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irina1246 [14]
3 years ago
14

Many scientists say, “There is no such thing as a closed system in the universe.” For example, even an unopened can of soda is a

n open system.
example explains why an unopened can of soda is an open system?
A.The unopened can of soda can become cold if placed in a refrigerator, allowing for heat to transfer between two systems.
B.The mass of the unopened can will eventually disappear.
C.Over time, chemical components of the unopened can of soda will change.
D.The unopened can of soda will become an open system If the can is opened.
Chemistry
1 answer:
tankabanditka [31]3 years ago
7 0
I believe the answer is "A."
You might be interested in
4 Al + 3 O2 --> 2 Al2O3
SIZIF [17.4K]

Answer:

The answer to your question is 7.4 moles of Aluminum

Explanation:

Data

moles of Al = ?

moles of Al₂O₃ = 3.7

Balanced chemical reaction

                4 Al  +  3 O₂  ⇒  2 Al₂O₃

To solve this problem use proportions and cross multiplication. Use the coefficients of the balanced chemical equation.

                4 moles of Aluminum ----------------- 2 moles of Al₂O₃

                 x                                  ----------------- 3.7 moles of Al₂O₃

                            x = (3.7 x 4) / 2

                            x = 14.8 / 2

                            x = 7.4 moles of Aluminum

5 0
3 years ago
Sodium chloride is an example of a
Andrei [34K]
C. Weak acid is the correct answer
6 0
3 years ago
Read 2 more answers
A sample of an ideal gas in a cylinder of volume 2.67 L at 298 K and 2.81 atm expands to 8.34 L by two different pathways. Path
Igoryamba

Explanation:

  • For path A, the calculation will be as follows.

As, for reversible isothermal expansion the formula is as follows.

            W = -2.303 nRT log(\frac{V_2}{V_1})

Since, we are not given the number of moles here. Therefore, we assume the number of moles, n = 1 mol.

As the given data is as follows.

              R = 8.314 J/(K mol),          T = 298 K ,

          V_{2} = 8.34 L,    V_{1} = 2.67 L

Now, putting the given values into the above formula as follows.

            W = -2.303 nRT log(\frac{V_2}{V_1})

   = -2.303 \times 1 \times 8.314 J/K mol \times 298 log(\frac{8.34}{2.67})

     = -2.303 \times 1 \times 8.314 J/K mol \times 298 \times 0.494

    = -2818.68 J

Hence, work for path A is -2818.68 J.

  • For path B, the calculation will be as follows.

Step 1: When there is no change in volume then W = 0

Hence, for step 1, W = 0

Step 2: As, the gas is allowed to expand against constant external pressure P_{external} = 1.00 atm.

So,              W = -P_{external} \times \Delta V

Now, putting the given values into the above formula as follows.

               W = -P_{external} \times \Delta V

                   = -1 atm \times (8.34 L - 2.67 L)  

                    = -5.67 atm L

As we known that, 1 atm L = 101.33 J

Hence, work will be calculated as follows.

       W = -\frac{101.33 J}{1 atm L} \times 5.67 atm L

            = -574.54 J

Therefore, total work done by path B = 0 + (-574.54 J)

                        W = -574.54 J

Hence, work for path B is -574.54 J.

3 0
3 years ago
The data in the table are the result of a paramecium being placed in a hypertonic salt solution.
kolezko [41]
<span>the answer is
C. The bar for very low concentration is twice the height of the bar for medium concentration.

proof
</span>
<span>
</span>
<span>Medium--------------------15

</span>Very Low-------------------30 = 2<span /> x 15 (<span>Medium)
</span>



3 0
3 years ago
Read 2 more answers
Chemical bonds contain energy that can be released when they are broken. True False
faust18 [17]
The answer ox Falseeeeeeee......
7 0
2 years ago
Read 2 more answers
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