1answer.
Ask question
Login Signup
Ask question
All categories
  • English
  • Mathematics
  • Social Studies
  • Business
  • History
  • Health
  • Geography
  • Biology
  • Physics
  • Chemistry
  • Computers and Technology
  • Arts
  • World Languages
  • Spanish
  • French
  • German
  • Advanced Placement (AP)
  • SAT
  • Medicine
  • Law
  • Engineering
Rasek [7]
3 years ago
8

A scientist digs up sample of arctic ice that is 458,000 years old. He takes it to his lab and finds that it contains 1.675 gram

s of krypton-81.
If the half-life of krypton-81 is 229,000 years, how much krypton-81 was present when the ice first formed?

Use the formula N = N0 .
Chemistry
1 answer:
Fiesta28 [93]3 years ago
3 0

Answer:

6.70 grams of krypton-81 was present when the ice first formed

Explanation:

Let use the below formula to find the amount of sample

N= N_0(\frac{1}{2})^n

where

n = \frac{t}{t_{\frac{1}{2}}}

here

t =  458,000 years

t_{\frac{1}{2}} = 229,000

\frac{t}{t_{\frac{1}{2}}} = \\frac{ 458,000}{229,000}

n = \frac{t}{t_{\frac{1}{2}}} = 2.000

Now substituting the values

1.675 = N_0(\frac{1}{2})^{2.000}}

1.675 = N_0\times (0.2500)

N_0= \frac{1.675}{0.2500}

N_0=6.70

You might be interested in
Which practice does the Montreal Protocol aim to reduce?
PolarNik [594]
Ozone which is present in the stratospheric region of atmosphere is helpful for preventing harmful UV rays from reaching the surface of earth. Due to human activity, several compounds  (specifically chlorofluorocarbons) are released in atmosphere. Due to inherent chemical stability of these compounds, the remain stable in lower region of atmosphere and slowly diffuse into stratosphere. On reaching the stratosphere, these compounds reacts with ozone and thereby depletes the effective concentration of ozone present in atmosphere. Hence, <span>the Montreal Protocol was signed in 1987 by major countries of the world. This aim of this protocol was to protect the stratospheric ozone layer by phasing out the production and consumption of ozone-depleting substances.</span>
6 0
3 years ago
Read 2 more answers
Apart from cases of nuclear processes (including fission), elements cannot be broken down to form other substances.
Shtirlitz [24]

Explanation:

<u>E</u><u>l</u><u>e</u><u>m</u><u>e</u><u>n</u><u>t</u> is ur answer

hope it helps u

plz mark as brainlist

7 0
3 years ago
From where do nuclear power plants get uranium?
Anvisha [2.4K]

Answer:

Uranium mines operate in many countries, but more than 85% of uranium is produced in six countries: Kazakhstan, Canada, Australia, Namibia, Niger, and Russia. Historically, conventional mines open pit or underground were the main source of uranium.

Explanation:

Hope this helps

4 0
3 years ago
, glucose is converted into what three products? science
Ivenika [448]

Answer:

carbon dioxide, water, sunlight

6 0
3 years ago
Read 2 more answers
Consider the reaction Mg(s) + I2 (s) → MgI2 (s) Identify the limiting reagent in each of the reaction mixtures below:
Lapatulllka [165]

Answer:

a) Nor Mg, neither I2 is the limiting reactant.

b) I2 is the limiting reactant

c) <u>Mg is the limiting reactant</u>

<u>d) Mg is the limiting reactant</u>

<u>e) Nor Mg, neither I2 is the limiting reactant.</u>

<u>f) I2 is the limiting reactant</u>

<u>g) Nor Mg, neither I2 is the limiting reactant.</u>

<u>h) I2 is the limiting reactant</u>

<u>i) Mg is the limiting reactant</u>

Explanation:

Step 1: The balanced equation:

Mg(s) + I2(s) → MgI2(s)

For 1 mol of Mg we need 1 mol of I2 to produce 1 mol of MgI2

a. 100 atoms of Mg and 100 molecules of I2

We'll have the following equation:

100 Mg(s) + 100 I2(s) → 100MgI2(s)

This is a stoichiometric mixture. <u>Nor Mg, neither I2 is the limiting reactant.</u>

b. 150 atoms of Mg and 100 molecules of I2

We'll have the following equation:

150 Mg(s) + 100 I2(s) → 100 MgI2(s)

<u>I2 is the limiting reactant</u>, and will be completely consumed. There will be consumed 100 Mg atoms. There will remain 50 Mg atoms.

