Radiometric isotope x decays to daughter isotope y with a half-life of 220,000 years. at present you have 1/4 gram of x in the rock. from the amount of daughter isotope y presently in the rock, you determine that the rock contained 16 grams of isotope x when it formed. how many half-lives have gone by? how old is the rock?

Answers

Answer 1

Answer:

The amount left after 1 half life = 16*1/2 = 8 g Then after each half life the amount of x will be 4 , 2 ,1 , 1/2, 1/4 grams

That's a total of 6 half-lives. Answer

Age s rock = 6*220,000 = 1,320,000 years Answer

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Rhodium has an atomic radius of 0.1345 nm and density of 12.41 gm/cm3 . Determine whether it has an FCC or BCC crystal structure.

Answers

Answer:

FCC.

Explanation:

Hello,

In this case, since the density is defined as:

$\rho =(n*M)/(Vc*N_A)$

Whereas n accounts for the number of atoms per units cell (2 for BCC and 4 for FCC), M the atomic mass of the element, Vc the volume of the cell and NA the Avogadro's number. Thus, for both BCC and FCC, the volume of the cell is:

$Vc_(BCC)=((4r)/(√(3) ) )^3=((4*0.1345x10^(-7)cm)/(√(3) ) )^3=2.997x10^(-23)cm^3\n\nVc_(FCC)=(2√(2)r)^(3) =(2√(2) *0.1345x10^(-7)cm)^3=5.506x10^(-23)cm^3$

Hence, we compute the density for each crystal structure:

$\rho _(BCC)=(n_(BCC)*M)/(Vc_(BCC)*N_A)=(2*102.9g/mol)/(2.337x10^(-23)cm^3*6.022x10^(23)/mol) =14.62g/cm^3\n\n\rho _(FCC)=(n_(FCC)*M)/(Vc_(FCC)*N_A)=(4*102.9g/mol)/(5.506x10^(-23)cm^3*6.022x10^(23)/mol) =12.41g/cm^3$

Therefore, since the density computed as a FCC crystal structure matches with the actual density, we conclude rhodium has a FCC crystal structure.

Regards.

Write the balanced chemical equation for each of these reactions. Include phases.1) When aqueous sodium hydroxide is added to a solution containing lead(II) nitrate, a solid precipitate forms.

2) However, when additional aqueous hydroxide is added the precipitate redissolves forming a soluble Pb(OH)42-(aq) complex ion

Answers

Answer: The chemical equations are given below.

Explanation:

For 1:

The chemical equation for the reaction of lead nitrate and sodium hydroxide follows:

$Pb(NO_3)_2(aq.)+2NaOH(aq.)\rightarrow Pb(OH)_2(s)+2NaNO_3(aq.)$

By Stoichiometry of the reaction:

1 mole of aqueous solution of lead nitrate reacts with 2 moles of aqueous solution of sodium hydroxide to produce 1 mole of solid lead hydroxide and 2 moles of aqueous solution of sodium nitrate.

For 2:

The chemical equation for the reaction of lead hydroxide and hydroxide ions follows:

$Pb(OH)_2(s)+2OH^-(aq.)\rightarrow [Pb(OH)_4]^(2-)(aq.)$

By Stoichiometry of the reaction:

1 mole of lead hydroxide reacts with 2 moles of aqueous solution of hydroxide ions to produce 1 mole of aqueous solution of tetra hydroxy lead (II) complex

Hence, the chemical equations are given above.

The balanced chemical equation for each of the reactions is as follows:

Pb(NO3)2(aq) + 2NaOH(aq) → 2 NaNO3(aq) + Pb(OH)2(s)
Pb(OH)2 + 2OH- → Pb(OH)42-

How to balance chemical equation?

A chemical equation is said to be balanced when the number of atoms of each element on both sides of the equation are the same.

According to this question, when aqueous sodium hydroxide is added to a solution containing lead(II) nitrate, a solid precipitate forms. The balanced equation are as follows:

Pb(NO3)2(aq) + 2NaOH(aq) → 2 NaNO3(aq) + Pb(OH)2(s)

Also, when additional aqueous hydroxide is added the precipitate redissolves forming a soluble Pb(OH)42-(aq) complex ion. The balanced equation is as follows:

Pb(OH)2 + 2OH- → Pb(OH)42-

Learn more about balanced equation at: brainly.com/question/1301642

This section of the periodic table is called a(n)

Answers

Answer:

Is it Group?

Explanation:

Group 2A (or IIA) of the periodic table are the alkaline earth metals: beryllium (Be), magnesium (Mg), calcium (Ca), strontium (Sr), barium (Ba), and radium (Ra).

22. What is the mass in grams of each of the following?a. 3.011 x 1023 atoms F
b. 1.50 x 1023 atoms Mg
c. 4.50 x 1012 atoms Cl
d. 8.42 x 1018 atoms Br
e. 25 atoms W
f. 1 atom Au

Answers

The mass in grams of 3.011 x 10²³ atoms of F is 9.5 g.

