Answer:
An atom of Al which has 13 protons and 10 electrons is Al cation (Al⁺³)
Explanation:
An atom consist of electron, protons and neutrons. Protons and neutrons are present with in nucleus while the electrons are present out side the nucleus.
All these three subatomic particles construct an atom. A neutral atom have equal number of proton and electron. In other words we can say that negative and positive charges are equal in magnitude and cancel the each other.
For example,
Al atom has 13 protons and 13 electrons. The number of positive and negative charge is equal thus it will be neutral atom.
While the atom of Al which have 13 proton and 10 electron is not neutral. The positive charge is greater than negative by 3. Which means 3 electrons are lose by Al atom and form cation "Al⁺³".
Thus an atom of Al which has 13 protons and 10 electrons is Al cation (Al⁺³)
45 molecules of chlorine gas (Cl₂) are needed to react with 30 atoms of aluminum (Al)
The balanced equation for the reaction is given below:
2Al + 3Cl₂ —> 2AlCl₃
From the balanced equation above,
2 atoms of Al required 3 molecules of Cl₂.
With the above information, we can determine the number of molecules of Cl₂ needed to react with 30 atoms of Al. This can be obtained as follow:
From the balanced equation above,
2 atoms of Al required 3 molecules of Cl₂.
Therefore,
30 atoms of Al will require =
= 45 molecules of Cl₂.
Thus, 45 molecules of chlorine gas (Cl₂) are needed to react with 30 atoms of aluminum (Al)
Learn more: brainly.com/question/24918379
Answer:

Explanation:
Firstly, write the expression for the equilibrium constant of this reaction:
![K_{eq} = \frac{[ADP][Pi]}{ATP}](https://tex.z-dn.net/?f=K_%7Beq%7D%20%3D%20%5Cfrac%7B%5BADP%5D%5BPi%5D%7D%7BATP%7D)
Secondly, we may relate the change in Gibbs free energy to the equilibrium constant using the equation below:

From here, rearrange the equation to solve for K:

Now we know from the initial equation that:
![K_{eq} = \frac{[ADP][Pi]}{ATP}](https://tex.z-dn.net/?f=K_%7Beq%7D%20%3D%20%5Cfrac%7B%5BADP%5D%5BPi%5D%7D%7BATP%7D)
Let's express the ratio of ADP to ATP:
![\frac{[ADP]}{[ATP]} = \frac{[Pi]}{K_{eq}}](https://tex.z-dn.net/?f=%5Cfrac%7B%5BADP%5D%7D%7B%5BATP%5D%7D%20%3D%20%5Cfrac%7B%5BPi%5D%7D%7BK_%7Beq%7D%7D)
Substitute the expression for K:
![\frac{[ADP]}{[ATP]} = \frac{[Pi]}{K_{eq}} = \frac{[Pi]}{e^{-\frac{\Delta G^o}{RT}}}](https://tex.z-dn.net/?f=%5Cfrac%7B%5BADP%5D%7D%7B%5BATP%5D%7D%20%3D%20%5Cfrac%7B%5BPi%5D%7D%7BK_%7Beq%7D%7D%20%3D%20%5Cfrac%7B%5BPi%5D%7D%7Be%5E%7B-%5Cfrac%7B%5CDelta%20G%5Eo%7D%7BRT%7D%7D%7D)
Now we may use the values given to solve:
![\frac{[ADP]}{[ATP]} = \frac{[Pi]}{K_{eq}} = \frac{[Pi]}{e^{-\frac{\Delta G^o}{RT}}} = [Pi]e^{\frac{\Delta G^o}{RT}} = 1.0 M\cdot e^{\frac{-30 kJ/mol}{2.5 kJ/mol}} = 6.14\cdot 10^{-6}](https://tex.z-dn.net/?f=%5Cfrac%7B%5BADP%5D%7D%7B%5BATP%5D%7D%20%3D%20%5Cfrac%7B%5BPi%5D%7D%7BK_%7Beq%7D%7D%20%3D%20%5Cfrac%7B%5BPi%5D%7D%7Be%5E%7B-%5Cfrac%7B%5CDelta%20G%5Eo%7D%7BRT%7D%7D%7D%20%3D%20%5BPi%5De%5E%7B%5Cfrac%7B%5CDelta%20G%5Eo%7D%7BRT%7D%7D%20%3D%201.0%20M%5Ccdot%20e%5E%7B%5Cfrac%7B-30%20kJ%2Fmol%7D%7B2.5%20kJ%2Fmol%7D%7D%20%3D%206.14%5Ccdot%2010%5E%7B-6%7D)
Answer:
Chromosphere
Explanation:
You see the middle layer of the sun’s atmosphere, the Chromosphere, at the start and end of a total eclipse.
Answer:
Hmm
Explanation:
Elemental antimony adults a layer structure (space group R3m No.166) in which layers consist of fused, ruffled, six-membered rings. The nearest and the next nearest neighbors form and irregular octahedral complex, with the 3 atoms in each layer slightly closer than 3 atoms in the next
Electric configuration : 4d^10 5s^2 5p^3