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Phoenix [80]
3 years ago
14

Phosphorus-32 has a half-life of 14.3 days. how many grams remain from a 10.0 gram sample after 30.0 days?

Chemistry
2 answers:
QveST [7]3 years ago
5 0

Answer: 2.23 grams

Explanation:

Radioactive decay follows first order kinetics.

Half-life of Phosphorus-32 = 14.3 days

\lambda =\frac{0.693}{t_{\frac{1}{2}}}=\frac{0.693}{14.3}= 0.05days^{-1}

N=N_o\times e^{-\lambda t}

N = amount left after time t= ?

N_0 = initial amount  = 10.0 g

\lambda = rate constant= 0.05days^{-1}

t= time  = 30 days

N=10\times e^{- 0.05 days^{-1}\times 30days}

N=2.23g

siniylev [52]3 years ago
3 0
When the nuclear half-life of the radioactive isotope is showing the time needed for the isotope to be half of its initial value of mass.

so with each half-life, the isotope will be halved of its initial value as example:

after the first half-life, the isotope will lose 50 % of its initial value

and after the second half-life, the isotope will lose 25% of its initial value 

and after the third half-life, the isotope will lose 12.5 % of its initial value

and so on,

So here to get how many numbers of half-lives we will use this formula:

numbers of half-lives = total time passed / the half-life of the isotope

                                    = 30 days / 14 days

                                    =2 days

∴remainig mass = initial mass / 2^numbers of half-lives

                            = 10 g / 2^2

                            = 2.5 g

 
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7 0
2 years ago
How many grams of aspartame (C14H18N2O5), also known as Nutrasweet, contain 1.98×1021molecules of aspartame?
Umnica [9.8K]

Answer:

0.97 grams

Explanation:

number of molecules=no of moles × Avogadros number

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number of moles=nu.of molecules ÷ 6.02×10power23 = 1.98×10power21 ÷ 6.02×10p23=0.0033 mol

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3 0
3 years ago
100.00 mL of 0.15 M nitrous acid (HNO2) are titrated with a 0.15 M NaOH solution. (a) Calculate the pH for the initial solution.
wolverine [178]

Answer:

a. pH = 2.04

b. pH = 3.85

c. pH = 8.06

d. pH = 11.56

Explanation:

The nitrous acid is a weak acid (Ka = 5.6x10⁻⁴) that reacts with NaOH as follows:

HNO₂ + NaOH → NaNO₂(aq) + H₂O(l)

a. At the beginning there is just a solution of 0.12M HNO₂. As Ka is:

Ka = [H⁺] [NO₂⁻] / [HNO₂]

Where [H⁺] and [NO₂⁻] ions comes from the same equilibrium ([H⁺] = [NO₂⁻] = X):

5.6x10⁻⁴ = X² / 0.15M

8.4x10⁻⁵ = X²

X = [H⁺] = 9.165x10⁻³M

As pH = -log [H⁺]

<h3>pH = 2.04</h3><h3 />

b. At this point we have HNO₂ and NaNO₂ (The weak acid and the conjugate base), a buffer. The pH of a buffer is obtained using H-H equation:

pH = pKa + log [NaNO₂] / [HNO₂]

<em>Where pH is the pH of the buffer,</em>

<em>pKa is -log Ka = 3.25</em>

<em>And [NaNO₂] [HNO₂] could be taken as the moles of each compound.</em>

<em />

The initial moles of HNO₂ are:

0.100L * (0.15mol / L) = 0.015moles

The moles of base added are:

0.0800L * (0.15mol / L) = 0.012moles

The moles of base added = Moles of NaNO₂ produced = 0.012moles.

And the moles of HNO₂ that remains are:

0.015moles - 0.012moles = 0.003moles

Replacing in H-H equation:

pH = 3.25 + log [0.012moles] / [0.003moles]

<h3>pH = 3.85</h3><h3 />

c. At equivalence point all HNO2 reacts producing NaNO₂. The volume added of NaOH must be 100mL. That means the concentration of the NaNO₂ is:

0.15M / 2 = 0.075M

The NaNO₂ is in equilibrium with water as follows:

NaNO₂(aq) + H₂O(l) ⇄ HNO₂(aq) + OH⁻(aq) + Na⁺

The equilibrium constant, kb, is:

Kb = Kw/Ka = 1x10⁻¹⁴ / 5.6x10⁻⁴ = 1.79x10⁻¹¹ = [OH⁻] [HNO₂] / [NaNO₂]

<em>Where [OH⁻] = [HNO₂] = x</em>

<em>[NaNO₂] = 0.075M</em>

<em />

1.79x10⁻¹¹ = [X] [X] / [0.075M]

1.34x10⁻¹² = X²

X = 1.16x10⁻⁶M = [OH⁻]

pOH = -log [OH-] = 5.94

pH = 14-pOH

<h3>pH = 8.06</h3><h3 />

d. At this point, 5mL of NaOH are added in excess, the moles are:

5mL = 5x10⁻³L * (0.15mol / L) =7.5x10⁻⁴moles NaOH

In 100mL + 105mL = 205mL = 0.205L. [NaOH] = 7.5x10⁻⁴moles NaOH / 0.205L =

3.66x10⁻³M = [OH⁻]

pOH = 2.44

pH = 14 - pOH

<h3>pH = 11.56</h3>
5 0
3 years ago
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