(3.5mol)(24.106 g/1mol c6h6) =84.371 g C6H6
Answer:
Explanation:
From the given information:
The concentration of metal ions are:
![[Ca^{2+}]= \dfrac{0.003474 \ M \times 20.49 \ mL}{10.0 \ mL}](https://tex.z-dn.net/?f=%5BCa%5E%7B2%2B%7D%5D%3D%20%5Cdfrac%7B0.003474%20%5C%20M%20%5Ctimes%2020.49%20%5C%20mL%7D%7B10.0%20%5C%20mL%7D)
![[Ca^{2+}]=0.007118 \ M](https://tex.z-dn.net/?f=%5BCa%5E%7B2%2B%7D%5D%3D0.007118%20%5C%20M)
![[Mg^2+] = \dfrac{0.003474 \ M\times (26.23 - 20.49 )mL}{10.0 \ mL}](https://tex.z-dn.net/?f=%5BMg%5E2%2B%5D%20%3D%20%5Cdfrac%7B0.003474%20%5C%20M%5Ctimes%20%2826.23%20%20-%2020.49%20%29mL%7D%7B10.0%20%5C%20mL%7D)
Mass of Ca²⁺ in 2.00 L urine sample is:
= 0.1598 g
Mass of Ca²⁺ = 159.0 mg
Mass of Mg²⁺ in 2.00 L urine sample is:
= 0.3461 g
Mass of Mg²⁺ = 346.1 mg
Answer:
True
Explanation:
In an uncompetitive inhibition, initially the substrate [S] binds to the active site of the enzyme [E] and forms an enzyme-substrate activated complex [ES].
The inhibitor molecule then binds to the enzyme- substrate complex [ES], resulting in the formation of [ESI] complex, thereby inhibiting the reaction.
This inhibition is called uncompetitive because the inhibitor does not compete with the substrate to bind on the active site of the enzyme.
Therefore, in an uncompetitive inhibition, the inhibitor molecule can not bind on the active site of the enzyme directly. The inhibitor can only bind to the enzyme-substrate complex formed.
