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2 years ago

8

a rock that originally had a mass of 1.00g of uranium_238 now only has .125g left. how old is the rock if the half life of urani

um_238 is 4.5 billion years?

1 answer:

Romashka [77]2 years ago

6 0

Answer:

no it would be 2.5 mill your welcome

Explanation:

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A wooden artifact from an ancient tomb contains 60% of the carbon-14 that is present in living trees. How long ago was the artif

noname [10]

Answer:

The age of the sample is 4224 years.

Explanation:

Let the age of the sample be t years old.

Initial mass percentage of carbon-14 in an artifact = 100%

Initial mass of carbon-14 in an artifact = $[A_o]$

Final mass percentage of carbon-14 in an artifact t years = 60%

Final mass of carbon-14 in an artifact = $[A]=0.06[A_o]$

Half life of the carbon-14 = $t_{1/2}=5730 years$

$k=\frac{0.693}{t_{1/2}}$

$[A]=[A_o]\times e^{-kt}$

$[A]=[A_o]\times e^{-\frac{0.693}{t_{1/2}}\times t}$

$0.60[A_o]=[A_o]\times e^{-\frac{0.693}{5730 year}\times t}$

Solving for t:

t = 4223.71 years ≈ 4224 years

The age of the sample is 4224 years.

8 0

2 years ago

Determine whether each of the descriptions of matter describes a heterogeneous mixture, a homogeneous mixture, or a pure substan

lorasvet [3.4K]

Answer: a. two substances present; two phases present : Heterogeneous mixture

b. two substances present; one phase present : Homogeneous mixture

c. one substance present; one phase present : pure substance.

d. one substance present; two phases present: Heterogeneous mixture

Explanation:

A pure substance is a substance which contains definite composition of only one type of component. Hence, it cannot be separated by physical means.

Mixture is a substance which contains two or more than two types of components and they can be separated by physical means as well.

Homogeneous mixtures: It is a mixture that has uniform composition throughout the solution and the particle size or shapes are not different. There is no physical boundary between the dispersed phase and dispersion medium.

Heterogeneous mixtures: It is a mixture that has non-uniform composition throughout the solution and the particle size or shapes are also different. There is a physical boundary between the dispersed phase and dispersion medium.

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2 years ago

In which substance does chlorine have an oxidation number of +1?

Anton [14]

The answer is (3) HClO. In the Cl2, chlorine has an oxidation number of zero. In HCl, the oxidation number is -1. In HClO2, the oxidation number is +3. In HClO, it is +1. You can calculate this by using O with oxidation number of -2 and H with +1.

8 0

2 years ago

1. Sam needs to know the volume of a cardboard box. Which tool

andreyandreev [35.5K]

Ruler is the answer of this question

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2 years ago

Determine the mole fractions and partial pressures of CO2, CH4, and He in a sample of gas that contains 1.20 moles of CO2, 1.79

BARSIC [14]

Answer : The mole fraction and partial pressure of $CH_4,CO_2$ and $He$ gases are, 0.267, 0.179, 0.554 and 1.54, 1.03 and 3.20 atm respectively.

Explanation : Given,

Moles of $CH_4$ = 1.79 mole

Moles of $CO_2$ = 1.20 mole

Moles of $He$ = 3.71 mole

Now we have to calculate the mole fraction of $CH_4,CO_2$ and $He$ gases.

$\text{Mole fraction of }CH_4=\frac{\text{Moles of }CH_4}{\text{Moles of }CH_4+\text{Moles of }CO_2+\text{Moles of }He}$

$\text{Mole fraction of }CH_4=\frac{1.79}{1.79+1.20+3.71}=0.267$

and,

$\text{Mole fraction of }CO_2=\frac{\text{Moles of }CO_2}{\text{Moles of }CH_4+\text{Moles of }CO_2+\text{Moles of }He}$

$\text{Mole fraction of }CO_2=\frac{1.20}{1.79+1.20+3.71}=0.179$

and,

$\text{Mole fraction of }He=\frac{\text{Moles of }He}{\text{Moles of }CH_4+\text{Moles of }CO_2+\text{Moles of }He}$

$\text{Mole fraction of }He=\frac{3.71}{1.79+1.20+3.71}=0.554$

Thus, the mole fraction of $CH_4,CO_2$ and $He$ gases are, 0.267, 0.179 and 0.554 respectively.

Now we have to calculate the partial pressure of $CH_4,CO_2$ and $He$ gases.

According to the Raoult's law,

$p_i=X_i\times p_T$

where,

$p_i$ = partial pressure of gas

$p_T$ = total pressure of gas = 5.78 atm

$X_i$ = mole fraction of gas

$p_{CH_4}=X_{CH_4}\times p_T$

$p_{CH_4}=0.267\times 5.78atm=1.54atm$

and,

$p_{CO_2}=X_{CO_2}\times p_T$

$p_{CO_2}=0.179\times 5.78atm=1.03atm$

and,

$p_{He}=X_{He}\times p_T$

$p_{He}=0.554\times 5.78atm=3.20atm$

Thus, the partial pressure of $CH_4,CO_2$ and $He$ gases are, 1.54, 1.03 and 3.20 atm respectively.

4 0

2 years ago