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victus00 [196]
3 years ago
13

Why is a thermos considered a (mostly) closed system? Question 3 options: different types of liquids stay hot thermoses come in

different shapes and sizes different types of hot or cold items can be stored in a thermos matter and energy cannot enter or leave the system when the thermos lid is on tightly
Chemistry
1 answer:
cestrela7 [59]3 years ago
4 0

Answer:

Different types of hot or cold items can be stored in a thermos and power cannot enter or exit the system when the thermos lid is tightly closed

Explanation:

Closed systems are those that do not interact or do not exchange energy with the environment that surrounds them, that is why internal temperatures and conditions are maintained.

The human body is an open system, that is, it would be the opposite of the thermos since we constantly exchange energy with the environment through sweating, emission of gases, urine, feces, and the ingestion of food.

Thermoses are systems specially created to maintain a medium, it will be maintained if its lid is hermetically closed to prevent heat leakage or entry in situations of cold fluids.

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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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