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Evgesh-ka [11]
3 years ago
9

Can anyone help me on this please?

Chemistry
1 answer:
Bad White [126]3 years ago
7 0
This question is testing to see how well you understand the "half-life" of radioactive elements, and how well you can manipulate and dance around them.  This is not an easy question.

The idea is that the "half-life" is a certain amount of time.  It's the time it takes for 'half' of the atoms in any sample of that particular unstable element to 'decay' ... their nuclei die, fall apart, and turn into nuclei of other elements.

Look over the table.  There are 4,500 atoms of this radioactive substance when the time is 12,000 seconds, and there are 2,250 atoms of it left when the time is ' y ' seconds.  Gosh ... 2,250 is exactly half of 4,500 !  So the length of time from 12,000 seconds until ' y ' is the half life of this substance !  But how can we find the length of the half-life ? ? ?

Maybe we can figure it out from other information in the table !

Here's what I found:

Do you see the time when there were 3,600 atoms of it ? 
That's 20,000 seconds.

... After one half-life, there were 1,800 atoms left.
... After another half-life, there were 900 atoms left.
... After another half-life, there were 450 atoms left. 

==>  450 is in the table !  That's at 95,000 seconds.

So the length of time from 20,000 seconds until 95,000 seconds
is three half-lifes.

The length of time is (95,000 - 20,000) = 75,000 sec

                                     3 half lifes = 75,000 sec

Divide each side by 3 :   1 half life = 25,000 seconds

There it is !  THAT's the number we need.  We can answer the question now.

==> 2,250 atoms is half of 4,500 atoms.

==> ' y ' is one half-life later than 12,000 seconds

==> ' y ' = 12,000 + 25,000

         y   = 37,000 seconds  .

Check: 
Look how nicely 37,000sec fits in between 20,000 and 60,000 in the table.

As I said earlier, this is not the simplest half-life problem I've seen.
You really have to know what you're doing on this one.  You can't
bluff through it.


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Rank the following fertilizers in decreasing order of mass percentage of nitrogen:
charle [14.2K]
<h3>Answer:</h3>

        NH₃ > NH₄NO₃ > (NH₄)₂HPO₄ > (NH₄)₂SO₄ > KNO₃ > (NH₄)H₂PO₄

<h3>Soution:</h3>

In (NH₄)₂HPO₄:

Mass of Nitrogen  =  N × 2  =  14 × 2  =  28 g.mol⁻¹

Molar Mass of (NH₄)₂HPO₄  =  132.06 g.mol⁻¹

Mass %age  =  Mass of N / M.Mass of (NH₄)₂HPO₄ × 100

Mass %age  =  28 g.mol⁻¹ / 132.06 g.mol⁻¹ × 100

Mass %age  =  21.20 %

In (NH₄)₂SO₄:

Mass of Nitrogen  =  N × 2  =  14 × 2  =  28 g.mol⁻¹

Molar Mass of (NH₄)₂SO₄  =  132.14 g.mol⁻¹

Mass %age  =  Mass of N / M.Mass of (NH₄)₂SO₄ × 100

Mass %age  =  28 g.mol⁻¹ / 132.14 g.mol⁻¹ × 100

Mass %age  =  21.18 %

In KNO₃:

Mass of Nitrogen  =  N × 1  =  14 × 1  =  14 g.mol⁻¹

Molar Mass of KNO₃  =  101.10 g.mol⁻¹

Mass %age  =  Mass of N / M.Mass of KNO₃ × 100

Mass %age  =  14 g.mol⁻¹ / 101.10 g.mol⁻¹ × 100

Mass %age  =  13.84 %

In (NH₄)H₂PO₄:

Mass of Nitrogen  =  N × 1  =  14 × 1  =  14 g.mol⁻¹

Molar Mass of (NH₄)H₂PO₄  =  115.03 g.mol⁻¹

Mass %age  =  Mass of N / M.Mass of (NH₄)H₂PO₄ × 100

Mass %age  =  14 g.mol⁻¹ / 115.03 g.mol⁻¹ × 100

Mass %age  =  12.17 %

In NH₃:

Mass of Nitrogen  =  N × 1  =  14 × 1  =  14 g.mol⁻¹

Molar Mass of NH₃  =  132.14 g.mol⁻¹

Mass %age  =  Mass of N / M.Mass of NH₃ × 100

Mass %age  =  14 g.mol⁻¹ / 17.03 g.mol⁻¹ × 100

Mass %age  =  82.20 %

In NH₄NO₃:

Mass of Nitrogen  =  N × 2  =  14 × 2  =  28 g.mol⁻¹

Molar Mass of NH₄NO₃  =  80.04 g.mol⁻¹

Mass %age  =  Mass of N / M.Mass of NH₄NO₃ × 100

Mass %age  =  28 g.mol⁻¹ / 80.04 g.mol⁻¹ × 100

Mass %age  =  34.98 %

5 0
3 years ago
Predict the following chemical formula for a compound between aluminum and chlorine. AlCl3 Al3Cl Al2Cl3 Al3Cl2
Tasya [4]
Both aluminum and chlorine have known charges, which are +3 and -1 respectively. To make them cancel each other out in charge, you would need 3 chlorine and for one aluminum, therefore AlCl_{3} would be correct
7 0
3 years ago
What is the molarity of a KF(aq) solution containing 3.0 mol of KF in 2.00L of solution?
sergeinik [125]

Answer : The molarity of KF in the solution is 1.5 M.

Explanation : Given,

Moles of KF = 3.0 mol

Volume of solution = 2.00 L

Molarity : It is defined as the number of moles of solute present in one liter of volume of solution.

Formula used :

\text{Molarity}=\frac{\text{Moles of }KF}{\text{Volume of solution (in L)}}

Now put all the given values in this formula, we get:

\text{Molarity}=\frac{3.0mol}{2.00L}=1.5mole/L=1.5M

Therefore, the molarity of KF in the solution is 1.5 M.

4 0
3 years ago
How many milliliters of 0.50 M KOH are needed
spayn [35]

Answer:

Option D. 30 mL.

Explanation:

Step 1:

The balanced equation for the reaction. This is given below:

HNO3 + KOH —> KNO3 + H2O

From the balanced equation above,

The mole ratio of the acid, nA = 1

The mole ratio of the base, nB = 1

Step 2:

Data obtained from the question. This include the following:

Volume of base, KOH (Vb) =.?

Molarity of base, KOH (Mb) = 0.5M

Volume of acid, HNO3 (Va) = 10mL

Molarity of acid, HNO3 (Ma) = 1.5M

Step 3:

Determination of the volume of the base, KOH needed for the reaction. This can be obtained as follow:

MaVa / MbVb = nA/nB

1.5 x 10 / 0.5 x Vb = 1

Cross multiply

0.5 x Vb = 1.5 x 10

Divide both side by 0.5

Vb = (1.5 x 10) /0.5

Vb = 30mL

Therefore, the volume of the base, KOH needed for the reaction is 30mL.

3 0
3 years ago
How many liters of a 2.0 M solution of HCl do you need to have 8.0 moles of HCI?
kakasveta [241]

Answer:

1M HCl: add 1mol/12M = 83 ml conc. HCl to 1L of water or 8.3ml to 100ml.

2M HCl: add 2mol/12M = 167 ml conc. HCl to 1L of water or 16.7ml to 100ml.

4 0
2 years ago
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