482VP I think is the correct answer.
What is your question please write properly>-< >_
pH=4.625
The classification of this sample of saliva : acid
<h3>Further explanation</h3>
The water equilibrium constant (Kw) is the product of concentration
the ions:
Kw = [H₃O⁺] [OH⁻]
Kw value at 25° C = 10⁻¹⁴
It is known [OH-] = 4.22 x 10⁻¹⁰ M
then the concentration of H₃O⁺:
![\tt 10^{-14}=4.22\times 10^{-10}\times [H_3O^+]\\\\(H_3O^+]=\dfrac{10^{-14}}{4.22\times 10^{-10}}=2.37\times 10^{-5}](https://tex.z-dn.net/?f=%5Ctt%2010%5E%7B-14%7D%3D4.22%5Ctimes%2010%5E%7B-10%7D%5Ctimes%20%5BH_3O%5E%2B%5D%5C%5C%5C%5C%28H_3O%5E%2B%5D%3D%5Cdfrac%7B10%5E%7B-14%7D%7D%7B4.22%5Ctimes%2010%5E%7B-10%7D%7D%3D2.37%5Ctimes%2010%5E%7B-5%7D)
pH=-log[H₃O⁺]
Saliva⇒acid(pH<7)
Answer:
P = 27.9 atm
Explanation:
Given data:
Mass of CO₂ = 25 g
Temperature = 25°C (25+273.15 K = 298.15 K)
Volume of gas = 0.50 L
Pressure of gas = ?
Solution:
Firs of all we will calculate the number of moles of gas,
Number of moles = mass/molar mass
Number of moles = 25 g/ 44 g/mol
Number of moles = 0.57 mol
Pressure of gas :
PV = nRT
P= Pressure
V = volume
n = number of moles
R = general gas constant = 0.0821 atm.L/ mol.K
T = temperature in kelvin
P × 0.50 L = 0.57 mol × 0.0821 atm.L/ mol.K × 298.15 K
P = 13.95 atm.L/ 0.50 L
P = 27.9 atm