In order for plants to go through photosynthesis, they must use <span>B. visible light.</span>
The function does residual Co2 plays big part and in
maintaining the body’s homeostasis. The addition of respiratory to reserve
volume or residual volume. The lung is
the one that who protect the organs, so in exchange of oxygen and carbon
dioxide.
Answer:
a) Ksp = 7.9x10⁻¹⁰
b) Solubility is 6.31x10⁻⁶M
Explanation:
a) InF₃ in water produce:
InF₃ ⇄ In⁺³ + 3F⁻
And Ksp is defined as:
Ksp = [In⁺³] [F⁻]³
4.0x10⁻²g / 100mL of InF₃ are:
4.0x10⁻²g / 100mL ₓ (1mol / 172g) ₓ (100mL / 0.1L) = <em>2.3x10⁻³M InF₃. </em>Thus:
[In⁺³] = 2.3x10⁻³M InF₃ × (1 mol In⁺³ / mol InF₃) = 2.3x10⁻³M In⁺³
[F⁻] = 2.3x10⁻³M InF₃ × (3 mol F⁻ / mol InF₃) = 7.0x10⁻³M F⁻
Replacing these values in Ksp formula:
Ksp = [2.3x10⁻³M In⁺³] × [7.0x10⁻³M F⁻]³ = <em>7.9x10⁻¹⁰</em>
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b) 0.05 moles of F⁻ produce solubility of InF₃ decrease to:
7.9x10⁻¹⁰ = [x] [0.05 + 3x]³
Where x are moles of In⁺³ produced from solid InF₃ and 3x are moles of F⁻ produced from the same source. That means x is solubility in mol / L
Solving from x:
x = -0.018 → False solution, there is no negative concentrations.
x = 6.31x10⁻⁶M → Right answer.
Thus, <em>solubility is 6.31x10⁻⁶M</em>
Answer : The electron configurations consistent with this fact is, (b) [Kr] 4d¹⁰
Explanation :
Electronic configuration : It is defined as the representation of electrons around the nucleus of an atom.
Number of electrons in an atom are determined by the electronic configuration.
Paramagnetic compounds : They have unpaired electrons.
Diamagnetic compounds : They have no unpaired electrons that means all are paired.
The given electron configurations of Palladium are:
(a) [Kr] 5s²4d⁸
In this, there are 2 electrons in 's' orbital and 8 electrons in 'd' orbital. From the partial orbital diagrams we conclude that 's' orbital are paired but 'd' orbital are not paired. So, this configuration shows paramagnetic.
(b) [Kr] 4d¹⁰
In this, there are 10 electrons in 'd' orbital. From the partial orbital diagrams we conclude that electrons in 'd' orbital are paired. So, this configuration shows diamagnetic.
(c) [Kr] 5s¹4d⁹
In this, there are 1 electron in 's' orbital and 9 electrons in 'd' orbital. From the partial orbital diagrams we conclude that 's' orbital and 'd' orbital are not paired. So, this configuration shows paramagnetic.