Answer:
1.87x10⁻³ M SO₄²⁻
Explanation:
The reaction of SO₄²⁻ with Ba²⁺ (From Ba(NO₃)₂) is:
SO₄²⁻(aq) + Ba²⁺(aq) → BaSO₄(s)
<em>Where 1 mole of SO₄²⁻ reacts per mole of Ba²⁺</em>
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To reach the end point in this titration, we need to add the same moles of Ba²⁺ that the moles that are of SO₄²⁻.
Thus, to find molarity of SO₄²⁻ we need to find first the moles of Ba²⁺ added (That will be the same of SO₄²⁻). And as the volume of the initial sample was 100mL we can find molarity (As ratio of moles of SO₄²⁻ per liter of solution).
<em>Moles Ba²⁺:</em>
7.48mL = 7.48x10⁻³L ₓ (0.0250moles / L) = 1.87x10⁻⁴ moles of Ba²⁺ = Moles of SO₄²⁻
<em>Molarity SO₄²⁻:</em>
As there are 1.87x10⁻⁴ moles of SO₄²⁻ in 100mL = 0.1L, molarity is:
1.87x10⁻⁴ moles of SO₄²⁻ / 0.1L =
<h3> 1.87x10⁻³ M SO₄²⁻</h3>
Answer:
The answer is
<h2>59.6 g </h2>
Explanation:
The mass of a substance when given the density and volume can be found by using the formula
<h3>mass = Density × volume</h3>
From the question
Density of aluminum = 2.00 g/mL
volume = 29.8 mL
The mass is
mass = 2 × 29.8
We have the final answer as
<h3>59.6 g</h3>
Hope this helps you
values of the quantum numbers: -6,-5,-4,-3,-2,-1,0,1,2,3,4,5,6
location of the electron: In the 7th energy level away from the nucleus.
Explanation:
From the description of the problem, the magnetic number is given is as -6,-5,-4,-3,-2,-1,0,1,2,3,4,5,6 and the electron is located in the 7th energy level away from the nucleus. Basically, the problem is testing for the understanding of the principal quantum numbers which gives the location of electrons and the magnetic quantum number that shows the spatial orientation of the orbitals.
The orbital designation of the describe electron is 7d
- Magnetic quantum number is limited by the azimuthal quantum number which is the quantum number describing the possible shapes. The azimuthal is given as L= n-1. "n" is the principal quantum number which is 7. Therefore L is 6 and the magnetic quantum numbers are -6,-5,-4,-3,-2,-1,0,1,2,3,4,5,6
- The position of the electron is given by the principal quantum number which represents the main energy level in which the orbital is located or the average distance from the nucleus. Here it is 7.
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