Answer:
Should be 0.6106 though i could be wrong
Explanation:
Answer:
0.7μM = 0.6 μM = 0.5 μM > 0.4 μM > 0.3 μM > 0.2 μM
Explanation:
An enzyme solution is saturated when all the active sites of the enzyme molecule are full. When an enzyme solution is saturated, the reaction is occurring at the maximum rate.
From the given information, an enzyme concentration of 1.0 μM Y can convert a maximum of 0.5 μM AB to the products A and B per second means that a 1.0 M Y solution is saturated when an AB concentration of 0.5 M or greater is present.
The addition of more substrate to a solution that contains the enzyme required for its catalysis will generally increase the rate of the reaction. However, if the enzyme is saturated with substrate, the addition of more substrate will have no effect on the rate of reaction.
<em>Therefore the reaction rates at substrate concentrations of 0.7μM, 0.6 μM, and 0.5 μM are equal. But the reaction rate at substrate concentrations of 0.2 μM is lower than at 0.3 μM, 0.3 μM is lower than 0.4 μM and 0.4 μM is lower than 0.5 μM, 0.6 μM and 0.7 μM.</em>
The reaction between N₂ and F₂ gives Nitrogen trifluoride as the product. The balanced equation is;
N₂ + 3F₂ → 2NF₃
The stoichiometric ratio between N₂ and NF₃ is 1 : 2
Hence,
moles of N₂ / moles of F₂ = 1 / 2
moles of N₂ / 25 mol = 0.5
moles of N₂ = 0.5 x 25 mol = 12.5 mol
Hence N₂ moles needed = 12.5 mol
At STP (273 K and 1 atm) 1 mol of gas = 22.4 L
Hence needed N₂ volume = 22.4 L mol⁻¹ x 12.5 mol
= 280 L
