Answer:
An example of these would be "I have always been interested in cars so I became a auto mechanic." or "I have always been interested in cars so I decided to work at a car dealership."
Explanation:
A big interest that you have that leds you to pick a job related to that interest.
Answer:
odorless, crystalline, white solid with a sour taste.
Explanation:
Answer:
The reactions free energy 
Explanation:
From the question we are told that
The pressure of (NO) is 
The pressure of (Cl) gas is 
The pressure of nitrosly chloride (NOCl) is 
The reaction is
⇆ 
From the reaction we can mathematically evaluate the
(Standard state free energy ) as

The Standard state free energy for NO is constant with a value

The Standard state free energy for
is constant with a value

The Standard state free energy for
is constant with a value

Now substituting this into the equation

The pressure constant is evaluated as

Substituting values


The free energy for this reaction is evaluated as

Where R is gas constant with a value of 
T is temperature in K with a given value of 
Substituting value
![\Delta G = -43 *10^{3} + 8.314 *298 * ln [0.0765]](https://tex.z-dn.net/?f=%5CDelta%20%20G%20%20%3D%20-43%20%2A10%5E%7B3%7D%20%2B%208.314%20%2A298%20%2A%20ln%20%5B0.0765%5D)


Answer:
yes
Explanation:
Solubility is an observation and no chemical reaction takes place. The composition of the compound/element is not changed.
- Hope that helped! Please let me know if you need further explanation.