There will be produced 100 MgI2 molecules.

c. 200 atoms of Mg and 300 molecules of I2

We'll have the following equation:

200 Mg(s) + 300 I2(s) →200 MgI2(s)

<u>Mg is the limiting reactant</u>, and will be completely consumed. There will be consumed 200 I2 molecules. There will remain 100 I2 molecules.

There will be produced 200 MgI2 molecules.

d. 0.16 mol Mg and 0.25 mol I2

We'll have the following equation:

Mg(s) + I2(s) → MgI2(s)

<u>Mg is the limiting reactant</u>, and will be completely consumed. There will be consumed 0.16 mol of I2. There will remain 0.09 mol of I2.

There will be produced 0.16 mol of MgI2.

e. 0.14 mol Mg and 0.14 mol I2

We'll have the following equation:

Mg(s) + I2(s) → MgI2(s)

This is a stoichiometric mixture. <u>Nor Mg, neither I2 is the limiting reactant.</u>

There will be consumed 0.14 mol of Mg and 0.14 mol of I2. there will be produced 0.14 mol of MgI2

f. 0.12 mol Mg and 0.08 mol I2

We'll have the following equation:

Mg(s) + I2(s) → MgI2(s)

<u>I2 is the limiting reactant</u>, and will be completely consumed. There will be consumed 0.08 moles of Mg. There will remain 0.04 moles of Mg.

There will be produced 0.08 moles of MgI2.

g. 6.078 g Mg and 63.455 g I2

We'll have the following equation:

Mg(s) + I2(s) → MgI2(s)

Number of moles of Mg = 6.078 grams / 24.31 g/mol = 0.250 moles

Number of moles I2 = 63.455 grams/ 253.8 g/mol = 0.250 moles

This is a stoichiometric mixture. <u>Nor Mg, neither I2 is the limiting reactant.</u>

There will be consumed 0.250 mol of Mg and 0.250 mol of I2. there will be produced 0.250 mol of MgI2

h. 1.00 g Mg and 2.00 g I2

We'll have the following equation:

Mg(s) + I2(s) → MgI2(s)

Number of moles of Mg = 1.00 grams / 24.31 g/mol = 0.0411 moles

Number of moles I2 = 2.00 grams/ 253.8 g/mol = 0.00788 moles

<u>I2 is the limiting reactant</u>, and will be completely consumed. There will be consumed 0.00788 moles of Mg. There will remain 0.03322 moles of Mg.

There will be produced 0.00788 moles of MgI2.

i. 1.00 g Mg and 2.00 g I2

We'll have the following equation:

Mg(s) + I2(s) → MgI2(s)

Number of moles of Mg = 1.00 grams / 24.31 g/mol = 0.0411 moles

Number of moles I2 = 20.00 grams/ 253.8 g/mol = 0.0788 moles

<u>Mg is the limiting reactant</u>, and will be completely consumed. There will be consumed 0.0411 moles of Mg. There will remain 0.0377 moles of I2.

There will be produced 0.0411 moles of MgI2.

4 0
3 years ago
Other questions:
  • Electronegativity increases as the size of the atom increases.<br> a. True<br> b. False
    13·1 answer
  • Explain what happens to energy and matter during a chemical reaction.
    15·1 answer
  • 3.74 ml of carbon dioxide (CO2) accidentally escaped during the experiment, what volume in liters would that be?
    13·2 answers
  • The majority of elements found in the periodic table can be classified as what
    9·1 answer
  • platinum is used as a catalyst to break down harmful gases in car exhaust into less harmful gases. Which statement best describe
    14·1 answer
  • What type of graph, that isn't a bar graph, would be used to show 3 different classes choices of a favourite program and how eac
    10·1 answer
  • What does a mass number mean and what does an atomic number mean?
    9·1 answer
  • This list of characteristics describes organisms in the ________ kingdom.
    5·1 answer
  • Which of the following statements BEST explains the data in the table?
    6·1 answer
  • How many atoms are in 3CaCO3
    10·1 answer
Add answer
Login
Not registered? Fast signup
Signup
Login Signup
Ask question!