Assume that the amount needed from the 5% acid is x and that the amount needed from the 6.5% acid is y.
We are given that:
The volume of the final solution is 200 ml
This means that:
x + y = 200
This can be rewritten as:
x = 200 - y .......> equation I
We are also given that:
The concentration of the final solution is 6%
This means that:
5%x + 6.5%y = 6% (x+y)
This can be rewritten as:
0.05 x + 0.065 y = 0.06 (x+y) ............> equation II
Substitute with equation I in equation II and solve for y as follows:
0.05 x + 0.065 y = 0.06 (x+y)
0.05 (200-y) + 0.065 y = 0.06 (200-y+y)
10 - 0.05 y + 0.065 y = 12
0.015y = 12-10 = 2
y = 2/0.015
y = 133.3334 ml
Substitute with the y in equation I to get the x as follows:
x = 200 - y
x = 200 - 133.3334
x = 66.6667 ml
Based on the above calculations:
The amount required from the 5% acid = x = 66.6667 ml
The amount required from the 6.5% acid = y = 133.3334 ml
Hope this helps :)
Answer:
MnO- Manganese Oxide
Explanation:
Empirical formula: This is the formula that shows the ratio of elements
present in a
compound.
How to determine Empirical formula
1. First arrange the symbols of the elements present in the compound
alphabetically to determine the real empirical formula. Although, there
are exceptions to this rule, E.g H2So4
2. Divide the percentage composition by the mass number.
3. Then divide through by the smallest number.
4. The resulting answer is the ratio attached to the elements present in
a compound.
Mn O
% composition 72.1 27.9
Divide by mass number 54.94 16
1.31 1.74
Divide by the smallest number 1.31 1.31
1 1.3
The resulting ratio is 1:1
Hence the Empirical formula is MnO, Manganese oxide
The equilibrium constant of a reaction is defined as:
"The ratio between equilibrium concentrations of products powered to their reaction quotient and equilibrium concentration of reactants powered to thier reaction quotient".
The reaction quotient, Q, has the same algebraic expressions but use the actual concentrations of reactants.
To solve this question we need this additional information:
<em>For this reaction, K = 6.0x10⁻² and the initial concentrations of the reactants are:</em>
<em>[N₂] = 4.0M; [NH₃] = 1.0x10⁻⁴M and [H₂] = 1.0x10⁻²M</em>
<em />
Thus, for the reaction:
N₂ + 3H₂ ⇄ 2NH₃
The equilibrium constant, K, of this reaction, is defined as:
![K = 6.0x10^{-2} = \frac{[NH_3]^2}{[N_2][H_2]^3}](https://tex.z-dn.net/?f=K%20%3D%206.0x10%5E%7B-2%7D%20%3D%20%5Cfrac%7B%5BNH_3%5D%5E2%7D%7B%5BN_2%5D%5BH_2%5D%5E3%7D)
And Q, is:
![Q = \frac{[NH_3]^2}{[N_2][H_2]^3}](https://tex.z-dn.net/?f=Q%20%3D%20%5Cfrac%7B%5BNH_3%5D%5E2%7D%7B%5BN_2%5D%5BH_2%5D%5E3%7D)
Where actual concentrations are:
[NH₃] = 1.0x10⁻⁴M
[N₂] = 4.0M
[H₂] = 2.5x10⁻¹M
Replacing:
![Q = \frac{[1.0x10^{-4}]^2}{[4.0][2.5x10^{-1}]^3}](https://tex.z-dn.net/?f=Q%20%3D%20%5Cfrac%7B%5B1.0x10%5E%7B-4%7D%5D%5E2%7D%7B%5B4.0%5D%5B2.5x10%5E%7B-1%7D%5D%5E3%7D)
<h3>Q = 1.6x10⁻⁷</h3>
As Q < K,
<h3>The chemical system will shift to the right in order to produce more NH₃</h3>
Learn more about chemical equililbrium in:
brainly.com/question/24301138
