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marta [7]
3 years ago
10

Please help me with these questions

Chemistry
1 answer:
zaharov [31]3 years ago
5 0
I’m just answering this so i can ask more questions and it has to be 20 words long so i hope you figure your problems out and merry christmas happy new year
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Answer:

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Explanation:

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An aluminum atom has a mass of 4.48 * 10-23 g and a small airplane has a mass of 5000 kg. Use this information to answer the que
Rudiy27

Answer:

a. 27g/mol

b. 1.85 x 10^5 moles

Explanation:Please see attachment for explanation

4 0
3 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
What kind of pattern can you see about elements in the same period with respect to electron
Fantom [35]

Answer: A pattern of same atomic orbitals can be seen about elements in the same period with respect to electron structures.

Explanation:

The horizontal rows in a period table are called periods.

Elements present in the same period will have same atomic orbitals.

For example, electronic distribution of Na is 2, 8, 1 and it is a third period element.

Similarly, electronic distribution of Cl is 2, 8, 7 and it is also a third period element.

Hence, both Na and Cl will have K, L, M shells, that is, they have three atomic orbitals.

Thus, we can conclude that a pattern of same atomic orbitals can be seen about elements in the same period with respect to electron structures.

5 0
3 years ago
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