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>
Answer : The structure of 
will be square-planar.
Explanation :
In the given molecule , 'Xe' is the central atom and 'H' is the terminal atom.
Xenon has 8 valence electrons and hydrogen has 1 valence electron. Therefore, the total number of valence electrons are 8 + 4(1) = 12 electrons.
The number of electrons used in Xe-H bonding = 8 electrons
The remaining electrons which are used as lone pair on central atom (Xe) = 12 - 8 = 4 electrons
There are 4 bonding pairs and 2 lone pairs of electrons, they will be arranged in the octahedral arrangement around the central atom with 2 lone pairs of electrons on central atom. The lone pairs are arranged linearly across the central atom. The resulting structure will be square-planar.
The structure of is shown below.
Answer:
They are multilayered
Spacesuits provide protection from radiation and offer normal air pressure inside
Explanation:
Answer:
the first energy level is closest to nuclear the second energy level is a little farther away than the first
Answer:
%yield of NH₃ = 30%
Explanation:
Actual yield of NH₃ = 40.8g
Theoretical yield = ?
Equation of reaction
N₂ + 3H₂ → 2NH₃
Molar mass of NH₃ = 17g/mol
Molarmass of N = 14.00
2 molecules of N = 2 * 14.00 = 28g/mol
Number of moles = mass / molar mass
Mass = number of moles * molar mass
Mass = 1 * 28.00 = 28g of N₂ (the number of moles of N₂ from the equation is 1).
From the equation of reaction,
28g of N₂ produce (2 * 17)g of NH₃
28g of N₂ = 34g of NH₃
112g of N₂ = x g of NH₃
X = (112 * 34) / 28
X = 136g of NH₃
Theoretical yield = 136g of NH₃
% yield = (actual yield / theoretical yield) * 100
% yield = (40.8 / 136) * 100
% yield = 0.3 * 100
% yield = 30%
