4.48
pH=pKa+log([A-/HA])
25% deprotonated tells us that A- is .25 and that the rest (75% is protonated) thats .75.
4 = pKa + log
![\frac{.25}{.75}](https://tex.z-dn.net/?f=%20%5Cfrac%7B.25%7D%7B.75%7D%20)
4 - log
![\frac{.25}{.75}](https://tex.z-dn.net/?f=%20%5Cfrac%7B.25%7D%7B.75%7D%20)
= pKa
4.48=pKa
Covalent compounds are generally not very hard because they are formed by two or more nonmetallic atoms.
<h3>COVALENT COMPOUNDS:</h3>
Covalent compounds are compounds whose constituent elements are joined together by covalent bonds.
Covalent bonding occurs when two or more nonmetallic atoms of an element share valence electrons. This means that covalent compounds will not be physically hard since they constitute non-metals.
Examples of covalent compounds are:
- H2 - hydrogen
- H2O - water
- HCl - hydrogen chloride
- CH4 - methane
Learn more about covalent compounds at: brainly.com/question/21505413
Answer:
Volume of sample after droping into the ocean=0.0234L
Explanation:
As given in the question that gas is idealso we can use ideal gas equation to solve this;
Assuming that temperature is constant;
Lets
and
are the initial gas parameter before dropping into the ocean
and
and
are the final gas parameter after dropping into the ocean
according to boyle 's law pressure is inversly proportional to the volume at constant temperature.
hence,
![P_1V_1=P_2V_2](https://tex.z-dn.net/?f=P_1V_1%3DP_2V_2)
P1=1 atm
V1=1.87L
P2=80atm
V2=?
After putting all values we get;
V2=0.0234L
Volume of sample after droping into the ocean=0.0234L
Answer:
V₂ = 1.92 L
Explanation:
Given data:
Initial volume = 0.500 L
Initial pressure =2911 mmHg (2911/760 = 3.83 atm)
Initial temperature = 0 °C (0 +273 = 273 K)
Final temperature = 273 K
Final volume = ?
Final pressure = 1 atm
Solution:
Formula:
P₁V₁/T₁ = P₂V₂/T₂
P₁ = Initial pressure
V₁ = Initial volume
T₁ = Initial temperature
P₂ = Final pressure
V₂ = Final volume
T₂ = Final temperature
by putting values,
V₂ = P₁V₁ T₂/ T₁ P₂
V₂ = 3.83 atm × 0.500 L × 273 K / 273 K × 1 atm
V₂ = 522.795 atm .L. K / 273 K.atm
V₂ = 1.92 L