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faltersainse [42]
3 years ago
9

Which direction does thermal energy flow in the following diagram? the picture is of someone holding an ice cream cone

Chemistry
2 answers:
Sliva [168]3 years ago
8 0
The answer is B. Heat flows from the hand to the ice cream. The state of equilibrium won't occur for this since the only reactant is heat.
Masja [62]3 years ago
5 0
<h2>Answer : Option B) Heat flows from the hand to the ice-cream.</h2><h3>Explanation : </h3>

The thermal energy flow would require a source to get transferred to an object which has less thermal energy than the source. Here, the energy source is a person and the object is ice-cream. The source of thermal energy will be in a person and the energy will flow from the hand of the person to the ice-cream as there is low thermal energy in ice-cream.

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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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