Answer:
Formula weight of H₂O molecule is 18.02 amu.
Explanation:
Given data:
Formula weight of H₂O = ?
Atomic mass of H = 1.008 amu
Atomic mass of O = 16.00 amu
Solution:
Formula weight:
"It is the sum of all the atomic weight of atoms present in given formula"
Formula weight of H₂O = 2×1.008 amu + 1×16.00 amu
Formula weight of H₂O = 18.02 amu
Thus, formula weight of H₂O molecule is 18.02 amu.
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 new pressure if the volume changes to 560.0 mL is, 280 mmHg
Explanation :
According to the Boyle's, law, the pressure of the gas is inversely proportional to the volume of gas at constant temperature and moles of gas.

or,

where,
= initial pressure = 560.00 mmHg
= final pressure = ?
= initial volume = 280 mL
= final volume = 560.0 mL
Now put all the given values in the above formula, we get:


Therefore, the new pressure if the volume changes to 560.0 mL is, 280 mmHg
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