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avanturin [10]
3 years ago
8

Dacă intr-o scoica se introduc câteva picături de acid se observa apariția efervescenței. Scoica conține carbonat de calciu.Daca

s-au introdus 1,46g de acid clorhidric ce cantități de produși de reacție se obțin?
Chemistry
1 answer:
Step2247 [10]3 years ago
5 0

Answer:

Explanation:

In English;

If a few drops of acid are introduced into a shell, the effervescence may be observed. The shell contains calcium carbonate.If 1.46g of hydrochloric acid have been introduced, what quantities of reaction products are obtained?

Given parameters:

Mass of HCl = 1.46g

Unknown:

Mass of CO₂ produced = ?

Mass of H₂O produced =?

Mass of CaCl₂ produced = ?

Solution

 The balanced reaction equation is given below:

             CaCO₃ + 2HCl → CaCl₂ + CO₂ + H₂O

The effervescence observed is due to the production of carbon dioxide gas.

In order to find the mass of each of the products obtain, we must first express the known mass of the acid in terms of moles:

 number of moles of HCl = \frac{mass of HCl}{molar mass of HCl}

  Molar mass of HCl = 1 + 35.5 = 36.5g/mol

  number of moles of HCl = \frac{1.46}{36.5} = 0.04mole

Now, we must find the number of moles of each of the products obtained

number of moles of CaCl₂:

      2 moles of HCl produced 1 mole of CaCl₂

       0.04 moles of HCl will produce  \frac{0.04}{2} moles = 0.02moles of CaCl₂

Also, number of mole of H₂O is 0.02moles

        number of mole of CO₂ is 0.02moles

Mass of the products:

  Mass of CaCl₂ = number of moles of CaCl₂ x molar mass of CaCl₂

        molar mass of CaCl₂ = 40 + 2(35.5) = 111g/mol

         mass of CaCl₂ = 0.02 x 111 = 2.22g

Mass of H₂O = number of moles of H₂O x molar mass of H₂O

        molar mass of H₂O = 2(1) + 16= 18g/mol

        mass of H₂O = 0.02 x 18 = 0.36g

Mass of CO₂ = number of moles of CO₂ x molar mass of CO₂

        molar mass of CO₂ = 12 + 2(16) = 44g/mol

         mass of CO₂ = 0.02 x 44 = 0.88g

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determine the ph of a buffer that is 0.55 M HNO2 and 0.75 M KNO2. tha value of Ka for HNO2 is 6.8*10^-4
Mariana [72]

Answer:

pH = 3.3

Explanation:

Buffer solutions minimize changes in pH when quantities of acid or base are added into the mix. The typical buffer composition is a weak electrolyte (wk acid or weak base) plus the salt of the weak electrolyte. On addition of acid or base to the buffer solution, the solution chemistry functions to remove the acid or base by reacting with the components of the buffer to shift the equilibrium of the weak electrolyte left or right to remove the excess hydronium ions or hydroxide ions is a way that results in very little change in pH of the system. One should note that buffer solutions do not prevent changes in pH but minimize changes in pH. If enough acid or base is added the buffer chemistry can be destroyed.

In this problem, the weak electrolyte is HNO₂(aq) and the salt is KNO₂(aq). In equation, the buffer solution is 0.55M HNO₂ ⇄ H⁺ + 0.75M KNO₂⁻ . The potassium ion is a spectator ion and does not enter into determination of the pH of the solution. The object is to determine the hydronium ion concentration (H⁺) and apply to the expression pH = -log[H⁺].

Solution using the I.C.E. table:

              HNO₂ ⇄    H⁺   +   KNO₂⁻

C(i)        0.55M       0M      0.75M

ΔC            -x            +x          +x

C(eq)  0.55M - x       x     0.75M + x    b/c [HNO₂] / Ka > 100, the x can be                                    

                                                             dropped giving ...

           ≅0.55M        x       ≅0.75M        

Ka = [H⁺][NO₂⁻]/[HNO₂] => [H⁺] = Ka · [HNO₂]/[NO₂⁻]

=> [H⁺] = 6.80x010⁻⁴(0.55) / (0.75) = 4.99 x 10⁻⁴M

pH = -log[H⁺] = -log(4.99 x 10⁻⁴) -(-3.3) = 3.3

Solution using the Henderson-Hasselbalch Equation:

pH = pKa + log[Base]/[Acid] = -log(Ka) + log[Base]/[Acid]

= -log(6.8 x 10⁻⁴) + log[(0.75M)/(0.55M)]

= -(-3.17) + 0.14 = 3.17 + 0.14 = 3.31 ≅ 3.3

3 0
3 years ago
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