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>
<em />
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>
Explanation:
The value of equilibrium constant doesn't change when a catalyst is added.
Equilibrium constant depends on Concentration of reactants , Pressure and Temperature.
Answer : The final temperature of the metal block is, 
Explanation :
As we know that,
![m_1\times c_1\times (T_{final}-T_1)=-[m_2\times c_2\times (T_{final}-T_2)]](https://tex.z-dn.net/?f=m_1%5Ctimes%20c_1%5Ctimes%20%28T_%7Bfinal%7D-T_1%29%3D-%5Bm_2%5Ctimes%20c_2%5Ctimes%20%28T_%7Bfinal%7D-T_2%29%5D)
.................(1)
where,
q = heat absorbed or released
= mass of aluminum = 55 g
= mass of water = 0.48 g
= final temperature = ?
= temperature of aluminum =
= temperature of water =
= specific heat of aluminum = 
= specific heat of water= 
Now put all the given values in equation (1), we get
![55g\times 0.900J/g^oC\times (T_{final}-25)^oC=-[0.48g\times 4.184J/g^oC\times (T_{final}-25)^oC]](https://tex.z-dn.net/?f=55g%5Ctimes%200.900J%2Fg%5EoC%5Ctimes%20%28T_%7Bfinal%7D-25%29%5EoC%3D-%5B0.48g%5Ctimes%204.184J%2Fg%5EoC%5Ctimes%20%28T_%7Bfinal%7D-25%29%5EoC%5D)
Thus, the final temperature of the metal block is,
Answer:
Different isotopes of the same element have the same number of protons in their atomic nuclei but differing numbers of neutrons. Radioisotopes are radioactive isotopes of an element. They can also be defined as atoms that contain an unstable combination of neutrons and protons, or excess energy in their nucleus.
Answer:
1.33 Å
Explanation:
Given that the edge length , a of the KCl which forms the FCC lattice = 6.28 Å
Also,
For the FCC lattice in which the anion-cation contact along the cell edge , the ratio of the radius of the cation to that of anion is 0.731.
Thus,
.................1
Also, the sum of the radius of the cation and the anion in FCC is equal to half of the edge length.
Thus,
...................2
Given that:
To find,
Using 1 and 2 , we get:
<u>Size of the potassium ion = 1.33 Å</u>
