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galina1969 [7]
3 years ago
5

The half life of oxygen is 2 minutes. What fraction of a sample of 0.15 will remain after 5 half lives?​

Chemistry
1 answer:
Natasha_Volkova [10]3 years ago
3 0

Answer:

3.13%.

Explanation:

The following data were obtained from the question:

Original amount (N₀) = 0.15

Half life (t½) = 2 mins

Number of half-life (n) = 5

Fraction of sample remaining =.?

Next, we shall determine the amount remaining (N) after 5 half-life. This can be obtained as follow:

Amount remaining (N) = 1/2ⁿ × original amount (N₀)

NOTE: n is the number of half-life.

N = 1/2ⁿ × N₀

N = 1/2⁵ × 0.15

N = 1/32 × 0.15

N = 0.15/32

N = 4.69×10¯³

Therefore, 4.69×10¯³ is remaining after 5 half-life.

Finally, we shall the fraction of the sample remaining after 5 half-life as follow:

Original amount (N₀) = 0.15

Amount remaining (N) = 4.69×10¯³

Fraction remaining = N/N₀ × 100

Fraction remaining = 4.69×10¯³/0.15 × 100

Fraction remaining = 3.13%

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Give the molecular geometry and number of electron groups for sf4. give the molecular geometry and number of electron groups for
Andrei [34K]

Answer:

5 electron groups, see saw

Explanation:

During the formation of SF4, the sulfur atom usually bonds with each of four fluorine atoms where 8 of valence electrons are used. The four fluorine atoms have 3 lone pairs of electrons in its octet which will further utilize 24 valence electrons. In addition, two electrons are present as a lone pair on the sulfur atom. We can determine sulfur’s hybridization state by counting of the number of regions of electron density on sulphur (the central atom in the molecule). When bonding takes place there is a formation of 4 single bonds to sulfur and it has 1 lone pair. Looking at this, we can say that the number of regions of electron density is 5. The hybridization state is sp3d.

SF4 molecular geometry is seesaw with one pair of valence electrons. The molecule is polar. The equatorial fluorine atoms have 102° bond angles instead of the actual 120° angle. The axial fluorine atom angle is 173° instead of the actual 180° bond angle.

3 0
3 years ago
If a sample of air initially occupies 240L at 2 atm how much pressure is required to compress it to 20L at constant temperature
IceJOKER [234]

Answer:

24 atm.

Explanation:

From the question given above, the following data were obtained:

Initial volume (V₁) = 240 L

Initial pressure (P₁) = 2 atm

Final volume (V₂) = 20 L

Temperature = constant

Final pressure (P₂) =?

The final pressure required, can be obtained by using the Boyle's law equation as shown below:

P₁V₁ = P₂V₂

2 × 240 = P₂ × 20

480 = P₂ × 20

Divide both side by 20

P₂ = 480 / 20

P₂ = 24 atm

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3 years ago
How much of a 400g sample remains after 4 years if a radioactive isotope has a half-life of 2 years?
timama [110]

Answer:

100 g

Explanation:

From the question given above, the following data were obtained:

Original amount (N₀) = 400 g

Time (t) = 4 years

Half-life (t½) = 2 years

Amount remaining (N) =?

Next, we shall determine the number of half-lives that has elapse. This can be obtained as follow:

Time (t) = 4 years

Half-life (t½) = 2 years

Number of half-lives (n) =?

n = t / t½

n = 4 / 2

n = 2

Thus, 2 half-lives has elapsed.

Finally, we shall determine the amount remaining of the radioactive isotope. This can be obtained as follow:

Original amount (N₀) = 400 g

Number of half-lives (n) = 2

Amount remaining (N) =?

N = 1/2ⁿ × N₀

N = 1/2² × 400

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3 0
3 years ago
Limiting Reactant Worksheet
faltersainse [42]

Answer:

a) 1.61 mol

b) Al is limiting reactant

c) HBr is in excess

Explanation:

Given data:

Moles of Al = 3.22 mol

Moles of HBr = 4.96 mol

Moles of H₂ formed = ?

What is limiting reactant =

What is excess reactant = ?

Solution:

Chemical equation:

2Al  + 2HBr  → 2AlBr + H₂

Now we will compare the moles:

                Al              :               H₂

                  2              :               1

                 3.22         :            1/2×3.22 = 1.61 mol

                HBr            :              H₂

                 2                :               1

                4.96            :          1/2×4.96 = 2.48 mol      

The number of moles of H₂ produced by Al are less it will be limiting reactant while HBr is present in excess.

Moles of H₂ :

Number of moles of H₂ = 1.61 mol

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