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:
See explanation and image attached
Explanation:
A bond line structure refers to any structure of a covalent molecule wherein the covalent bonds present in the molecule are represented with a single line for each level of bond order.
The bond-line structure of CH3CH2O(CH2)2CH(CH3)2 has been shown in the image attached. We know that oxygen has a lone pair of electrons and this has been clearly shown also in the image attached.
Answer:
23.92 g
Explanation:
Molar mass of H2SO4 = (2×1)+32+(16×4)= 2+32+48= 82g/mol
H2SO4 + 2NaOH ---> Na2SO4 + 2H2O
I mole of H2SO4 = 2 moles of NaOH
24.5/82 = 24.5/82 × 2
= 0.598 moles of NaOH will neutralize
Mass= mole× molar mass
Molar mass of NaOH= 23+16+1 = 40g/mol
Mass= 0.598 × 40 = 23.92g of NaOH