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tester [92]
3 years ago
10

Uestion 1

Chemistry
1 answer:
Bingel [31]3 years ago
5 0

Answer:

3.14 \times 10^{23} \text{ S atoms}

Explanation:

1. Calculate the moles of Al₂S₃

\text{Moles of Al$_{2}$S}_{3} = \text{26.1 g} \times \dfrac{\text{1 mol}}{\text{150.16 g}} = \text{0.1738 mol Al$_{2}$S}_{3}

2. Calculate the moles of S atoms

\text{Moles of S atoms} = \text{0.1738 mol Al$_{2}$S}_{3} \times \dfrac{\text{3 mol S atoms }}{\text{1 mol Al$_{2}$S}_{3}} = \text{0.5214 mol S atoms}

3. Calculate the number of S atoms

\text{No. of S atoms} = \text{0.5214 mol S atoms} \times \dfrac{6.022 \times 10^{23} \text{ S atoms}}{\text{1 mol S atoms}} = \mathbf{3.14 \times 10^{23}} \textbf{ S atoms}

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Explanation:

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Calculate the change in pH when 71.0 mL of a 0.760 M solution of NaOH is added to 1.00 L of a solution that is 1.0O M in sodium
Eddi Din [679]

Explanation:

It is known that pK_{a} value of acetic acid is 4.74. And, relation between pH and pK_{a} is as follows.

                    pH = pK_{a} + log \frac{[CH_{3}COOH]}{[CH_{3}COONa]}

                          = 4.74 + log \frac{1.00}{1.00}

So, number of moles of NaOH = Volume × Molarity

                                                   = 71.0 ml × 0.760 M

                                                    = 0.05396 mol

Also, moles of  CH_{3}COOH = moles of CH_{3}COONa

                                          = Molarity × Volume

                                          = 1.00 M × 1.00 L

                                          = 1.00 mol

Hence, addition of sodium acetate in NaOH will lead to the formation of acetic acid as follows.

            CH_{3}COONa + NaOH \rightarrow CH_{3}COOH

Initial :    1.00 mol                                  1.00 mol

NaoH addition:               0.05396 mol

Equilibrium : (1 - 0.05396 mol)    0           (1.00 + 0.05396 mol)

                    = 0.94604 mol                       = 1.05396 mol

As, pH = pK_{a} + log \frac{[CH_{3}COONa]}{[CH_{3}COOH]}

               = 4.74 +  log \frac{0.94604}{1.05396}

               = 4.69

Therefore, change in pH will be calculated as follows.

                         pH = 4.74 - 4.69

                               = 0.05

Thus, we can conclude that change in pH of the given solution is 0.05.

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