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3 years ago

What is the approximate time of death if the body temperature was 15.6°C?

Chemistry

1 answer:

WARRIOR [948]3 years ago

6 0

Answer:

This stiffening process, called Rigor Mortis, has a roughly known time of occurrence and can therefore be used to estimate time of death. In general: If the body feels warm and no rigor is present, death occurred under 3 hours before. If the body feels warm and stiff, death occurred 3-8 hours earlier.

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a calorimeter contained 75.0 g of water at 16.95 C. A 93.3-g sample of iron at 65.58 C was placed in it, giving a final temperat

Nostrana [21]

Answer:- $\frac{382.69J}{0C}$ .

Solution:- Mass of Iron added to water is 93.3 g. Initial temperature of iron metal is 65.58 degree C and final temperature of the system is 19.68 degree C.

temperature change, $\Delta T$ for iron metal = 65.58 - 19.68 = 45.9 degree C

specific heat for the metal is given as 0.444 J per g per degree C.

let's calculate the heat lost by iron metal using the equation:

$q=mc\Delta T$

where, q is the heat energy, m is mass, c is specific heat and delta T is change in temperature. let's plug in the values and calculate q for iron metal:

$q=93.3g(45.9^0C)(\frac{0.444J}{g.^0C})$

q = 1901.42 J

Using same equation we will calculate the heat gained by water.

mass of water is 75.0 g.

$\Delta T$ for water = 19.68 - 16.95 = 2.73 degree C

specific heat for water is 4.184 J pr g per degree C. Let's plug in the values:

$q=75.0g(\frac{4.184J}{g.^0C})(2.73^0C)$

q = 856.674 J

Total heat lost by iron metal is the sum of heat gained by water and calorimeter.

So, heat gained by calorimeter = heat lost by iron metal - geat gained by water

heat gained by calorimeter = 1901.42 J - 856.674 J = 1044.746 J

Change in temperature for calrimeter is same as for water that is 2.73 degree C

For calorimeter, $q=C.\Delta T$

$C=\frac{q}{\Delta T}$

$C=\frac{1044.746J}{2.73^0C}$

$C=\frac{382.69J}{0C}$

So, the heat capacity of calorimeter is $\frac{382.69J}{0C}$ .

4 0

3 years ago

What would happen if a neutron was added to Lithium (7)

Ainat [17]

Answer:

if you add a neutron to lithium-7 it becomes a compound nucleus of lithium-8

8 0

3 years ago

Use the standard reduction potentials located in the 'Tables' linked above to calculate the equilibrium constant for the reactio

Paladinen [302]

Answer:

Check the explanation

Explanation:

cell CuE Ecell 0.337 (-0.14) Ecl0.477 V

Since $E^o_{ cell } > 0$ , the value of \Delta G^o will be negative.

$\Delta G^o < 0$

$\Delta G^o =-nFE^o_{ cell }.$ .....(1)

But

$\Delta G^o =-RT ln K$ ......(2)

From (1) and (2)

$\Delta G^o =-RT ln K=-nFE^o_{ cell }$

ln K = $\frac{nFE^o_{ cell } }{RT }$

ln K = $\frac{ 2 \times 96500 \times 0.477 }{8.314 \times \left ( 25+273.15 \right ) }$

ln K =37.139

K = $1.3468 \times 10^{16}$

Hence, the value of the equilibrium constant is $1.35 \times 10^{16}$

6 0

3 years ago

Can someone pls help me answer e-j given the instructions above, thank youu

vladimir1956 [14]

Answer:

e. Not balanced C and O

f. Balanced

g. Balanced

h. Unbalanced Al and O

i. Balanced

j. Balanced

Second part

a. 2

b. 2 and 2

c. For O is 5 while H20 is 4

Explanation:

Please mark as brainliest

4 0

3 years ago

PLEASE HELP, I REALLY NEED HELP WITH THIS!!!

Mkey [24]

Answer:

Explanation:

There are 6 32S atoms and 2 36S atoms

Their respective contributions to the average aomic mass is (6*32)=192, and (2*36)=72. Added together, this makes 264 total. Divide by the total number of atoms, 8. (264/8) = 33.00

The weighted atomic mass of these S atoms is 33.00 amu (atomic mass units)

3 0

3 years ago