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lianna [129]
3 years ago
12

Fermium-253 is a radioactive isotope of fermium that has a half-life of 3.0 days. A scientist obtained a sample that contained 2

16 micrograms of fermium-253.
Complete the table to show how much fermium-253 should remain in the sample at the indicated times after the scientist obtained the sample.

Chemistry
2 answers:
Dahasolnce [82]3 years ago
5 0

Problem 2

You start out with 216 ugrams of Fermium - 253. After 3 days, you will have 1/2 as much. 108 ugrams is what you have.

Another 3 days goes by. You started with 108 ugrams. That gets cut in 1/2 again. Now you have 54 ugrams.

Finally another 3 days goes by. You started with 54 ugrams. you now have 1/2 as much which would be 27 ugrams

#days              Amount in micrograms

0                              216

3                               108

6                                54

9                                27

Problem One

You are using Nitrogen as your base example. The first thing you should do is fill in the table. Then you should try and make some rules. You need the rules in case the exam you are preparing for picks a different element to talk about these bond tendencies. In any event, it's handy to think this way.

<em><u>Table</u></em>

Bond               Energy Kj/Mol               Bond Length pico meters

N - N                 167                                                145

N=N                  418                                                125

N≡N                  942                                               110

<em><u>Rules</u></em>

As the number of bonds INCREASES, the energy contained in the bond goes UP

As the number of bonds INCREASES, the length of the bond goes DOWN.

Lera25 [3.4K]3 years ago
5 0

1. <em>Radioactive decay </em>

The <em>half-life</em> of Fm-198 (3.0 days) is the time it takes for half the Fm to decay.  

After one half-life, half (50 %) of the original amount will remain.

After a second half-life, half of that amount (25 %) will remain, and so on.

We can construct a table as follows:

 No. of                   Fraction         Amount

<u>half-lives</u>  <em><u>t</u></em><u>/days</u>  <u>remaining</u>   <u>remaining/µg</u>

       1          3.0             ½                  216

       2         6.0             ¼                  108

       3         9.0             ⅛                    54

2. <em>Bond length and bond energy </em>

The <em>bond order</em> of a bond is the number of bonding electrons between the two nuclei.

As the <em>number of bonding electrons increases</em>, the nuclei are pulled more tightly together (the <em>bond energy increases</em>) and the <em>bond length decreases</em>.

We can construct a table as follows:

<u>Bond</u>  <u>Bond energy/kJ·mol⁻¹</u><u> </u>  <u>Bond length/pm </u>

N-N                 167                                 145

N=N                 418                                 125

N≡N                 942                                 110

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6 0
3 years ago
A gas has a density of 1.57 g/L at 40.0 °C and 2.00 atm of pressure. What is the identity of the gas?
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Answer:

Neon

Explanation:

Step 1: Given and required data

  • Density of the gas (ρ): 1.57 g/L
  • Temperature (T): 40.0°C
  • Pressure (P): 2.00 atm
  • Ideal gas constant (R): 0.08206 atm.L/mol.K

Step 2: Convert T to Kelvin

We will use the following expression.

K = °C + 273.15 = 40.0 + 273.15 = 313.2 K

Step 3: Calculate the molar mass of the gas (M)

For an ideal gas, we will use the following expression.

ρ = P × M/R × T

M = ρ × R × T/P

M = 1.57 g/L × 0.08206 atm.L/mol.K × 313.2 K/2.00 atm

M = 20.17 g/mol

The gas with a molar mass of 20.17 g/mol is Neon.

6 0
2 years ago
The kinetic molecular theory describes the behavior of gases in terms of particles in?
Dominik [7]
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6 0
3 years ago
Which of the following statements does NOT describe the anomeric carbon? A. This carbon is attached to two oxygens. B. This carb
MaRussiya [10]

Answer:

The answer is E. All of the statements describe the anomeric carbon.

Explanation:

When a sugar switches from its open form to its ring form, the carbon from the carbonyl (aldehyde if it is an aldose, or a ketone in the case of a ketose) suffers a nucleophilic addition by one of the hydroxyls in the chain, preferably one that will form a 5 or 6 membered ring after the reaction.

As such, the anomeric carbon will have two oxygens attached (The original one and the one that bonded when the ring closed).

It will be chiral, given that it has 4 different groups attached. (-OR,-OH,-H and -R, where R is the carbon chain).

The hydroxyl group can be in any position (Above of below the ring), depending on with side the addition took place. (See attachment)

It is the carbon of the carbonyl in the open-chain form of the sugar, because it is the only one that can react with the Hydroxyls.

8 0
3 years ago
The half-life of nitrogen-13 is 10.0 minutes. if you begin with 53.3 mg of this isotope, what mass remains after 25.9 minutes ha
zimovet [89]

Hello!

The half-life is the time of half-disintegration, it is the time in which half of the atoms of an isotope disintegrate.

We have the following data:

mo (initial mass) = 53.3 mg

m (final mass after time T) = ? (in mg)

x (number of periods elapsed) = ?

P (Half-life) = 10.0 minutes

T (Elapsed time for sample reduction) = 25.9 minutes

Let's find the number of periods elapsed (x), let us see:

T = x*P

25.9 = x*10.0

25.9 = 10.0\:x

10.0\:x = 25.9

x = \dfrac{25.9}{10.0}

\boxed{x = 2.59}

Now, let's find the final mass (m) of this isotope after the elapsed time, let's see:

m =  \dfrac{m_o}{2^x}

m =  \dfrac{53.3}{2^{2.59}}

m \approx \dfrac{53.3}{6.021}

\boxed{\boxed{m \approx 8.85\:mg}}\end{array}}\qquad\checkmark

I Hope this helps, greetings ... DexteR! =)

3 0
3 years ago
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