The empirical formula for the compound is - P₂O₅
the empirical formula is the simplest ratio of whole numbers of components in a compound.
molecular formula is the actual ratio of components in a compound.
we have to first find the number of empirical units in the molecular formula
molecular mass - 283.89 g/mol
mass of empirical formula - 283.8 g
number of empirical units - 283.89 g/mol / 283.8 g
number of empirical units - 1.000
therefore empirical formula = molecular formula
molecular formula - P₂O₅
Answer:
The value of dissociation constant of the monoprotic acid is
.
Explanation:
The pH of the solution = 2.46
![pH=-\log[H^+]](https://tex.z-dn.net/?f=pH%3D-%5Clog%5BH%5E%2B%5D)
![2.46=-\log[H^+]](https://tex.z-dn.net/?f=2.46%3D-%5Clog%5BH%5E%2B%5D)
![[H^+]=0.003467 M](https://tex.z-dn.net/?f=%5BH%5E%2B%5D%3D0.003467%20M)
![HA\rightleftharpoons H^++A^-](https://tex.z-dn.net/?f=HA%5Crightleftharpoons%20H%5E%2B%2BA%5E-)
Initially
0.0144 0 0
At equilibrium
(0.0144-x) x x
The expression if an dissociation constant is given by :
![K_a=\frac{[A^-][H^+]}{[HA]}](https://tex.z-dn.net/?f=K_a%3D%5Cfrac%7B%5BA%5E-%5D%5BH%5E%2B%5D%7D%7B%5BHA%5D%7D)
![K_a=\frac{x\times x}{(0.0144-x)}](https://tex.z-dn.net/?f=K_a%3D%5Cfrac%7Bx%5Ctimes%20x%7D%7B%280.0144-x%29%7D)
![x=[H^+]=0.003467 M](https://tex.z-dn.net/?f=x%3D%5BH%5E%2B%5D%3D0.003467%20M)
![K_a=\frac{0.003467 \times 0.003467 }{(0.0144-0.003467 )}](https://tex.z-dn.net/?f=K_a%3D%5Cfrac%7B0.003467%20%5Ctimes%200.003467%20%7D%7B%280.0144-0.003467%20%29%7D)
![K_a=1.099\times 10^{-3}](https://tex.z-dn.net/?f=K_a%3D1.099%5Ctimes%2010%5E%7B-3%7D)
The value of dissociation constant of the monoprotic acid is
.
The partial pressure of oxygen in a sample of air increases if the temperature is increased.
Answer: Option 1
<u>Explanation:
</u>
According to Guy-Lussac's law, at constant volume, pressure exhibited by the gas molecules will be directly proportional to the temperature of the gas molecules. It is also known that pressure of mixture of gas molecules is the sum of partial pressure of each gas molecule in the mixture.
If the temperature increases, the partial pressure and the pressure of the mixture of gas also tend to increase. As it can be seen that at higher altitudes, the low temperature leads to the decrease in oxygen's partial pressure in the air.
So, it can also be concluded that temperature increases the oxygen's partial pressure in air increases.
Answer : The new volume of the air is, 6.83 L
Explanation :
Charles' Law : It states that volume of the gas is directly proportional to the temperature of the gas at constant pressure.
Mathematically,
![\frac{V_1}{T_1}=\frac{V_2}{T_2}](https://tex.z-dn.net/?f=%5Cfrac%7BV_1%7D%7BT_1%7D%3D%5Cfrac%7BV_2%7D%7BT_2%7D)
where,
are the initial volume and temperature of the gas.
are the final volume and temperature of the gas.
We are given:
![V_1=5.00L\\T_1=0^oC=(0+273)K=273K\\V_2=?\\T_2=100^oC=(100+273)K=373K](https://tex.z-dn.net/?f=V_1%3D5.00L%5C%5CT_1%3D0%5EoC%3D%280%2B273%29K%3D273K%5C%5CV_2%3D%3F%5C%5CT_2%3D100%5EoC%3D%28100%2B273%29K%3D373K)
Putting values in above equation, we get:
![\frac{5.00L}{273K}=\frac{V_2}{373K}\\\\V_2=6.83L](https://tex.z-dn.net/?f=%5Cfrac%7B5.00L%7D%7B273K%7D%3D%5Cfrac%7BV_2%7D%7B373K%7D%5C%5C%5C%5CV_2%3D6.83L)
Therefore, the new volume of the air is, 6.83 L