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Anika [276]
3 years ago
7

10) Explain what arsenic had to do with preserving bodies in the past and how that related to Sylvester’s body. Tell me how the

solution was also used in the Civil War.
9) The scientist would like to study Sylvester’s body again and look at the contents of his body to try to determine when he may have died. How would studying food help determine the time of history a person died? (For instance, if they found chicken McNuggets, this may indicate the mummy lived recently)
Chemistry
1 answer:
Salsk061 [2.6K]3 years ago
5 0
10) Arsenic is a chemical agent which was used in preserving dead bodies in the past. It is mainly a formaldehyde mixture with coloring agents to give a dead body the look of life. Sylvester was a third mummy that was embalmed with arsenic in the late 1800s and is now on exhibit at Ye Olde Curiosity Shop in Seattle, Washington. His mummified body showed an extremely unusual since the remains weighed approximately 80 lbs when a body is composed of 70-80% water, after dehydration, the weight should be about 20-30% the premortem weight. The solution was also used during Civil War to preserve dead bodies <span>as the dead were being shipped home from the battlefield.
</span>
9) R<span>esearch can yield dates when certain plants/animals became domesticated and entered the standard diets of people, meaning you would not see masses eating beef before the cow became domesticated. To further this you can look at when a specific food was introduced into the diets of people in the geographic area of the person you are studying. </span>


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I NEED HELP PLEASE, THANKS!
Crazy boy [7]

Answer:

Here's what I get.

Explanation:

1. Brønsted-Lowry theory

An acid is a substance that can donate a proton to another substance.

A  base is a substance that can accept a proton from another substance.

2. pH of ammonia

The chemical equation is

\rm NH$_{3}$ + \text{H}$_{2}$O \, \rightleftharpoons \,$ NH$_{4}^{+}$ + \text{OH}$^{-}$

For simplicity, let's re-write this as

\rm B + H$_{2}$O \, \rightleftharpoons\,$ BH$^{+}$ + OH$^{-}$

(a) Set up an ICE table.

                     B + H₂O ⇌ BH⁺ + OH⁻

I/mol·L⁻¹:     0.335             0        0

C/mol·L⁻¹:       -x                +x       +x

E/mol·L⁻¹:  0.335 + x          x         x

\rm K_{\text{b}} = \dfrac{\text{[BH}^{+}]\text{[OH}^{-}]}{\text{[B]}} = 1.8 \times 10^{-5}\\\\\dfrac{x^{2}}{0.335 - x} = 1.8 \times 10^{-5}

Check for negligibility:

\dfrac{0.335}{1.8 \times 10^{-5}} = 28 000 > 400\\\\x \ll 0.335

(b) Solve for [OH⁻]

\dfrac{x^{2}}{0.335} = 1.8 \times 10^{-5}\\\\x^{2} = 0.335 \times 1.8 \times 10^{-5}\\x^{2} = 6.03 \times 10^{-6}\\x = \sqrt{6.03 \times 10^{-6}}\\x = \text{[OH]}^{-} = \mathbf{2.46 \times 10^{-3}} \textbf{ mol/L}

(c) Calculate the pOH

\text{pOH} = -\log \text{[OH}^{-}] = -\log(2.46 \times 10^{-3}) = 2.61

(d) Calculate the pH

pH = 14.00 - pOH = 14.00 - 2.61 = 11.39

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