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Ahat [919]
3 years ago
8

You have 25.0 mL HCI of unknown concentration. It takes 12.5 of 0.2 M NaOH to neutralize the acid. Determine the concentration o

f HCI in the following steps. Write the balanced equation for the reaction.
Chemistry
1 answer:
Fittoniya [83]3 years ago
8 0
The balanced equation for the reaction between NaOH and HCl is 
NaOH + HCl ---> NaCl + H₂O
stoichiometry of NaOH to HCl is 1:1
during neutralisation, number of H⁺ ions = OH⁻ ions. Together they form water molecules and neutralise the solution as water is neutral.
Number of NaOH moles in 1000 ml - 0.2 mol
therefore in 12.5 ml -0.2/1000 x 12.5
                                = 0.0025 mol
at neutralisation point , number of OH⁻ mol = number of H⁺ ions
    since stoichiometry of base to acid is 1:1
number of HCl moles reacted = 0.0025 mol
concentration is the number of moles in 1000 ml
in 25 ml number of HCl moles - 0.0025 mol
Therefore in 1000 ml - 0.0025 mol /25 x 1000
  concentration of HCl - 0.1 M
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At a given temperature, the elementary reaction A − ⇀ ↽ − B , A↽−−⇀B, in the forward direction, is first order in A A with a rat
Svetllana [295]

Answer:

The equilibrium constant for the reversible reaction = 0.0164

Explanation:

At equilibrium the rate of forward reaction is equal to the rate of backwards reaction.

The reaction is given as

A ⇌ B

Rate of forward reaction is first order in [A] and the rate of backward reaction is also first order in [B]

The rate of forward reaction = |r₁| = k₁ [A]

The rate of backward reaction = |r₂| = k₂ [B]

(Taking only the magnitudes)

where k₁ and k₂ are the forward and backward rate constants respectively.

k₁ = 0.010 s⁻¹

k₂ = 0.0610 s⁻¹

|r₁| = 0.010 [A]

|r₂| = 0.016 [B]

At equilibrium, the rate of forward and backward reactions are equal

|r₁| = |r₂|

k₁ [A] = k₂ [B] (eqn 1)

Note that equilibrium constant, K, is given as

K = [B]/[A]

So, from eqn 1

k₁ [A] = k₂ [B]

[B]/[A] = (k₁/k₂) = (0.01/0.0610) = 0.0163934426 = 0.0164

K = [B]/[A] = (k₁/k₂) = 0.0164

Hope this Helps!!!

5 0
3 years ago
How many grams of iki would it take to obtain a 100 ml solution of 0.300 m iki? how many grams of iki would it take to create a
GalinKa [24]

0.300 M IKI represents the concentration which is in molarity of a potassium iodide solution. This means that for every liter of solution there are 0.300 moles of potassium iodide. Knowing that molarity is a ratio of solute to solution.

By using a conversion factor:

100 ml x (1L / 1000 mL) x (0.300 mol Kl / 1 L) x (166.0g / 1 mol Kl) = 4.98 g

Therefore, in the first conversion by simply converting the unit of volume to liter, Molarity is in L where the volume is in liters. The next step is converted in moles from volume by using molarity as a conversion factor which is similar to how density can be used to convert between volume and mass. After converting to moles it is simply used as molar mass of Kl which is obtained from periodic table to convert from mole to grams.

In order to get the grams of IKI to create a 100 mL solution of 0.600 M IKI, use the same formula as above:

100 ml x (1L / 1000 mL) x (0.600 mol Kl / 1 L) x (166.0g / 1 mol Kl) = 9.96 g

3 0
3 years ago
Calculate E ° for the half‑reaction, AgCl ( s ) + e − − ⇀ ↽ − Ag ( s ) + Cl − ( aq ) given that the solubility product constant
antoniya [11.8K]

Answer: The value of E^{o} for the half-cell reaction is 0.222 V.

Explanation:

Equation for solubility equilibrium is as follows.

          AgCl(s) \rightleftharpoons Ag^{+}(aq) + Cl^{-}(aq)

Its solubility product will be as follows.

       K_{sp} = [Ag^{+}][Cl^{-}]

Cell reaction for this equation is as follows.

     Ag(s)| AgCl(s)|Cl^{-}(0.1 M)|| Ag^{+}(1.0 M)| Ag(s)

Reduction half-reaction: Ag^{+} + 1e^{-} \rightarrow Ag(s),  E^{o}_{Ag^{+}/Ag} = 0.799 V

Oxidation half-reaction: Ag(s) + Cl^{-}(aq) \rightarrow AgCl(s) + 1e^{-},   E^{o}_{AgCl/Ag} = ?

Cell reaction: Ag^{+}(aq) + Cl^{-}(aq) \rightarrow AgCl(s)

So, for this cell reaction the number of moles of electrons transferred are n = 1.

    Solubility product, K_{sp} = [Ag^{+}][Cl^{-}]

                                               = 1.77 \times 10^{-10}

Therefore, according to the Nernst equation

           E_{cell} = E^{o}_{cell} - \frac{0.0592 V}{n} log \frac{[AgCl]}{[Ag^{+}][Cl^{-}]}

At equilibrium, E_{cell} = 0.00 V

Putting the given values into the above formula as follows.

         E_{cell} = E^{o}_{cell} - \frac{0.0592 V}{n} log \frac{[AgCl]}{[Ag^{+}][Cl^{-}]}

        0.00 = E^{o}_{cell} - \frac{0.0592 V}{1} log \frac{1}{[Ag^{+}][Cl^{-}]}    

       E^{o}_{cell} = \frac{0.0592}{1} log \frac{1}{K_{sp}}

                  = 0.0591 V \times log \frac{1}{1.77 \times 10^{-10}}

                  = 0.577 V

Hence, we will calculate the standard cell potential as follows.

           E^{o}_{cell} = E^{o}_{cathode} - E^{o}_{anode}

       0.577 V = E^{o}_{Ag^{+}/Ag} - E^{o}_{AgCl/Ag}

       0.577 V = 0.799 V - E^{o}_{AgCl/Ag}

       E^{o}_{AgCl/Ag} = 0.222 V

Thus, we can conclude that value of E^{o} for the half-cell reaction is 0.222 V.

3 0
3 years ago
A loaf of bread has a mass of 300 grams and the volume of the loaf of bread is 10 cm by 30 cm and 10 cm what is the density of t
Free_Kalibri [48]

Answer:

0.1

Explanation:

300/(30*10*10)=0.1

6 0
3 years ago
Is the answer less dense?
xxMikexx [17]
I believe it is less dense.. that is why ice floats on water
5 0
3 years ago
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