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DIA [1.3K]
2 years ago
13

If excess caso4(s) is mixed with water at 25 ∘c to produce a saturated solution of caso4, what is the equilibrium concentration

of ca2+?
Chemistry
1 answer:
Ray Of Light [21]2 years ago
3 0

<em>Answer:</em>

The equlibrium concentration sof Ca+2 ion willl be 4.9×10∧-3 M

<em>Data Given:</em>

              Ksp of CaSO4 = 2.4 × 10∧-5

              CaSO4 ⇔ Ca+2   +  SO4∧-2

<em>Solution:</em>

                Ksp = [Ca+2].[ SO4∧-2]

                 2.4 × 10∧-5 = [x].[x]= x²

                 x =  4.9×10∧-3 M

<em>Result:</em>

  • The conc. of Ca+2 ion is 4.9×10∧-3 M
You might be interested in
Fe3+(aq)+6H2O(l)⇌Fe(H2O)63+(aq) : F e 3 + ( a q ) + 6 H 2 O ( l ) ⇌ F e ( H 2 O ) 6 3 + ( a q ) : blank is the Lewis acid and bl
Tasya [4]

Answer:

Lewis acid- Fe3+

Lewis base- water molecule

Explanation:

Acids and bases have been defined in diverse ways. There have been definitions put forward by Arrhenius, Brownstead and Lowry as well as Lewis. Each definition his useful in its own way.

Lewis acids are lone pair acceptors such as metal ions. This implies that in the particular instance of this reaction, Fe3+ is the lewis acid.

Similarly, a Lewis base is a lone pair donor, all ligands are lone pair donors since they donate one or more lone pairs of electrons to Lewis acids. In the particular instance of this reaction, the Lewis base is the water molecule.

6 0
2 years ago
A student is given a sample of CuSO4(s) that contains a solid impurity that is soluble and colorless. The student wants to deter
DochEvi [55]

Answer:

The impurity which is present in the solution of copper sulphate (CuSO4) is determined by the an instrument known as spectrophotometer.

Explanation:

Spectrophotometer is a device or an instrument which is used to determine the concentration of a chemical by measuring the detection of light intensity that is coming from the solution. If the solution of copper sulphate is checked through spectrophotometer, we can can determined or measure the amount of copper sulphate and the impurity in the solution.

3 0
2 years ago
Determine whether or not the mixing of each of the two solutions indicated below will result in a buffer.
WARRIOR [948]
Part A

75.0 mL of 0.10 M HF; 55.0 mL of 0.15 M NaF

This combination will form a buffer.

Explanation

Here, weak acid HF and its conjugate base F- is available in the solution

Part B

150.0 mL of 0.10 M HF; 135.0 mL of 0.175 M HCl

This combination cannot form a buffer.

Explanation

Here, moles of HF = 0.15 x 0.1 = 0.015 moles

Moles of HCl = 0.135 x 0.175 = 0.023

Since HCl is a strong acid and the number of HCl is higher than HF. This prevents the dissociation of HF and the conjugate base F- will not be available in the solution

Part C

165.0 mL of 0.10 M HF; 135.0 mL of 0.050 M KOH

This combination will form a buffer.

Explanation

Moles of HF = 0.165 x 0.1 = 0.0165 moles

Moles of KOH = 0.135 x 0.05 = 0.00675 moles

Moles of KOH is not sufficient for the complete neutralization of HF. Thus weak acid HF and its conjugate base F- is available in the solution and form a buffer

Part D

125.0 mL of 0.15 M CH3NH2; 120.0 mL of 0.25 M CH3NH3Cl

This combination will form a buffer

Explanation

Here, weak acid CH3NH3+ and its conjugate base CH3NH2 is available in the solution and form a buffer

Part E

105.0 mL of 0.15 M CH3NH2; 95.0 mL of 0.10 M HCl

This combination will form a buffer

Explanation

Moles of CH3NH2 = 0.105 x 0.15 = 0.01575 moles

Moles of HCl = 0.095 x 0.1 = 0.0095 moles

Thus the HCl completely reacts with CH3NH2 and converts a part of the CH3NH2 to CH3NH3+. This results weak acid CH3NH3+ and its conjugate base CH3NH2 is in the solution and form a buffer
5 0
2 years ago
In Universe L, recently discovered by an intrepid team of chemists who also happen to have studied interdimensional travel, quan
Advocard [28]

Answer:

Manganese, Fifth transition element

[X] 3d⁶ 4s¹

Iron, Sixth transition element

[X] 3d⁶ 4s²

Explanation:

Complete Question

In Universe L, recently discovered by an intrepid team of chemists who also happen to have studied interdimensional travel, quantum mechanics works as it does in our universe, except that there are six d orbitals instead of the usual number we observe here. Use these facts to write the ground-state electron configurations of the sixth and seventh elements in the first transition series in Universe L. Note; you may use [X] to stand for the electron configuration of the noble gas at the end of the row before the first transition series.

Solution

In our universe, there are 5 d orbitals.

And according to Aufbau's principles that electrons fill the lower energy orbitals before they fill higher energy orbitals and Hund's Rule that states that electrons are fed singly to all the orbitals of a subshell before pairing occurs.

The fifth and sixth transition elements in our universe is then Manganese and Iron respectively.

Manganese - [Ar] 3d⁵ 4s²

Iron - [Ar] 3d⁶ 4s²

So, in the new universe L, where there are six d orbitals, for manganese, the fifth transition metal, because half filled orbitals are more stable than partially filled orbitals (that woukd have been rhe case if we leave 5 electrons on the 3d orbital), the 4s orbital is filled to half of its capacity and the one electron removed from the 4s is used to fill the six 3d orbital to half of its capacity too.

For the sixth transition element, the new extra electron just fills the lower energy 4s orbital, leaving the six 3d orbitals all half-filled.

Hence, they both have ground state configurations of

- Manganese, Fifth transition element

[X] 3d⁶ 4s¹

- Iron, Sixth transition element

[X] 3d⁶ 4s²

Hope this Helps!!!

7 0
2 years ago
The molar concentration of a sugar solution in an open beaker has been determined to be 0.3M. Calculate the solute potential at
antoniya [11.8K]

Answer:

- 7.48

Explanation:

Given:

Concentration of the sugar solution, C = 0.3 M

Temperature, T = 27° C = 273 + 27 = 300 K

Now,

The solute potential is given as:

solute potential = - iCRT

where,

i is the number of particles the particular molecule will make in water

i = 1 for sugar

R is the universal gas constant = 0.0831 liter bar/mole-K

on substituting the respective values, we get

solute potential = - 1 × 0.3 × 0.0831 × 300

or

The solute potential = - 7.479 ≈ - 7.48

8 0
2 years ago
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