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GenaCL600 [577]
3 years ago
5

A woman is found dead on a boat floating in the middle of a lake at 4:30am. There is blunt force trauma to her head but no other

discernable damage. She is approximately 30 pounds overweight. She is wearing thermal clothing and a winter jacket, gloves, and hat. Climatological data indicate it was a clear, cold, and windy night with no precipitation. Calculate the time of death based only on a body temperature of 15.6°C (60.8°F)? Now consider the other factors presented. How would each affect this estimation?
Chemistry
1 answer:
Olenka [21]3 years ago
5 0

Answer:

Dead bodies lose one degree Celsius per hour, which we could deduce that the person died 12 hours ago.

Explanation:

On the other hand, it is very important to take into account that this number of hours is very approximate, since not only does it vary greatly with the environment and the acclimatization of the environment in which the person dies, but also in this specific case, insulating clothing was presented , which made the loss of body heat much more difficult and could easily be more than 12 hours.

The minimum for this particular situation is 12 hours to 24 hours possible postmortem.

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Cu + 2 AgNO3 + Cu(NO3)2 + 2Ag
Allushta [10]

Answer:

7.28 moles Ag°

Explanation:

              Cu°          +    2 AgNO₃  => Cu(NO₃)₂ + 2Ag°

Given 7.28 moles     7.28 moles

To determine limiting reactant, divide the mole values by the respective coefficient of balanced equation. The resulting smallest value is the limiting reactant. Note: this is a short cut method for determining limiting reactant only. Once the limiting reactant is determined one must use the given mole values of the limiting reactant to solve problem. That is ...

Limiting reactant determination:

              Cu°          +    2 AgNO₃  => Cu(NO₃)₂ + 2Ag°

Cu: 7.28 / 1  = 7.28

AgNO₃ : 7.28 / 2 = 3.64 => Limiting Reactant is AgNO₃

Solving Problem depends on AgNO₃; Cu will be in excess.

Since coefficients of AgNO₃ & Ag° are equal, then the moles AgNO₃ used equals moles Ag° produced and is therefore 7.28 moles Ag°.

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