Answer:
The molarity of HCl is 0.138 M
Explanation:
The titration reaction is as follows:
2HCl + Ca(OH)₂ → CaCl₂ + 2H₂O
When no more HCl is left, the small excess of Ca(OH)₂ added will cause the pH to rise and the indicator will turn. At this point, the number of moles of Ca(OH)₂ added will be the same as half the number of moles of HCl since 1 mol Ca(OH)₂ reacts with 2 moles HCl. Then:
At the endpoint:
moles Ca(OH)₂ = moles HCl / 2
Knowing the number of moles of Ca(OH)₂ added, we can calculate the number of moles of the acid:
mol Ca(OH)₂ = Volume added * concentration of Ca(OH)₂
mol Ca(OH)₂ = 0.0265 l * 0.130 mol/l = 3.45 x 10⁻³ mol Ca(OH)₂
The number of moles of HCl will be:
mol HCl = 2 * 3.45 x 10⁻³ mol = 6.89 x 10⁻³ mol HCl
This number of moles was present in 50.0 ml, then, in 1000 ml:
mol of HCl in 1000 ml = 6.89 x 10⁻³ mol HCl * (1000ml / 50ml) = 0.138 mol
Then:
Molarity HCl = 0.138 M
Answer:
-41. 47
Explanation:
m = q / Cp x T
m = Mass
q = Energy (or joules)
Cp = Heat Capacity
T = Change in Temperature
Water's heat capacity is always 4.18.
This is the formula you'll need for change in temperature:
Final - Initial
So, 33 - 78 = -45
m = 7800 / 4.18 x -45
= -41.47
Answer:
The
expression for the weak base equilibrium is:
![K_b=\frac{[(CH_3)_3NH^+][OH^-]}{[(CH_3)_3N]}](https://tex.z-dn.net/?f=K_b%3D%5Cfrac%7B%5B%28CH_3%29_3NH%5E%2B%5D%5BOH%5E-%5D%7D%7B%5B%28CH_3%29_3N%5D%7D)
Explanation:

The expression of the equilibrium constant of base
can be given as:
![K_c=\frac{[(CH_3)_3NH^+][OH^-]}{[(CH_3)_3N][H_2O]}](https://tex.z-dn.net/?f=K_c%3D%5Cfrac%7B%5B%28CH_3%29_3NH%5E%2B%5D%5BOH%5E-%5D%7D%7B%5B%28CH_3%29_3N%5D%5BH_2O%5D%7D)
]![K_b=K_c\times [H_2O]=\frac{[(CH_3)_3NH^+][OH^-]}{[(CH_3)_3N]}](https://tex.z-dn.net/?f=K_b%3DK_c%5Ctimes%20%5BH_2O%5D%3D%5Cfrac%7B%5B%28CH_3%29_3NH%5E%2B%5D%5BOH%5E-%5D%7D%7B%5B%28CH_3%29_3N%5D%7D)
As we know, water is pure solvent, we can put ![[H_2O]=1](https://tex.z-dn.net/?f=%5BH_2O%5D%3D1)
![K_b=K_c\times 1=\frac{[(CH_3)_3NH^+][OH^-]}{[(CH_3)_3N]}](https://tex.z-dn.net/?f=K_b%3DK_c%5Ctimes%201%3D%5Cfrac%7B%5B%28CH_3%29_3NH%5E%2B%5D%5BOH%5E-%5D%7D%7B%5B%28CH_3%29_3N%5D%7D)
So, the the
expression for the weak base equilibrium is:
![K_b=\frac{[(CH_3)_3NH^+][OH^-]}{[(CH_3)_3N]}](https://tex.z-dn.net/?f=K_b%3D%5Cfrac%7B%5B%28CH_3%29_3NH%5E%2B%5D%5BOH%5E-%5D%7D%7B%5B%28CH_3%29_3N%5D%7D)
Answer:
In Cl
, the 2 is a subscript because it indicates there are 2 of the same elements. The Lewis structure would display it as Cl-Cl.
On the other hand, a superscript would indicate a specific charge.
All subscripts show the amount of the specific element there is.
An example would be O
or N
, they both show that there are 2 of the same elements.
If the subscript is outside a parenthesis such as
it indicates there are 2
molecules.
0.1 mL of the stock solution of the enzyme is taken and made up to 5.0 mL with 0.001M HCl in order to prepare a 50-fold diluted enzyme.
<h3>What is dilution?</h3>
Dilution is a process of making a solution of lower concentration from a solution of higher concentration by the addition of solvent to a given volume of the solution of higher concentration.
Dilution of solutions is done using the dilution formula in order to determine the given volume of diluent or stock solution required. The dilution formula is given below:
where:
- C1 = Initial concentration of enzyme
- C2 = Final concentration of enzyme
- V1 = Initial volume
- V2 = Final volume
For the enzyme dilution;
C1 = 1 mg/mL
C2 = 1/50 mg/ml = 0.02 mg/ml
V= ?
V2 = 5 ml
V1 = C2V2/C1
V1 = 0.02 * 5/1 = 0.1 mL
Therefore, 0.1 mL of the stock solution of the enzyme is taken and made up to 5.0 mL with 0.001M HCl in order to prepare a 50-fold diluted enzyme.
Learn more about dilution at: brainly.com/question/24881505
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