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tankabanditka [31]
3 years ago
14

A rock sample from the moon includes a mineral that contains small amounts of the radioactive isotope Potassium-40 and its daugh

ter element Argon-40 (half-life of 1.3 billion years). This mineral would not form with any Argon-40. Consider a crystal with 7 atoms of Argon-40 for every 1 atom of Potassium-40. How many atoms of Potassium-40 were present when the crystal formed for each atom of Potassium-40 that exists today
Chemistry
1 answer:
Simora [160]3 years ago
6 0

Answer:

There were originally 8 atoms of Potassium-40.

Explanation:

The half-life of a radioactive material is the time taken for half the original material to decay or the time required for a quantity of the radioactive substance to reduce to half of its initial value.

If the original material formed without any Argon-40, it means that the atoms originally present were Potassium-40 atoms.

Presently, there are 7 Argon-40 atoms for every 1 of Potassium-40, we can deduce the number of half-lifes the Potassium-40 has undergone as follows :

After one half-life, (1/2) there will be one Potassium-40 atom for every Argon-40 atom.

After a second half life, 1/2 × 1/2 = 1/4: there will be one Potassium-40 atom for every three atoms of Argon-40.

After a third half-life, 1/4 × 1/2 = 1/8: there will be one Potassium-40 atom for every 7 atoms of Argon-40.

Since there are 1/8 atoms of Potassium-40 presently, there were originally 8 atoms of Potassium-40.

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fenix001 [56]
Step 1: Write the unbalanced equation,

                                 C₂H₆  +  O₂    →    CO₂  +  H₂<span>O

There are 2 C at left hand side and 1 carbon at right hand side. So, multiply CO</span>₂ by 2 to balance C atoms at both side. So,

                                 C₂H₆  +  O₂    →   2 CO₂  +  H₂O

Now, count number of H atoms at both sides. There are 6 H atoms at left hand side and 2 at right hand side. Multiply H₂O by 3 to balance H atoms.


                                 C₂H₆  +  O₂    →   2 CO₂  +  3 H₂O

At last, balance O atoms. There are 2 O atoms at left hand side and 3 O atoms at right hand side. Multiply O₂ with 1.5 (i.e. 3/2) to balance O atoms. i.e.

                                 C₂H₆  +  3/2 O₂    →   2 CO₂  +  3 H₂O

Hence, the equation is balanced. If you want to make equation fraction free then multiply all equation with 2. i.e.

                           ( C₂H₆  +  3/2 O₂    →   2 CO₂  +  3 H₂O ) × 2

                                2 C₂H₆  +  3 O₂    →   4 CO₂  +  6 H₂O
5 0
4 years ago
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4 AlF3 + 3 O2 ----------&gt; 2Al2O3 + 6 F2
anyanavicka [17]

Explanation:

6 F2------->4 AlF3

F2-----------> 4/6 AlF3

8.25 F2 ---------> 4×8.25/6 AlF3

so 5.5 moles

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3 years ago
What is the distinction between magma and lava​
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Answer:

Scientists use the term magma for molten rock that is underground and lava for molten rock that breaks through the Earth's surface.

7 0
3 years ago
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How does nuclear fission of Uranium - 234 result in electricity being generated?
nevsk [136]

Answer:

See explanation

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The use of Uranium - 234 to generate electricity depends on a fission reaction. The uranium nuclide is bombarded by fast moving neutrons leading to a chain reaction. Control rods and moderators are used to keep the nuclear reaction under control.

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7 0
3 years ago
2. Calculate the mass of 3.47x1023 gold atoms.
lapo4ka [179]

3.47 x 10^{23} atoms of gold have mass of 113.44 grams.

Explanation:

Data given:

number of atoms of gold = 3.47 x 10^{23}

mass of the gold in given number of atoms = ?

atomic mass of gold =196.96 grams/mole

Avagadro's number = 6.022 X 10^{23}

from the relation,

1 mole of element contains 6.022 x 10^{23} atoms.

so no of moles of gold given = \frac{3.47 X 10^{23}  }{6.022 X 10^{23} }

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from the relation:

number of moles = \frac{mass}{atomic mass of 1 mole}

rearranging the equation,

mass = number of moles x atomic mass

mass = 0.57 x 196.96

mass = 113.44 grams

thus, 3.47 x 10^{23} atoms of gold have mass of 113.44 grams

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