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4 years ago

Which two substances do geologists use in radiocarbon dating?

Chemistry

2 answers:

Vadim26 [7]4 years ago

7 0

The answer would be, Carbon 12, and Carbon 14.

mr_godi [17]4 years ago

6 0

Geologists use $\boxed{{\text{carbon - 12 and carbon - 14}}}$ in radiocarbon dating.

Further Explanation:

Radiocarbon dating is a technique that is used to determine the age of a substance with the help of radioactive carbon isotope. It is also known as carbon dating or carbon-14 dating. This is primarily used in archaeology to determine the age of fossil fuels.

Naturally, carbon exists in two stable nonradioactive forms and these are carbon-12 and carbon-13. It has one unstable radioactive isotope of carbon-14.

Carbon-14 is also known as radiocarbon. It is formed continuously in the lower stratosphere and upper troposphere. When neutrons are made to strike on nitrogen-14 atom, the formation of carbon-14 takes place. The formation reaction of carbon-14 is as follows:

$_{\text{7}}^{{\text{14}}}{\text{N}} + {\text{n}} \to _{\text{6}}^{{\text{14}}}{\text{C}} + {\text{p}}$

Here,

$_{\text{7}}^{{\text{14}}}{\text{N}}$ is an atom of nitrogen-14.

n is a neutron.

p is a proton.

$_{\text{6}}^{{\text{14}}}{\text{C}}$ is an atom of carbon-14.

The expression to determine the age of any substance is as follows:

${\text{Age}} = - 8033 \cdot \ln \left( {Fm} \right)$

Here, Fm is the fraction modern.

The fraction modern (Fm) is defined as the ratio of carbon-14 to that of carbon-12 present in the sample $\left( {^{{\text{14}}}{\text{C}}{{\text{/}}^{12}}{\text{C}}} \right)$ .

Since both carbon-12 and carbon-14 are used in the expression to calculate the age of any substance, both are used in radiocarbon dating.

Learn more:

What nuclide will be produced in the given reaction? brainly.com/question/3433940
Calculate the nuclear binding energy: brainly.com/question/5822604

Answer details:

Grade: Senior School

Chapter: Nuclear chemistry

Subject: Chemistry

Keywords: radiocarbon dating, carbon dating, carbon-14, age, nitrogen-14, carbon-14 dating, neutron, proton, carbon-12, carbon-13, unstable, stable, radioactive, non-radioactive, radiocarbon, upper troposphere, lower stratosphere, carbon, substance.

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4. Which two compounds contain the same total number of atoms?

Alja [10]

Answer:

Explanation:

Atom is the smallest particle of an element

7 0

4 years ago

Read 2 more answers

153 mL of 2.5 M HF is reacted with an excess of Ca(OH)2. How many grams of CaF2 will be produced?

Delvig [45]

Answer:

15 g

Explanation:

Data given:

amount of HF = 153 mL 2.5 M HF

amount of Ca(OH)₂ = Excess

grams of CaF₂ = ?

Reaction Given:

2HF + Ca(OH)₂ ------→ 2H₂O + CaF₂

Solution:

First we have to find number of moles of HF in 153 mL of 2.5 M HF

For this we will use following formula

Molarity = moles of solute / liter of solution

Rearrange above equation

moles of solute = Molarity x liter of solution . . . . . (1)

Put values in above equation (1)

moles of solute = 2.5 x 1 L

moles of solute = 2.5

So,

we come to know that there are 2.5 moles of solute (HF) in 1 L of solution

Now how many moles of solute will be present in 153 ml of solution

Convert 153 mL to Liter

1000 mL = 1 L

153 mL = 153/1000 = 0.153 L

Apply Unity Formula

2.5 moles HF ≅ 1 L solution

X moles of HF ≅ 0.153 L solution

moles of HF = 2.5 moles x 0.153 mL solution / 1 L solution

moles of HF = 0.383 moles

So, 153 mL contains 0.383 moles of HF

Now Look at the reaction:

2HF + Ca(OH)₂ ------→ 2H₂O + CaF₂

2 mol 1 mol

From the reaction we come to know that 2 moles of HF gives 1 mole of CaF₂ then how many moles of CaF₂ will be produced from o.383 moles of HF

Apply Unity Formula

2 moles HF ≅ 1 mole of CaF₂

0.383 moles of HF ≅ X moles of CaF₂

moles of CaF₂ = 0.383 moles x 1 mole / 2 mol

moles of CaF₂ = 0.192 moles

So, 0.192 moles of CaF₂ will be produced by 0.383 moles of HF

Now we will find mass of 0.192 moles of CaF₂

Formula will be used

mass in grams = no. of moles x molar mass . . . . . . . (2)

molar mass of CaF₂ = 40 + 2(19)

molar mass of CaF₂ = 40 + 38 = 78 g/mol

Put values in eq. 2

mass in grams = 0.192 x 78 g/mol

mass in grams = 14.976 g

rounding the value

mass in grams = 15 g

So,153 mL of 2.5 M HF is reacted with an excess of Ca(OH)₂ will produce 15 g of CaF₂.

6 0

4 years ago

Morphine is an effective pain killer but is also highly addictive. calculate the ph of a 0.115 m solution of morphine if its pkb

VMariaS [17]

Answer is: pH value of morphine is 10.64.
Chemical reaction: C₁₇H₁₉NO₃ + H₂O → C₁₇H₁₉NO₃H⁺ +OH⁻.
Kb(C₁₇H₁₉NO₃) = 10∧(-5,79) = 1,62·10⁻⁶.
c(C₁₇H₁₉NO₃) = 0,115 M.
Kb(C₁₇H₁₉NO₃) = c(OH⁻) · c(C₁₇H₁₉NO₃H⁺) ÷ c(C₁₇H₁₉NO₃).
c(OH⁻) = c(C₁₇H₁₉NO₃H⁺) = x.
1,62·10⁻⁶ = x² ÷ (0,115 - x).
Solve quadratic equation: x = c(OH⁻) = 0,000431 M.
pOH = -log(0,000431 M) = 3,36.
pH = 14 - 3,36 = 10,64.

5 0

3 years ago

Percent of water in Na2CO3 • 10H2O

Troyanec [42]

Hey there!

In order to solve for the percentage of water in the compound, you will first need to find its total molar mass. You can do this by adding up the molar masses of each individual element in the compound. Then, you will divide the mass that you find of the water molecules by the total mass to get the percentage.

→ Na₂CO₃ × 10 H₂O

→ Na₂ = 22.9898 × 2 = 45.9796
→ C = 12.0107
→ O₃ = 15.999 × 3 = 47.997
→ 10 H₂O = 18.015 × 10 = 180.15

Now, just add all of those numbers up for the total molar mass.

→ 45.9796 + 12.0107 + 47.997 + 180.15 = 286.1373

The last step is to divide the molar mass of the 10 water molecules by the total mass.

→ 180.15 ÷ 286.1373 = 0.62959 ≈ 0.63

Your answer will be about 63%.

Hope this helped you out! :-)

5 0

3 years ago

15 moles of an ideal gas at STP has a volume of ____liters

S_A_V [24]

Answer:

A. 336 L

Explanation:

STP is 0 °C and 1 bar.

At STP, the molar volume of a gas is 22.71 L.

Volume = 15 mol × 22.71 L/mol

Volume = 341 L

===============

The above answer is not in your list of options.

It appears that you are using the OLD (pre-1982) definition of STP, at which the molar volume of a gas is 22.4 L.

Then

V = 15 mol × 22.4 L/mol = 336 L

4 0

3 years ago