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Rom4ik [11]
3 years ago
5

8)

Chemistry
1 answer:
Rina8888 [55]3 years ago
7 0
B mass but not of charge
You might be interested in
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
Automobile bodies contain significant amounts of iron. The iron is protected by the addition of zinc. This is called galvanizati
Oksana_A [137]

<u>Answer:</u> The balanced chemical equation is written below.

<u>Explanation:</u>

Galvanization is defined as the process in which a protective layer of zinc is applied to iron or steel to prevent the metal from rusting.

Zinc prevents the oxidation of iron and acts as a reducing agent in the process.

The half reaction for the process follows:

<u>Oxidation half reaction:</u>  Zn\rightarrow Zn^{2+}+2e^-

<u>Reduction half reaction:</u>  Fe^{2+}+2e^-\rightarrow Fe

Net chemical equation:  Zn+Fe^{2+}\rightarrow Zn^{2+}+Fe

Hence, the balanced chemical equation is written above.

6 0
3 years ago
Use the given data at 500 K to calculate ΔG°for the reaction
Anton [14]

Answer : The  value of \Delta G^o for the reaction is -959.1 kJ

Explanation :

The given balanced chemical reaction is,

2H_2S(g)+3O_2(g)\rightarrow 2H_2O(g)+2SO_2(g)

First we have to calculate the enthalpy of reaction (\Delta H^o).

\Delta H^o=H_f_{product}-H_f_{reactant}

\Delta H^o=[n_{H_2O}\times \Delta H_f^0_{(H_2O)}+n_{SO_2}\times \Delta H_f^0_{(SO_2)}]-[n_{H_2S}\times \Delta H_f^0_{(H_2S)}+n_{O_2}\times \Delta H_f^0_{(O_2)}]

where,

\Delta H^o = enthalpy of reaction = ?

n = number of moles

\Delta H_f^0 = standard enthalpy of formation

Now put all the given values in this expression, we get:

\Delta H^o=[2mole\times (-242kJ/mol)+2mole\times (-296.8kJ/mol)}]-[2mole\times (-21kJ/mol)+3mole\times (0kJ/mol)]

\Delta H^o=-1035.6kJ=-1035600J

conversion used : (1 kJ = 1000 J)

Now we have to calculate the entropy of reaction (\Delta S^o).

\Delta S^o=S_f_{product}-S_f_{reactant}

\Delta S^o=[n_{H_2O}\times \Delta S_f^0_{(H_2O)}+n_{SO_2}\times \Delta S_f^0_{(SO_2)}]-[n_{H_2S}\times \Delta S_f^0_{(H_2S)}+n_{O_2}\times \Delta S_f^0_{(O_2)}]

where,

\Delta S^o = entropy of reaction = ?

n = number of moles

\Delta S_f^0 = standard entropy of formation

Now put all the given values in this expression, we get:

\Delta S^o=[2mole\times (189J/K.mol)+2mole\times (248J/K.mol)}]-[2mole\times (206J/K.mol)+3mole\times (205J/K.mol)]

\Delta S^o=-153J/K

Now we have to calculate the Gibbs free energy of reaction (\Delta G^o).

As we know that,

\Delta G^o=\Delta H^o-T\Delta S^o

At room temperature, the temperature is 500 K.

\Delta G^o=(-1035600J)-(500K\times -153J/K)

\Delta G^o=-959100J=-959.1kJ

Therefore, the value of \Delta G^o for the reaction is -959.1 kJ

3 0
3 years ago
Which of these elements is unlikely to have a reaction with any element or compound?
Kay [80]
Argon is a noble gas. Argon has a full outer shell. This makes it so that it does not need to react with any of the other elements to be stable.


With Rubidium and Cobalt its a whole different story.

I hope that helps!
8 0
3 years ago
How can you tell which ionic compounds will react in an aqueous solution
Darina [25.2K]
1.  Determine if the ionic substances can break apart into ions.
 - e.g. CaCO3 isn't very soluble, do it can't dissolve and dissociate.  If it can't pop apart, no ions.
2.  Swap the partners for all the other ions that you can get from step 1.  You can skip pairings with the same charge - a + can't get close to another + to react.
3.  Use solubility, acid/base, and redox rules to see if anything will happen with the ions in solution.<span />
3 0
3 years ago
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