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LiRa [457]
3 years ago
8

How many milliliters of 5.50 m hcl(aq) are required to react with 9.55 g of zn(s)?

Chemistry
1 answer:
Ganezh [65]3 years ago
6 0
Zn(s) + 2HCl(aq) --> ZnCl2(aq) + H2(g)  From the balanced chemical equation, 2 moles of HCl is required to react with 1 mole of Zn   Mass of Zn = 9.55 g  Molar mass of Zinc = 65.41 g/mol   Number of moles = mass / molar mass   Moles of Zn = 9.55 / 65.41   = 0.146 mol  Moles of HCl required = 2 x moles of Zn   = 2 x 0.146   = 0.292 mol
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5 0
3 years ago
As a chemist for an agricultural products company, you have just developed a new herbicide,"Herbigon," that you think has the po
Ganezh [65]

Answer:

pH = 2.03

Explanation:

The pH can be calculated using the following equation:

pH = -log [H_{3}O^{+}]  (1)

The concentration of H₃O⁺ is calculated using the dissociation constant of the next reaction:

CH₃COOH + H₂O ⇄  CH₃COO⁻ + H₃O⁺    

   1.00 M    

K_{a} = \frac{[CH_{3}COO^{-}][H_{3}O^{+}]}{[CH_{3}COOH]}

Solving the above equation for H₃O⁺, we have:    

[H_{3}O^{+}] = \frac{Ka*[CH_{3}COOH]}{[CH_{3}COO^{-}]}    (2)    

The dissociation constant is equal to:    

pKa = -log(Ka) \rightarrow Ka = 10^{-pKa} = 10^{-4.76} = 1.74 \cdot 10^{-5}    

Now, by solving the equation of the solubility product for Herbigon, we can find [CH₃COO⁻]:

CH₃COOX  ⇄  CH₃COO⁻ +  X⁺  

                                             5.00x10⁻³ M

K_{sp} = [CH_{3}COO^{-}][X^{+}]

[CH_{3}COO^{-}] = \frac{K_{sp}}{[X^{+}]} = \frac{9.40 \cdot 10^{-6}}{5.00 \cdot 10^{-3}} = 1.88 \cdot 10^{-3} M

By entering the values of [CH₃COO⁻] and Ka, into equation (2) we can calculate [H₃O⁺]:

[H_{3}O^{+}] = \frac{1.74 \cdot 10^{-5}*[1.00]}{[1.88 \cdot 10^{-3}]} = 9.26 \cdot 10^{-3} M

Hence, the pH is:

pH = -log [H_{3}O^{+}] = -log [9.26 \cdot 10^{-3}] = 2.03

Therefore, the pH must be 2.03 to yield a solution in which the concentration of X⁺ is 5.00x10⁻³M.

I hope it helps you!  

6 0
4 years ago
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kirill115 [55]

Answer:

option d is right

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