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lidiya [134]
3 years ago
15

The smallest whole unit into which a compound can be divided and still be that same compound is a(n)

Chemistry
2 answers:
Alborosie3 years ago
8 0
The smallest whole unit into which a compound can be divided and still be that same compound is a molecule. 
elena55 [62]3 years ago
5 0

The smallest whole unit into which a compound can be divided and still be that same compound is A MOLECULE.

A compound is a substance that is formed by mixing two or more elements together chemically. The atoms of the elements in the compound have been combined in specific ratio, so they have become molecules. A molecule refers to the smallest part of a compound. So, the molecule represent the smallest unit in which a compound can exist.

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why do we heat the test tube in a glass beaker during a chemistry experiment and not directly over a flame?
bearhunter [10]
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7 0
3 years ago
Scientists use patterns in the periodic table to what?
Luba_88 [7]
I found this on google

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I hope this helps
4 0
2 years ago
Which of the following is an example of convection?
Dovator [93]

Answer:

<h3>C. hot air rising and cooler air failing. </h3>

Explanation:

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3 0
2 years ago
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
How do Hydrogen-1, Hydrogen-2, and Hydrogen-3 differ from each other?
Anna71 [15]

Answer:

different from the number of their neutrons

Explanation:

They each have one single proton (Z = 1), but differ in the number of their neutrons. Hydrogen has no neutron, deuterium has one, and tritium has two neutrons. ... Their nuclear symbols are therefore 1H, 2H, and 3H. The atoms of these isotopes have one electron to balance the charge of the one proton.

7 0
2 years ago
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