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3 years ago

The half life of Po-218 is three minutes. How much of a 200 atom sample will remain after 15 minutes

Chemistry

1 answer:

MakcuM [25]3 years ago

6 0

<h3>Answer:</h3>

6.25 atoms

<h3>Explanation:</h3>

<u>We are given</u>;

The half life of Po-218 is 3 minutes

Initial sample is 200 atom
Time of decay is 15 minutes

We are required to calculate the remaining mass after decay;

Half life refers to the time taken for original amount of a radioactive sample to decay to a half.

To calculate the remaining mass we use the formula;

N = N₀ × 0.5^n where n is the number of half lives, N is the new amount and N₀ is the original amount.

n = 15 min ÷ 3 min

= 5

Therefore;

New amount = 200 atom × 0.5^5

= 6.25 atoms

Therefore; the amount of the sample that will remain after 15 minutes is 6.25 atoms.

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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

N = 1/4 × 400

N = 0.25 × 400

N = 100 g

Thus, the amount of the radioactive isotope remaing is the 100 g.

3 0

3 years ago

Hydrogen cyanide, HCN, can be made by a two-step process. First, ammonia reacts with O2 to give nitric oxide, NO.

stealth61 [152]

Answer:

The mass of HCN is 79.65 g.

The mass of reactant which remain at the end of both reactions is 88.5 g.

Explanation:

Given that,

Mass of ammonia = 50.2 g

Mass of methane = 48.4 g

Hydrogen cyanide, HCN, can be made by a two-step process

Ammonia reacts with O₂ to give nitric oxide NO.

The reaction is,

$4NH_{3}+5O_{2}\Rightarrow 4NO+6H_{2}O$

We need to calculate the mole of NO

Using given data,

2.25 g NH_{3}=\dfrac{50.2}{17}= 2.95\ mole\ NH_{3} [/tex]

$4\ mole NH_{3}\ glose 4\ mol NO$

2.95 mol NH₃ will produced 2.95 mol NO

Then nitric oxide reacts with methane,

The reaction is,

$2NO+2CH_{4}\Rightarrow 2HCN+2H_{2}O+H_{2}$

We need to calculate the mole of methane

Using given data,

$mole\ of\ methane=\dfrac{48.4}{16}$

$mole\ of\ methane = 3.03\ moles$

2 mole NO produced 2 mole HCN

2.95 mol NO will produced $\dfrac{2.95\times3.03}{3.03}$ = 2.95 mol HCN

We need to calculate the mass of HCN

Using formula of mass

$m=N\times M$

Where, N = number of mole

M = molecular mass

Put the value into the formula

$m=2.95\times27$

$m= 79.65\ g$

The mass of HCN is 79.65 g.

We need to calculate the mass of NO

Using formula of mass

$m=N\times M$

Where, N = number of mole

M = molecular mass

Put the value into the formula

$m=2.95\times30$

$m= 88.5\ g$

Hence, The mass of HCN is 79.65 g.

The mass of reactant which remain at the end of both reactions is 88.5 g.

4 0

3 years ago

What are some possible explanations you think a researcher might find to explain why the fish are washing up on the shore of Lak

8_murik_8 [283]

Answer:

the temperature or weather ,

5 0

3 years ago

A 265-mL flask contains pure helium at a pressure of 751 torrs. A second flask with a volume of 465 mL contains pure argon at a

Nadya [2.5K]

Answer:

Total Pressure = 745.6 torr

Partial Pressure of He = 272.8 torr

Partial Pressure of Ar = 472.8 torr

Explanation:

Step 1: Data given

Volume of the flask helium = 265 mL

Pressure in the helium flask = 751 torr = 751/760 atm

Volume of the flask argon = 465 mL

Pressure in the argon flask = 727 torr = 727/760 atm

The total pressure exerted by a gaseous mixture is equal to the sum of the partial pressures of each individual component in a gas mixture.

Step 2: Calculate total volume

Total volume = 265 mL + 465 mL = 730 mL = 0.730 L

Step 3: Boyle's Law:

P1V1=P2V2

⇒ with P1 = total pressure gas exerts in its own flask

⇒ with V1 = volume of flask with stopcock valve closed

⇒ with P2 = partial pressure of gas exerts on total volume of both flasks when stopcock valve is opened

⇒ with V2 = total volume of both flasks with stopcock valve opened

Helium using Boyle's Law equation from above:

P1V1=P2V2

⇒ with P1 = Pressure of helium = 751 /760 = 0.98816 atm

⇒ with V1 = volume of helium = 0.265 L

⇒ with P2 = The new partial pressure of helium

⇒ with V2 = total volume = 0.730 L

(0.98816 atm)(0.265L)=P2(0.730L)

P2=0.359 atm

Argon using Boyle's Law equation from above:

P1V1=P2V2

⇒ with P1 = Pressure of argon = 727/760 = 0.95658 atm

⇒ with V1 = volume of argon = 0.465 L

⇒ with P2 = The new partial pressure of argon

⇒ with V2 = total volume = 0.730 L

(0.95658 atm)(0.465L)=P2(0.730L)

P2=0.609 atm

Step 4: Convert pressure in atm to torr

Pressure helium = 0.359 atm = 272.8 torr

Pressure argon = 0.609 atm = 472.8 torr

Step 5: Calculate Total pressure

Ptotal = P(He)+P(Ar)

⇒ Pt = total pressure of the gas mixture

⇒ P(He) = partial pressure of Helium

⇒ P(Ar) = partial pressure of Argon

Pt = 272.8 torr + 472.8 torr

Pt = 745.6 torr

Total Pressure = 745.6 torr

Partial Pressure of He = 272.8 torr

Partial Pressure of Ar = 472.8 torr

5 0

3 years ago

A substance has a solubility of 350 ppm.how many grams of the substance are present in 1.0l of a saturated solution

sergeinik [125]

Since the given solubility is 350 ppm, convert it first with fraction of solubility. by dividing the solubility with 10^6
S = 350 / 10^6
s = 3.5 x 10^-4
the multiply it to the total solution to calculate the amount of substance present
m = ( 3.5 x 10^-4 ) ( 1.01 )
m = 3.535 x 10^-4 g of the substance present

6 0

3 years ago

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