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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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AURORKA [14]

the answer is c. Gas molecules will never collide with the walls of the container

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3 years ago
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If 1.02 g of nickel reacted with 750. mL of 0.112 M hydrobromic acid, how much of each will be present at the end of the reactio
kati45 [8]

Answer:

35.1% is percent yield

Explanation:

<em>Full question: Assume no volume change.  If you formed 0.0910 atm of gas, what is the percent yield?</em>

<em />

The reaction that is occurring is:

Ni + 3HBr → NiBr₃ + 3/2H₂(g)

First, we will determine moles of Ni and HBr to determine limiting reactant and theoretical yield

Using ideal gas law, we can determine the moles of hydrogen formed. Thus, we can find percent yield:

<em>Moles Ni (Molar mass: 58.69g/mol):</em>

1.02g * (1mol / 58.69g) = 0.01738moles Ni

<em>Moles HBr:</em>

0.750L * (0.112mol/L) = 0.084 moles of HBr.

For a complete reaction of the 0.084 moles of HBr you need:

0.084mol HBr * (1 mole Ni / 3 moles HBr) = 0.028 moles of Ni.

As there are just 0.01738 moles of Ni, the Ni is limiting reactant. Assuming a theoretical yield, moles of H₂ produced are:

0.01738moles Ni * (3/2 H₂ / 1 mol Ni) = 0.02607 moles H₂

Now, moles of H₂ produced are:

PV = nRT

PV/RT = n

<em>Where P is pressure (0.0910atm)</em>

<em>V is volume (2.50L)</em>

<em>R is gas constant (0.082atmL/molK)</em>

<em>T is absolute temperature in Kelvin (30°C + 273.15 = 303.15K)</em>

<em>And n are moles</em>

PV/RT = n

0.0910atm*2.50L/0.082atmL/molK*303.15K = n

0.00915 moles = n

<em />

And percent yield (Produced moles / Theoretical moles * 100) is:

0.00915 moles / 0.02607moles =

<h3>35.1% is percent yield</h3>
8 0
2 years ago
If you dilute 19.0 mL of the stock solution to a final volume of 0.310 L , what will be the concentration of the diluted solutio
Dahasolnce [82]

Answer:

M_2=0.613M_1

Explanation:

M_1 = Concentration of stock solution

M_2 = Concentration of solution

V_1 = Volume of stock solution = 19 mL

V_2 = Volume of solution = 0.31 L= 310 mL

We have the relation

M_1V_1=M_2V_2\\\Rightarrow M_2=\dfrac{M_1V_1}{V_2}\\\Rightarrow M_2=\dfrac{M_119}{310}\\\Rightarrow M_2=M_1\times\dfrac{19}{310}\\\Rightarrow M_2=0.613M_1

\boldsymbol{\therefore M_2=0.613M_1}

The concentration of the diluted solution will be 0.613 times the concentration of the stock solution.

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Answer:

\large\boxed{\text{28 mol of O$_{2}$}}

Explanation:

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