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adoni [48]
3 years ago
13

Which postmortem parameter is fixed by 8 hours?

Chemistry
1 answer:
zlopas [31]3 years ago
4 0
The postmortem cooling or algor mortis gets fixed by 8- 10 hours after death. The temperature of the body after death can be used to determine the time since death.
You might be interested in
PLEASE HELP QUICK AS POSSIBLE!!!!
GalinKa [24]

Answer:

4180J

Explanation:

(25.0g)(4.184J/g°C)(75°C-35.0°C)

(25.0g)(40.0°C)(4.184J/g°C)

(1.00*10³g°C)(4.184J/g°C) = 4184J

use sig figs:

4180J

8 0
3 years ago
Nitrogen gas reacts with hydrogen gas to produce ammonia. How many liters of hydrogen gas at 95kPa and 15∘C are required to prod
viktelen [127]

Answer:

222.30 L

Explanation:

We'll begin by calculating the number of mole in 100 g of ammonia (NH₃). This can be obtained as follow:

Mass of NH₃ = 100 g

Molar mass of NH₃ = 14 + (3×1)

= 14 + 3

= 17 g/mol

Mole of NH₃ =?

Mole = mass /molar mass

Mole of NH₃ = 100 / 17

Mole of NH₃ = 5.88 moles

Next, we shall determine the number of mole of Hydrogen needed to produce 5.88 moles of NH₃. This can be obtained as follow:

N₂ + 3H₂ —> 2NH₃

From the balanced equation above,

3 moles of H₂ reacted to produce 2 moles NH₃.

Therefore, Xmol of H₂ is required to p 5.88 moles of NH₃ i.e

Xmol of H₂ = (3 × 5.88)/2

Xmol of H₂ = 8.82 moles

Finally, we shall determine the volume (in litre) of Hydrogen needed to produce 100 g (i.e 5.88 moles) of NH₃. This can be obtained as follow:

Pressure (P) = 95 KPa

Temperature (T) = 15 °C = 15 + 273 = 288 K

Number of mole of H₂ (n) = 8.82 moles

Gas constant (R) = 8.314 KPa.L/Kmol

Volume (V) =?

PV = nRT

95 × V = 8.82 × 8.314 × 288

95 × V = 21118.89024

Divide both side by 95

V = 21118.89024 / 95

V = 222.30 L

Thus the volume of Hydrogen needed for the reaction is 222.30 L

8 0
3 years ago
Elements are made up of one type of atom. Compounds are particles that have more than one atom joined together. They can be ....
Agata [3.3K]

Answer:

metal or non metal

Explanation:

this is the answer

4 0
2 years ago
In a chemical formula such as C6H12O6 and H2O, what do you call the little numbers below the normal line of text.
Morgarella [4.7K]

These are called subscript number.

That is the number below the normal line of test are called subscript number.

This number indicate the indicate the number of atoms of the element present in the chemical formula.

In both of these C₆H₁₂O₆ and H₂O, the number written below the line of the text are called subscript numbers.

3 0
3 years ago
Read 2 more answers
The compound known as diethyl ether, commonly referred to as ether, contains carbon, hydrogen, and oxygen. A 1.376 g sample of e
hoa [83]

Answer:

The answer to your question is: C₄H₁₀O

Explanation:

Data

          CxHyOz

mass sample : 1.376 g

mass CO₂ = 3.268 g

mass H₂O = 1.672 g

Process

Reaction

                      CxHyOz  + O₂ ⇒   CO₂  +  H₂O

1.- Calculate the moles and mass of carbon

Molecular mass CO₂ = 44g

                      44 g of CO₂ --------------  12 g of C

                      3.268 g of CO₂  --------    x

                         x = (3.268 x 12) / 44

                        x = 0.891 g of Carbon

                       12 g of carbon -----------  1 mol

                       0.891 g of C     ----------   x

                       x = (0.891 x 1) / 12

                       x = 0.0743 moles of carbon

2.- Calculate the moles and mass of hydrogen

                      18 g of water --------------- 2 g of H

                      1.672 g of H₂O ------------  x

                      x = (1.672 x 2) / 18

                      x = 0.186 g of hydrogen

                      1 g of hydrogen ------------  1 mol of H

                      0.186 g of H       ------------  x

                      x = (0.186 x 1) / 1

                      x = 0.186 moles of H

3.- Calculate the mass of Oxygen and its moles

Mass of Oxygen = 1.376 - 0.891 - 0.186

                           = 0.299 g of O₂

Moles of Oxygen

                             16 g of Oxygen ---------------- 1 mol

                             0.299 g of O    -----------------  x

                             x = (0.299 x 1) / 16

                             x = 0.019 moles of Oxygen

4.- Divide by the lowest number of moles

Carbon         0.0743/ 0.019 = 3.9 ≈ 4.0

Hydrogen     0.186/ 0.019 = 9.7 = 10

Oxygen         0.019/ 0.019 = 1

5.- Write the empirical formula

                              C₄H₁₀O                  

4 0
3 years ago
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