Major Plates
Africa Plate
Antarctic Plate
Indo-Australian Plate
Australian Plate
Eurasian Plate
North American Plate
South American Plate
<span>Pacific Plate
Minor Plates
There are dozens of smaller plates, the seven largest of which are:
</span>Arabian Plate
Caribbean Plate
Juan de Fuca Plate
Cocos Plate
Nazca Plate
Philippine Sea Plate
<span>Scotia Plate</span>
the answer of these question is weight
Answer:
kJ/mol
Explanation: <u>Enthalpy</u> <u>Change</u> is the amount of energy in a reaction - absorption or release - at a constant pressure. So, <u>Standard</u> <u>Enthalpy</u> <u>of</u> <u>Formation</u> is how much energy is necessary to form a substance.
The standard enthalpy of formation of HCl is calculated as:

→ 
Standard Enthalpy of formation for the other compounds are:
Calcium Hydroxide:
-1002.82 kJ/mol
Calcium chloride:
-795.8 kJ/mol
Water:
-285.83 kJ/mol
Enthalpy is given per mol, which means we have to multiply by the mols in the balanced equation.
Calculating:
![-17.2=[-795.8+2(285.85)]-[-1002.82+2\Delta H]](https://tex.z-dn.net/?f=-17.2%3D%5B-795.8%2B2%28285.85%29%5D-%5B-1002.82%2B2%5CDelta%20H%5D)



So, the standard enthalpy of formation of HCl is -173.72 kJ/mol
The integrated rate law for a second-order reaction is given by:
![\frac{1}{[A]t} = \frac{1}{[A]0} + kt](https://tex.z-dn.net/?f=%20%5Cfrac%7B1%7D%7B%5BA%5Dt%7D%20%3D%20%20%20%5Cfrac%7B1%7D%7B%5BA%5D0%7D%20%2B%20kt%20)
where, [A]t= the concentration of A at time t,
[A]0= the concentration of A at time t=0
<span>k =</span> the rate constant for the reaction
<u>Given</u>: [A]0= 4 M, k = 0.0265 m–1min–1 and t = 180.0 min
Hence, ![\frac{1}{[A]t} = \frac{1}{4} + (0.0265 X 180)](https://tex.z-dn.net/?f=%20%5Cfrac%7B1%7D%7B%5BA%5Dt%7D%20%3D%20%5Cfrac%7B1%7D%7B4%7D%20%2B%20%280.0265%20X%20180%29%20)
<span> = 4.858</span>
<span><span><span>Therefore, [A]</span>t</span>= 0.2058 M.</span>
<span>
</span>
<span>Answer: C</span>oncentration of A, after 180 min, is 0.2058 M
Carbonates
OPTION A is the correct answer