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ehidna [41]
4 years ago
13

Give the mathematical relationship for an unsaturated solution in comparing Q with K sp:__________.

Chemistry
1 answer:
e-lub [12.9K]4 years ago
3 0

Answer:

Q < Ksp

Explanation:

The general equilibrium of a constant product solubility, ksp, is:

AB ⇄ A⁺ + B⁻

<em>Where Ksp is defined as:</em>

Ksp = [A⁺] [B⁻]

When [A⁺] [B⁻] = Ksp, the solution is saturated or oversaturated because there are the maximum amount of ions that solution can dissolve.

When the solution is oversaturated, AB is produced.

Now, in a unsaturated solution, the [A⁺] [B⁻] is less than the maximum amount that can be dissolved. That means:

[A⁺] [B⁻] = Q < Ksp

Q is defined in the same way than Ksp, just in Q the system is not in equilibrium.

Right answer is:

<h3>Q < Ksp</h3>
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A 7.50-g sample of iron is heated in oxygen to form an iron oxide. if 10.71 g of the oxide is formed, what is its empirical form
Zielflug [23.3K]
Answer is: empirical formula is Fe₂O₃.
m(Fe) = 7,50 g.
m(iron oxide) = 10,71 g.
n(Fe) = m(Fe) ÷ M(Fe).
n(Fe) = 7,50 g ÷ 55,85 g/mol = 0,134 mol.
m(O) = m(iron oxide) - m(Fe).
m(O) = 10,71 g - 7,50 g = 3,21 g.
n(O) = 3,21 g ÷ 16 g/mol = 0,20 mol.
n(Fe) : n(O) = 0,134 mol : 0,2 mol = 2 : 3.
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4 years ago
2 Nobr +heat &gt; 2 no2+br2 what happens when you remove nobr
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Which method of separation would be most appropriate for separating a mixture of water and alcohol?(1 point)
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fractional distillation  is the correct answer

Explanation:

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8 0
2 years ago
Three Stoichiometry Questions
andrezito [222]

Answer:

Explanation:

7)

Given data:

Mass of aluminium = 2.5 g

Mass of oxygen = 2.5 g

Mass of aluminium oxide = 3.5 g

Percent yield = ?

Solution:

Chemical equation:

4Al + 3O₂   →   2Al₂O₃

Number of moles of Al:

Number of moles = mass/ molar mass

Number of moles = 2.5 g/ 27 g/mol

Number of moles = 0.09 mol

Number of moles of oxygen:

Number of moles = mass/ molar mass

Number of moles = 2.5 g/ 32 g/mol

Number of moles = 0.08 mol

Now we will compare the moles of aluminium oxide with aluminium and oxygen.

                          Al         ;       Al₂O₃

                           4         :        2

                        0.09      :       2/4×0.09 = 0.045

                          O₂       :        Al₂O₃

                          3         :          2

                         0.08    :        2/3 ×0.08 = 0.053

The number of moles of aluminium oxide produced by Al are less so it will limiting reactant.

Mass of aluminium oxide:

Mass = number of moles × molar mass

Mass = 0.045  × 101.96 g/mol

Mass = 4.6 g

Percent yield:

Percent yield = actual yield / theoretical yield ×100

Percent yield = 3.5 g / 4.6 ×100

Percent yield = 76.1%

8)

Given data:

Mass of copper produced = 3.47 g

Mass of aluminium = 1.87 g

Percent yield = ?

Solution:

Chemical equation:

2Al + 3CuSO₄   →   Al₂(SO₄)₃ + 3Cu

Number of moles of Al:

Number of moles = mass/ molar mass

Number of moles = 1.87 g/ 27 g/mol

Number of moles = 0.07 mol

Now we will compare the moles of copper with aluminium.

                          Al         ;       Cu

                           2         :        3

                        0.07      :       3/2×0.09 = 0.105

             

Mass of copper:

Mass = number of moles × molar mass

Mass = 0.105  × 63.55 g/mol

Mass = 6.67 g

Percent yield:

Percent yield = actual yield / theoretical yield ×100

Percent yield =  3.47 g / 6.67 × 100

Percent yield = 52%

                       

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