Answer is: pH value of weak is 3.35.
Chemical reaction (dissociation): HA(aq) → H⁺(aq) + A⁻(aq).
c(HA) = 0.0055 M.
α = 8.2% ÷ 100% = 0.082.
[H⁺] = c(HA) · α.
[H⁺] = 0.0055 M · 0.082.
[H⁺] = 0.000451 M.
pH = -log[H⁺].
pH = -log(0.000451 M).
pH = 3.35.
pH (potential of
hydrogen) is a numeric scale used to specify the acidity or basicity <span>an aqueous solution.</span>
Protons and neutrons are located in the nucleus of the atom while the electrons move in the trajectory of the shell
<h3>Further explanation
</h3>
Isotopes are atoms whose no-atom has the same number of protons while still having a different number of neutrons.
So Isotopes are elements that have the same Atomic Number (Proton)
Isotopes of Helium : helium-3 and helium-4
protons = 2
electrons=protons=2
neutron=mass number-atomic number=3-2=1
protons = 2
electrons=protons=2
neutron=mass number-atomic number=3-2=1
protons = 2
electrons=protons=2
neutron=mass number-atomic number=4-2=2
Protons and neutron in the nucleus, electrons in the shell
To find the net ionic equation we must first write the balanced equation for the reaction. We must bear in mind that the reagents Ca(NO3)2 and Na2S are in the aqueous state and as product we will have CaS in the solid state, since it is not soluble in water and NaNO3 in the aqueous state.
The balanced equation of the reaction will be:
Ca(NO3)2(aq) + → Ca(aq) + 2Na(s)NO3Now, c(aq)ompounds in the aqueous state can be written in their ionic form, so the reaction will transform into:Na2S +
So, the answer will be option A
No it does not ALWAYS go through all the stages because each rock forms differently at different rates. Igneous rock can weather to sediement rock, or melt to magma, however this does not necessarily mean it does this every time. It all depends on the temperature, wind, and pressure/heat.
There are two ways to solve this problem. We can use the ICE method which is tedious and lengthy or use the Henderson–Hasselbalch equation. This equation relates pH and the concentration of the ions in the solution. It is expressed as
pH = pKa + log [A]/[HA]
where pKa = - log [Ka]
[A] is the concentration of the conjugate base
[HA] is the concentration of the acid
Given:
Ka = 1.8x10^-5
NaOH added = 0.015 mol
HC2H3O2 = 0.1 mol
NaC2H3O2 = 0.1 mol
Solution:
pKa = - log ( 1.8x10^-5) = 4.74
[A] = 0.015 mol + 0.100 mol = .115 moles
[HA] = .1 - 0.015 = 0.085 moles
pH = 4.74 + log (.115/0.085)
pH = 4.87
