is true.
b). The Michaelis -Menten kinetics equation states that :
![$V_0 = \frac{V_{max} \times [S]}{Km+[S]}$](https://tex.z-dn.net/?f=%24V_0%20%3D%20%5Cfrac%7BV_%7Bmax%7D%20%5Ctimes%20%5BS%5D%7D%7BKm%2B%5BS%5D%7D%24)
Here the initial velocity changes directly with the substrate concentration as is directly proportional to [S]. But 
is same for any particular concentration of the enzymes. Thus, option (b) is true.
c). As the substrate concentration increases, the initial velocity also increases. Thus option (c) is false.
d). Option (d) explains the procedures to estimate the initial velocity which is correct. Thus, option (d) is true.
Answer:
I think that it it correctly balanced that is my opinion and, because the way it is set up, that the answer will tell you weather or not it is correctly balanced or not.