The mass in grams of 1.50 x 10²³ atoms of Mg is 5.98 g.

The mass in grams of 4.50 x 10¹² atoms of Cl is 2.65 x 10⁻¹⁰ g.

The mass in grams of 8.42 x 10¹⁸ atoms of Br is 1.12 x 10⁻³ g.

The mass in grams of 25 atoms of W is 3.1 x 10⁻²¹ g.

The mass in grams of 1 atom of Au is 3.27 x 10⁻²² g.

What is the mass in grams of 3.011 x 10²³ atoms F?

The mass in grams of 3.011 x 10²³ atoms of F is calculated as follows;

6.023 x 10²³ atoms = 19 g of F

3.011 x 10²³ atoms F = ?

= (3.011 x 10²³ x 19 g)/(6.023 x 10²³)

= 9.5 g

The mass in grams of 1.50 x 10²³ atoms of Mg is calculated as follows;

6.023 x 10²³ atoms = 24g of Mg

1.5 x 10²³ atoms F = ?

= (1.5 x 10²³ x 24 g)/(6.023 x 10²³)

= 5.98 g

The mass in grams of 4.50 x 10¹² atoms of Cl is calculated as follows;

6.023 x 10²³ atoms = 35.5 g of Cl

4.5 x 10²³ atoms Cl = ?

= (4.5 x 10¹² x 35.5 g)/(6.023 x 10²³)

= 2.65 x 10⁻¹⁰ g

The mass in grams of 8.42 x 10¹⁸ atoms of Br is calculated as follows;

6.023 x 10²³ atoms = 80 g of Br

8.42 x 10¹⁸ atoms Br = ?

= (8.42 x 10¹⁸ x 80 g)/(6.023 x 10²³)

= 1.12 x 10⁻³ g

The mass in grams of 25 atoms of W is calculated as follows;

6.023 x 10²³ atoms = 74 g of W

25 atoms W = ?

= (25 x 74 g)/(6.023 x 10²³)

= 3.1 x 10⁻²¹ g

The mass in grams of 1 atom of Au is calculated as follows;

6.023 x 10²³ atoms = 197 g of Au

1 atom of Au = ?

= (1 x 197 g)/(6.023 x 10²³)

= 3.27 x 10⁻²² g

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Final answer:

This solution provides the calculations necessary to convert the number of atoms of various elements (smallest particle of an element) to grams. It does so by using the molar mass of each element and Avogadro's number.

Explanation:

The mass of atoms can be determined by using Avogadro's number (6.022 x 1023 atoms/mol) and the molar mass of the specific element (g/mol). We use these to create a conversion factor and multiply by the number of atoms given.

For F (fluorine), which has a molar mass of about 18.9984 g/mol, 3.011 x 1023 atoms F is 9.00 g F.
For Mg (magnesium), with molar mass of about 24.3050 g/mol, 1.5 x 1023 atoms Mg is 6.07 g Mg.
For Cl (chlorine), with molar mass of about 35.453 g/mol, 4.50 x 1012 atoms Cl is 2.67 x 10-10 g Cl.
For Br (bromine), with molar mass about 79.904 g/mol, 8.42 x 1018 atoms Br is 0.12 g Br.
For W (tungsten), with molar mass about 183.84 g/mol, 25 atoms W is 7.65 x 10-22 g W.
For Au (gold), with molar mass about 197.0 g/mol, 1 atom Au is 3.28 x 10-22 g Au.

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In an endothermic reaction what is true of the enthalpya) is has increased
b) is has decreased
c)it has remained uncharged
d) it has at minimum been halved

Answers

Answer is: a) is has increased.

There are two types of reaction:

1) endothermic reaction (chemical reaction that absorbs more energy than it releases).

For example, the breakdown of ozone is an endothermic process. Ozone has lower energy than molecular oxygen (O₂) and oxygen atom, so ozone need energy to break bond between oxygen atoms.

2) exothermic reaction (chemical reaction that releases more energy than it absorbs).

For example, ΔH(reaction) = -225 kJ/mol; this is exothermic reaction.

What is an extensive property? *A property that changes if temperature changes
A property that will NOT change if temperature changes
A property that changes if the amount of substance changes
A property that does NOT change if the amount of substance changes
Help :( pls

Answers

Answer:

A property that changes if the amount of substance changes

Explanation:

An extensive property is a property that depends on the amount of matter in a sample.

Final answer:

An extensive property changes if the amount of substance changes. For instance, mass and volume are extensive properties as they would vary depending on the amount of substance.

Explanation:

An extensive property is a property that changes if the amount of substance changes. For example, mass and volume are extensive properties. If you have two separate samples of a substance, each with a different amount, their mass and volume would be different. On the other hand, the melting point or boiling point of the substance, which are examples of intensive properties, would not change regardless of the amount of substance.

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