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erastovalidia [21]
3 years ago
14

PLEASE HELP 30 POINTS

Chemistry
2 answers:
Annette [7]3 years ago
5 0

Answer:

1) Only Organism Z is alive, and Organism X has been dead longer than Organism Y.

2) S (sulfur) was oxidized and N (nitrogen) was reduced.

3) Chlorine is reduced in reaction 1 and iron is oxidized in reaction 2.

Explanation:

1) After death of an organism there is radioactive disintegration of C-14 allotrope of carbon, which increases the ratio of C-12 to C-14 in a dead organism as compared to a living organism.

Longer the time of death, more the ratio of the two isotopes.

So as the ratio of isotopes is same in Z, it is alive.

The ratio of isotopes if higher in X than Y,X has been dead longer than Organism Y.

2) oxidation is increase in  oxidation state due to loss of electrons and reduction is decrease in oxidation sate due to gain of electrons.

Here oxidation state of hydrogen is same on both the sides of equation.

The oxidation state of sulfur increases and that of nitrogen decreases so sulfur has undergone oxidation while nitrogen reduction.

3) As mentioned above, the oxidation state of iron increases (0 to +2) so it is undergoing oxidation while that of chlorine decreases (0  to -1) so it is undergoing reduction.

posledela3 years ago
4 0

The answears are in the attached photo.

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The average atomic mass of carbon is 12.01 amu. Based on the atomic
zvonat [6]

Answer: There is a very large percentage of C-12.

Explanation:

The atomic mass is must closer to that of C-12 than of C-13, meaning that there is a much larger percentage of C-12.

7 0
2 years ago
Can someone please help me?
Svet_ta [14]

Answer:

D,E, and F are correct

Explanation:

Longer wavelengths = lower frequencies = lower energy

Shorter wavelength = high frequency = higher energy

Gamma rays have shortest wavelength and therefore highest frequency and energy

Radio waves equally have a shorter wavelength and therefore high frequency and energy

4 0
3 years ago
Create a chart that compares physical and chemical properties. Give two examples for each type of property
Zinaida [17]

<em>Answer:</em>

<em>Chemical properties:</em>

Those properties which change the chemical nature of matter.

<em>Example:</em>  

  •       Heat of combustion
  •       Enthalpy of formation

<em>Physical properties:</em>

Those properties which do not change the chemical nature of matter.

<em>Example</em>

  • B.P
  • M.P
  • F.P

<em>Differences between chemical and physical properties:</em>

       Chemical properties                                       Physical properties

1. Observed after the change bringing           1. Observed with out being

the change                                                            change

2. These changes the molecules                    2. only change physical state

3. Chemical identity changes                          3.Chemical identity not changes

4. Structure of material changes                     4.Structure of material not change            

5. Chemical reaction is needed                       5. No need of Chemical reaction

6. depend on composition                           6. Does not depend on composition

3 0
3 years ago
What is the equilibrium membrane potential due to na ions if the extracellular concentration of na ions is 142 mm and the intrac
Ganezh [65]

The equilibrium membrane potential is 41.9 mV.

To calculate the membrane potential, we use the <em>Nernst Equation</em>:

<em>V</em>_Na = (<em>RT</em>)/(<em>zF</em>) ln{[Na]_o/[Na]_ i}

where

• <em>V</em>_Na = the equilibrium membrane potential due to the sodium ions

• <em>R</em> = the universal gas constant [8.314 J·K^(-1)mol^(-1)]

• <em>T</em> = the Kelvin temperature

• <em>z</em> = the charge on the ion (+1)

• <em>F </em>= the Faraday constant [96 485  C·mol^(-1) = 96 485 J·V^(-1)mol^(-1)]

• [Na]_o = the concentration of Na^(+) outside the cell

• [Na]_i = the concentration of Na^(+) inside the cell

∴ <em>V</em>_Na =

[8.314 J·K^(-1)mol^(-1) × 293.15 K]/[1 × 96 485 J·V^(-1)mol^(-1)] ln(142 mM/27 mM) = 0.025 26 V × ln5.26 = 1.66× 25.26 mV = 41.9 mV

4 0
3 years ago
what is an empirical formulaWhat is the percent composition by mass of nitrogen in (NH4)2CO3 (gram-formula mass = 96.0 g/mol)?
Margarita [4]

Answer:

Percentage composition = 14.583%

Explanation:

In chemistry, the emprical formular of a compound is the simplest formular a compound can have. It shows the simplest ratio in which the elements are combined in the compound.

Percentage composition by mass of Nitrogen

Nitrogen = 14g/mol

In one mole of the compound;

Mass of Nitrogen = 1 mol * 14g/mol = 14g

Mass of compound = 1 mol * 96.0 g/mol = 96

Percentage composition of Nitrogen = (Mass of Nitrogen /  Mass of compound) * 100

percentage composition = 14/96   * 100

Percentage composition = 0.14583 * 100

Percentage composition = 14.583%

3 0
3 years ago
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