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3 years ago

Which element is most important to scientists doing radiometric dating? nitrogen carbon -12 carbon - 14 hydrogen next?

Chemistry

2 answers:

Novosadov [1.4K]3 years ago

4 0

We know that we use carbon-14 for radiometric dating of any fossil or old substance containing carbon

. The choice of Carbon-14 for dating purpose is due to its radioactive nature

So after a certain time period the amount of C-14 decreases from a body after death and by measuring the amount of Carbon 14 remained in a body we can assess the age of that particular body

A living organism continuously exchange Carbon -14 from atmosphere and this stops after death.

zmey [24]3 years ago

4 0

Carbon-14.

<h3>Further explanation </h3>

Living things like animals and plants have a known proportion of carbon-14 in their tissues. Carbon-14 is one of the radioisotopes of carbon.
When they die, they stop taking carbon in, then the amount of carbon-14 decreases at a known rate.
The age of ancient organic materials can be obtained by measuring the amount of carbon-14 that remains. This activity is what we call radioactive dating.
Carbon-14 has a half-life of 5,700 years.

It can be concluded that the element of carbon-14 is most important to scientists doing radiometric dating.

Uses of radioactivity

Radiography. A picture can be obtained on film with the help of radioactivity emissions. This is known in medicine as a roentgen film. X-rays are used which are produced by electronic devices.
Sterilizing. Gamma rays can be used to kill bacteria, mold, and insects in food even after it has been packaged.
PET or Positron Emission Tomography is a technique that uses an isotope of carbon, especially carbon-11, as the radioactive tracer to measure the changes in blood flow within the brain.
Radioactive tracers. The most common tracer is technetium-99 and is very safe because it only emits gamma rays and doesn't cause much ionization. A small amount of iodine-123 can be used for medical checkings, like checking for a blocked kidney.
Radiotherapy as a treatment for cancerous and tumors. Tracers like cobalt-60, technetium-99, iodine-131, and phosphor-32 are used to detect, locate, and diagnose tumors and treat cancers.

And much more.

<h3>Learn more</h3>

The characteristics of electromagnetic waves brainly.com/question/727976
Find out the fraction of the space within the atom is occupied by the nucleus brainly.com/question/10818405
Carbon atoms can form four covalent bonds brainly.com/question/3064597

Keywords: element is most important, radiometric dating, nitrogen, carbon-11, carbon-14, hydrogen, treatment for cancerous, tumors, radiography, half-life, radiotherapy, the age, living things, animals and plants

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Complete the following table of neutralization reaction and also write down the names of acid base and salt in it.

scZoUnD [109]

Answer:

SO4

Explanation:

HNO3+K

SO4+NH4OH

Br+KOH

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3 years ago

How does the amount of carbon dioxide affect the rate of photosynthesis?

swat32

Answer:

An increase in the carbon dioxide concentration increases the rate at which carbon is incorporated into carbohydrate in the light-independent reaction, and so the rate of photosynthesis generally increases until limited by another factor.

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2 years ago

Read 2 more answers

1. Calculate the energy change (q) of the surroundings (water) using the enthalpy equation

Rasek [7]

Answer:

Q1: 728.6 J.

Q2:

a) 668.8 J.

b) 0.3495 J/g°C.

Explanation:

Q1: Calculate the energy change (q) of the surroundings (water) using the enthalpy equation:

The amount of heat absorbed by water = Q = m.c.ΔT.

where, m is the mass of water (m = d x V = (1.0 g/mL)(24.9 mL) = 24.9 g).

c is the specific heat capacity of liquid water = 4.18 J/g°C.

ΔT is the temperature difference = (final T - initial T = 32.2°C - 25.2°C = 7.0°C).

∴ The amount of heat absorbed by water = Q = m.c.ΔT = (24.9 g)(4.18 J/g°C)(7.0°C) = 728.6 J.

Q2: Calculate the energy change (q) of the surroundings (water) using the enthalpy equation

qwater = m × c × ΔT.

We can assume that the specific heat capacity of water is 4.18 J / (g × °C) and the density of water is 1.00 g/mL. calculate the specific heat of the metal. Use the data from your experiment for the unknown metal in your calculation.

a) First part: the energy change (q) of the surroundings (water):

The amount of heat absorbed by water = Q = m.c.ΔT.

where, m is the mass of water (m = d x V = (1.0 g/mL)(25 mL) = 25 g).

c is the specific heat capacity of liquid water = 4.18 J/g°C.

ΔT is the temperature difference = (final T - initial T = 31.6°C - 25.2°C = 6.4°C).

∴ The amount of heat absorbed by water = Q = m.c.ΔT = (25 g)(4.18 J/g°C)(6.4°C) = 668.8 J.

b) second part:

Q water = Q unknown metal.

Q unknown metal = - 668.8 J. (negative sign due to the heat is released from the metal to the surrounding water).

Q unknown metal = - 668.8 J = m.c.ΔT.

m = 27.776 g, c = ??? J/g°C, ΔT = (final T - initial T = 31.6°C - 100.5°C = - 68.9°C).

- 668.8 J = m.c.ΔT = (27.776 g)(c)( - 68.9°C) = - 1914 c.

∴ c = (- 668.8)/(- 1914) = 0.3495 J/g°C.

3 0

3 years ago

A compound is 40.0% c, 6.70% h, and 53.3% o by mass. assume that we have a 100.-g sample of this compound. the molecular formula

atroni [7]

When you have 100 g of compound, then based on the percentages given, there are 40.0 g C, 6.70 g H, and 53.3 g O. Convert those to moles:

C: 40.0 g / 12.0 = 3.33 moles of C
H: 6.70 g / 1.01 = 6.63 moles of H 
O: 53.3 / 16.0 = 3.33 moles of O

Dividing by the smallest (3.33), we get a C:H:O mole ratio of 1:2:1
So, The empirical formula is CH2O.
Now, That formula has a molar mass of [12.0 + 2(1.0) + 16.0] = 30.0

And we are given it's molar mass is = 240

So, no. of units of CH2O = 240 / 30 = 8

8 x CH2O = C8H16O8, and that is the molecular formula.

$C_8H_{16}O_8$

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3 years ago

A reaction produces 0.801 moles of H2O. How many molecules of H20 are produced?

timama [110]

Answer:

1 mole of molecules = 6.022 × 10²³ molecules

1 = 6.022 × 10²³ molecules / mole

Multiplying a number with [6.022 × 10²³ molecules / mole] is like multiplying the number with 1 so 0.877 mole × 6.022 × 10²³ molecules / mole = 5.28 × 10²³ molecules

5.28 × 10²³ molecules of water is produced.

Explanation:

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2 years ago