I need hellppp please.

Which ligand binds tightest? ligand A, with a dissociation constant ( Kd ) of 10−9 M ligand D, with a percent occupancy of 80% a

Delvig [45]

Answer:

The answer will be Ligand A with a dissociation constant (Kd) of $10^{-9}$ M

Explanation:

When the dissociation constant in the ligand is small (in order of nano) ( $10^{-9}$ ) it will be more tied. Due to a dissociation constant measures how much a ligand can be able to be separated from the protein so if the number is small it means that the ligand is highly binded to the protein.

On the other hand, the occupancy percentage of the ligand does not imply binding. Conversely, a High-affinity ligand binding with the proteins implies that a relatively low concentration of a ligand is adequate to occupy the maximum ligand-binding site.

6 0

3 years ago

Why is it important to know which category of microorganism has caused an infection?

patriot [66]

Ima say a cause it sounds smart realistically

7 0

3 years ago

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The equilibrium constant Kp for the reaction (CH3),CCI (g) = (CH3),C=CH, (g) + HCl (g) is 3.45 at 500. K. (5.00 x 10K) Calculate

Karolina [17]

<u>Answer:</u> The value of $K_p$ for the reaction is 6.32 and concentrations of $(CH_3)_2C=CH,HCl\text{ and }(CH_3)_3CCl$ is 0.094 M, 0.094 M and 0.106 M respectively.

<u>Explanation:</u>

Relation of $K_p$ with $K_c$ is given by the formula:

$K_p=K_c(RT)^{\Delta ng}$

where,

$K_p$ = equilibrium constant in terms of partial pressure = 3.45

$K_c$ = equilibrium constant in terms of concentration = ?

R = Gas constant = $0.0821\text{ L atm }mol^{-1}K^{-1}$

T = temperature = 500 K

$\Delta n_g$ = change in number of moles of gas particles = $n_{products}-n_{reactants}=2-1=1$

Putting values in above equation, we get:

$3.45=K_c\times (0.0821\times 500)^{1}\\\\K_c=\frac{3.45}{0.0821\times 500}=0.084$

The equation used to calculate concentration of a solution is:

$\text{Molarity}=\frac{\text{Moles}}{\text{Volume (in L)}}$

Initial moles of $(CH_3)_3CCl(g)$ = 1.00 mol

Volume of the flask = 5.00 L

So, $\text{Concentration of }(CH_3)_3CCl=\frac{1.00mol}{5.00L}=0.2M$

For the given chemical reaction:

$(CH_3)_3CCl(g)\rightarrow (CH_3)_2C=CH(g)+HCl(g)$

Initial: 0.2 - -

At Eqllm: 0.2 - x x x

The expression of $K_c$ for above reaction follows:

$K_c=\frac{[(CH_3)_2C=CH]\times [HCl]}{[(CH_3)_3CCl]}$

Putting values in above equation, we get:

$0.084=\frac{x\times x}{0.2-x}\\\\x^2+0.084x-0.0168=0\\\\x=0.094,-0.178$

Negative value of 'x' is neglected because initial concentration cannot be more than the given concentration

Calculating the concentration of reactants and products:

$[(CH_3)_2C=CH]=x=0.094M$

$[HCl]=x=0.094M$

$[(CH_3)_3CCl]=(0.2-x)=(0.2-0.094)=0.106M$

Hence, the value of $K_p$ for the reaction is 6.32 and concentrations of $(CH_3)_2C=CH,HCl\text{ and }(CH_3)_3CCl$ is 0.094 M, 0.094 M and 0.106 M respectively.

8 0

3 years ago

The product side of a chemical reaction is shown.

Inga [223]

The number that represents the coefficient on the product side of the chemical reaction, $-- > 7Ti_2(SO_4)_3$ is 7.

<h3>Coefficients of chemical equations</h3>

In equations representing chemical reactions, the coefficient of each reactant or product of a reaction is the number that comes on the left-hand side just before the chemical formula.

The coefficient of each species in a chemical reaction is obtainable when the equation of the reaction is balanced.

For example, in the following equation: 2A + B = 3C + D

The coefficients of A, B, C, and D are 2, 1, 3, and 1 respectively.

Applying this to the product side of a chemical reaction; $-- > 7Ti_2(SO_4)_3$

It means that the coefficient of the product is 7.

More on coefficients of chemical equations can be found here: brainly.com/question/28294176

#SPJ1

7 0

1 year ago

Which substance is the reducing agent in this reaction? 2KMnO4+3Na2SO3+H2O→2MnO2+3Na2SO4+2KOH

olganol [36]

I believe <span>Na2SO3 is the solution to the problem.</span>

6 0

3 years ago

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