Answer:
Please answer my question
Answer:
Aluminum
Explanation:
Given




<em>See attachment for chart</em>
Required
Identify the unknown substance
To do this, we simply calculate the specific heat capacity from the given parameters using:

This gives:

So, we have:




From the attached chart, we have:
--- The specific heat capacity of Aluminum
<em>Hence, the unknown substance is Aluminum</em>
Mass spectrum of Dodecane will give following information.
1 ) Molecular Peak or Parent Peak:
The Parent peak will appear at m/z = 170. The intensity of this peak will be very weak.
2) Fragments:
Usually the fragments of such long chain alkanes appear with spacing of 14 amu, Hence, the peaks in dodecane will be as follow,
170 - 156 - 142 - 128 - 114 - 100 - 86 - 72 - 58 - 44 - 30 - 16
3) Base Peak:
Most probably the Base peak will appear at m/z = 57. This peak is due to the formation of tertiary butyl cation as the intensity mainly depends upon the stability of cation. So this cation might form due to rearrangment giving the intensity of 100%.
Answer:


Explanation:
first write the equilibrium equaion ,
⇄ 
assuming degree of dissociation
=1/10;
and initial concentraion of
=c;
At equlibrium ;
concentration of
![[C_3H_5O_3^{-} ]= c\alpha](https://tex.z-dn.net/?f=%5BC_3H_5O_3%5E%7B-%7D%20%20%5D%3D%20c%5Calpha)
![[H^{+}] = c\alpha](https://tex.z-dn.net/?f=%5BH%5E%7B%2B%7D%5D%20%3D%20c%5Calpha)

is very small so
can be neglected
and equation is;

= 
![P_H =- log[H^{+} ]](https://tex.z-dn.net/?f=P_H%20%3D-%20log%5BH%5E%7B%2B%7D%20%5D)





composiion ;
![c=\frac{1}{\alpha} \times [H^{+}]](https://tex.z-dn.net/?f=c%3D%5Cfrac%7B1%7D%7B%5Calpha%7D%20%5Ctimes%20%5BH%5E%7B%2B%7D%5D)
![[H^{+}] =antilog(-P_H)](https://tex.z-dn.net/?f=%5BH%5E%7B%2B%7D%5D%20%3Dantilog%28-P_H%29)
![[H^{+} ] =0.0014](https://tex.z-dn.net/?f=%5BH%5E%7B%2B%7D%20%5D%20%3D0.0014)


Answer:
The final balanced equation is :

Explanation:

Balancing in acidic medium:
First we will determine the oxidation and reduction reaction from the givne reaction :
Oxidation:

Balance the charge by adding 2 electrons on product side:
....[1]
Reduction :

Balance O by adding water on required side:

Now, balance H by adding
on the required side:

At last balance the charge by adding electrons on the side where positive charge is more:
..[2]
Adding [1] and [2]:

The final balanced equation is :
