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3 years ago

Compare and contrast absolute dating and relative dating.

Chemistry

2 answers:

ankoles [38]3 years ago

8 0

Answer:

Absolute dating is based on calculations of the age of rock strata based on half lives of minerals, relative dating is based on the assumed age of fossils found in the strata and the laws of super imposition.

Explanation:

PilotLPTM [1.2K]3 years ago

3 0

Answer:

Absolute dating and relative dating are both methods used to determine the age of a fossil. Absolute dating gives the actual age of a fossil in years based on the amount of radioactive and stable elements in the fossil. Relative dating gives an estimate of the age of a fossil based on the location of the fossil in relation to other fossils.

Explanation:

sample response

You might be interested in

A chemistry student needs of acetic acid for an experiment. He has available of a w/w solution of acetic acid in acetone. Calcul

Volgvan

Answer:

49.4 g Solution

Explanation:

There is some info missing. I think this is the original question.

A chemistry student needs 20.0g of acetic acid for an experiment. He has 400.g available of a 40.5 % w/w solution of acetic acid in acetone.

Calculate the mass of solution the student should use. If there's not enough solution, press the "No solution" button. Round your answer to 3 significant digits.

We have 400 g of solution and there are 40.5 g of solute (acetic acid) per 100 grams of solution. We can use this info to find the mass of acetic acid in the solution.

$400gSolution \times \frac{40.5gSolute}{100gSolution} = 162 g Solute$

Since we only need 20.0 g of acetic acid, there is enough of it in the solution. The mass of solution that contains 20.0 g of solute is:

$20.0gSolute \times \frac{100gSolution}{40.5gSolute} = 49.4g Solution$

4 0

3 years ago

1. Pewien tlenek azotu o masie cząsteczkowej 108 u zawiera 74,07% tlenu. Wykonaj stosowne obliczenia i napisz wzór sumaryczny te

love history [14]

Answer:

1. Stąd empiryczny wzór substancji to N₂O₅

2. W związku z tym ilość w gramach chlorku sodu NaCl is 114,4 g.

Explanation:

1. Mamy tutaj;

Masa molowa tlenku azotu = 108u

Masa azotu = 14,0067u

Masa tlenu = 15,999 u

74,07% masy tlenku azotu to tlen

Dlatego masa obecnego tlenu = 108 × 74,07 / 100 = 79,9956 u

Masa obecnego azotu = 108 - 79,9956 = 28,0044u

Liczba moli tlenu = 79,9956 / 15,999 = 5,00003 ≈ 5

Liczba moli azotu = 28,0044 / 14,0067 = 1,99935 ≈ 2

Stąd empiryczny wzór substancji to N₂O₅.

2. Kiedy sód reaguje z chlorem, mamy;

2Na (s) + Cl₂ (g) → 2NaCl (s)

Dlatego 2 mole sodu Na reaguje z 1 molem chloru gazowego Cl₂, z wytworzeniem 2 moli chlorku sodu NaCl

W związku z tym 1 mol sodu Na reaguje z 1/2 molem chloru gazowego Cl₂ z wytworzeniem 1 mola chlorku sodu NaCl

Masa Na obecnego w reakcji = 45 g

Masa molowa sodu = 22,989769u

Liczbę moli sodu w 45 g sodu podano w następujący sposób;

$Liczba \, \, moli \, \, Na= \frac{Mass \, of \, Na}{Molowy \, masa \, z \, Na} = \frac{45}{22.989769} = 1.96 \, mole$

Z czego 1,96 moli sodu Na reaguje z 1/2 × 1,96 mola chloru gazowego Cl₂ z wytworzeniem 1,96 mola chlorku sodu NaCl

Masa molowa NaCl = 58,44 g / mol

Dlatego masa NaCl = liczba moli NaCl × masa molowa NaCl

Masa NaCl = 1,96 × 58,44 = 114,39001 g ≈ 114,4 g

W związku z tym ilość w gramach chlorku sodu NaCl = 114,4 g.

4 0

3 years ago

A gas stream containing n-hexane in nitrogen with a relative saturation of 0.58 (as a fraction, multiply by 100% if you prefer %

kondor19780726 [428]

This problem is describing a gas mixture whose mole fraction of hexane in nitrogen is 0.58 and which is being fed to a condenser at 75 °C and 3.0 atm, obtaining a product at 3.0 atm and 20 °C, so that the removed heat from the system is required.

In this case, it is recommended to write the enthalpy for each substance as follows:

$H_{C-6}=y_{C-6}C_v(T_b-Ti)+\Delta _vH+C_v(T_f-Tb)\\\\H_{N_2}=y_{N_2}C_v(T_f-Ti)$

Whereas the specific heat of liquid and gaseous n-hexane are about 200 J/(mol*K) and 160 J/(mol*K) respectively, its condensation enthalpy is 31.5 kJ/mol, boiling point is 69 °C and the specific heat of gaseous nitrogen is about 29.1 J/(mol*K) according to the NIST data tables and $y_{C-6}$ and $y_{N_2}$ are the mole fractions in the gaseous mixture. Next, we proceed to the calculation of both heat terms as shown below:

$H_{C-6}=0.58*200(69-75)+(-31500)+160(20-69)=-40036J/mol\\\\H_{N_2}=0.42*29.1(20-75)=-672.21J/mol$

It is seen that the heat released by the nitrogen is neglectable in comparison to n-hexanes, however, a rigorous calculation is being presented. Then, we add the previously calculated enthalpies to compute the amount of heat that is removed by the condenser:

$Q=-40036+(-672.21)=-40708.21J$

Finally we convert this result to kJ:

$Q=-40708.21J*\frac{1kJ}{1000J}\\\\Q=-40.7kJ$

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6 0

2 years ago

What is the molarity of a 10 L solution containing 5.0 moles of solute?

Setler79 [48]

Molarity = Moles of solute/ L(liters) of solution

So let's plug in the information.

5.0 moles/10L = 0.5 M

3 0

3 years ago

Read 2 more answers

Why is it important to note the temperature when determining the density of a substance?

Andru [333]

It is important to take note of th temperature in determining the density of a substance because this will set as a basis and will likely be a variable in the experiment because this will also contribute on the effects of the experiment and a basis of how the experiment has turned to be that way.

7 0

3 years ago