Strontium sulfate becomes less soluble in an aqueous solution when sodium sulfator is added because
2 answers:
Answer:
The addition of sulfate ions shifts equilibrium to the left.
Explanation:
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In this case, according to the following ionization of strontium sulfate:
It is evidenced that when sodium sulfate is added, sulfate, is actually added in to the solution, which causes the equilibrium to shift leftwards according to the Le Ch athelier's principle. Thus, the answer in this case would be:
The addition of sulfate ions shifts equilibrium to the left.
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Answer:
The wavelength of the emitted photon will be approximately 655 nm, which corresponds to the visible spectrum.
Explanation:
In order to answer this question, we need to recall Bohr's formula for the energy of each of the orbitals in the hydrogen atom:
, where:
[tex]m_{e}[tex] = electron mass
e = electron charge
[tex]\epsilon_{0}[tex] = vacuum permittivity
[tex]\hbar[tex] = Planck's constant over 2pi
n = quantum number
[tex]E_{1}[tex] = hydrogen's ground state = -13.6 eV
Therefore, the energy of the emitted photon is given by the difference of the energy in the 3d orbital minus the energy in the 2nd orbital:
[tex]E_{3} - E_{2} = -13.6 eV(\frac{1}{3^{2}} - \frac{1}{2^{2}})=1.89 eV[tex]
Now, knowing the energy of the photon, we can calculate its wavelength using the equation:
[tex]E = \frac{hc}{\lambda}[tex], where:
E = Photon's energy
h = Planck's constant
c = speed of light in vacuum
[tex]\lambda[tex] = wavelength
Solving for [tex]\lambda[tex] and substituting the required values:
[tex]\lambda = \frac{hc}{E} = \frac{1.239 eV\mu m}{1.89 eV}=0.655\mu m = 655 nm[tex], which correspond to the visible spectrum (The visible spectrum includes wavelengths between 400 nm and 750 nm).