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s2008m [1.1K]
3 years ago
8

Explain how a mummy decreases in mass from 200 to 170 kg

Chemistry
1 answer:
kiruha [24]3 years ago
5 0
While the laws of thermodynamics state that matter cannot be created or destroyed, in the mummification process, bodies are capable of losing mass through several ways. One way is in the beginnings of the mummification process itself, where internal organs were removed. While the total weight of internal organs varies, it's generally around 20 pounds, or 9.07 kilograms. Another way the body loses mass in mummification is through the dehydration of bodily tissue through the application of salt to the mummy, both internally and externally. Humans average around 60% of their total body weight being composed of water. Through completely effective, which is nearly impossible, total dehydration of the body would leave only around 40% of the body's initial mass present, (0.4(remaining mass))(200kg)=80 kg, but that is nearly impossible without highly sophisticated instruments with exceptional climate control. However, this dehydration would also decrease a mummy's mass greatly. 
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What element has 80 protons, 81 neutrons, 79 electrons?
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Answer:

The element will be ^{161}_{80}X

Explanation:

Given that,

Number of proton = 80

Number of neutron = 81

Number of electron = 79

We know that,

The atomic number is equal to the number of proton.

So, the atomic number is 80.

According to atomic number,

The element will be mercury.

We need to calculate the atomic mass

Using formula of atomic mass

atomic\ mass=atomic\ number+number\ of\ neutron

Put the value into the formula

atomic\ mass=80+81

atomic\ mass=161

We need to find the element

Using atomic mass and atomic number

atomic\ mass\ number A=161

atomic\ number Z=80

So, the element will be

^{A}_{Z}X

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Hence, The element will be ^{161}_{80}X

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Explanation:

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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
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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.

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Explanation:

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