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disa [49]
3 years ago
5

**PLEASE SHOW YOUR WORK**Calculate how much energy would be necessary to convert 250.0 grams of ice at -30°C to steam at 145°C.

Chemistry
1 answer:
garri49 [273]3 years ago
7 0

Answer:

Q= mc∆T(ice) + mLF(ice) + mc∆T(water) + mLV(water) + mc∆T(steam)

m=250 g = 0.25 kg = ¼ kg c(ice)= 2100 J/kg.K c(water)= 4200 J/kg.K LF(ice)= 333.7 kJ/kg LV(water)= 2256 kJ/kg c(steam)= 2080 J/kg.K

Explanation:

Q= ¼ × 2100 × (0°-(-30°)) + ¼ × 333700 + ¼ × 4200 × (100°-0°) + ¼ × 2256000 + ¼ × 2080 × (145°-100°)

Q= 15750 + 83425 + 105000 + 564000 + 23400

Q= 791575 J

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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
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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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