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Alekssandra [29.7K]
3 years ago
15

A particular radioactive nuclide has a half-life of 1000 years. What percentage of an initial population of this nuclide has dec

ayed after 3500 years?
Chemistry
1 answer:
Delicious77 [7]3 years ago
3 0

Answer:

91.16% has decayed & 8.84% remains

Explanation:

A = A₀e⁻ᵏᵗ => ln(A/A₀) = ln(e⁻ᵏᵗ) => lnA - lnA₀ = -kt => lnA = lnA₀ - kt

Rate Constant (k) = 0.693/half-life = 0.693/10³yrs = 6.93 x 10ˉ⁴yrsˉ¹

Time (t) = 1000yrs  

A = fraction of nuclide remaining after 1000yrs

A₀ = original amount of nuclide = 1.00 (= 100%)  

lnA = lnA₀ - kt

lnA = ln(1) – (6.93 x 10ˉ⁴yrsˉ¹)(3500yrs) = -2.426

A = eˉ²∙⁴²⁶ = 0.0884 = fraction of nuclide remaining after 3500 years

Amount of nuclide decayed = 1 – 0.0884 = 0.9116 or 91.16% has decayed.

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Using spectroscopic notation write the complete electron configuration for the fluorine atom. Using noble gas notation write the
prohojiy [21]

Explanation:

Flourine has atomic number of 9 and hence 9 electrons in its neutral state. The full electronic configuration is given as;

1s2 2s2 2p5

Carbon has atomic number of 6 and hence 6 electrons in it's neutral state. The noble gas notation as the following format;

[closest noble gas before the element] remaining electrons

The nearest noble gas to carbon is Helium, the noble gas notation is given as;

[He] 2s4

3 0
3 years ago
A 20.0 gram sample of an element contains 4.95 x 1023 atoms. what is the element?
Sonja [21]
You will want to find how many grams are in a whole mole so you know which element it is. To do this, find out how much of a mole you have.

4.95 x 10^23 atoms / 6.022 x 10^23 atoms (one whole mole of any element) = .8219860511 or ~82% of 1 mole

Now we know that, find what to multiply 20 g by to get the rest of the mole.

1 mole / .8219860511 mole = 1.216565657

20 g x 1.216565657 = ~24.33 g / mol

Now that you have grams per mole, you can look at the periodic table and the molar masses to see which this number is closely aligned.

Your answer is Magnesium (Mg), which has a molar mass of 24.305 g


7 0
2 years ago
There are two binary compounds of mercury and oxygen. heating either of them results in the decomposition of the compound, with
grandymaker [24]

\text{Hg} \text{O} and \text{Hg}_{2} \text{O}.

Assuming complete decomposition of both samples,

  • m(\text{Hg}) = m(\text{residure})
  • m(\text{O}) = m(\text{loss})

First compound:

  • m(\text{O}) = m(\text{loss}) = 0.6498 - 0.6018 = 0.048 \; g
  • m(\text{Hg}) = m(\text{residure}) = 0.6018 \; g

n = m/M; 0.6498 \; g of the first compound would contain

  • n(\text{O atoms}) = 0.048 \; g  / 16 \; g \cdot mol^{-1}= 0.003 \; mol
  • n(\text{Hg atoms}) = 0.6018 \; g  / 200.58 \; g \cdot mol^{-1}= 0.003 \; mol

Oxygen and mercury atoms seemingly exist in the first compound at a 1:1 ratio; thus the empirical formula for this compound would be \text{Hg} \text{O} where the subscript "1" is omitted.

Similarly, for the second compound

  • m(\text{O}) = m(\text{loss}) = 0.016 \; g
  • m(\text{Hg}) = m(\text{residure}) = 0.4172 - 0.016 = 0.4012  \; g

n = m/M; 0.4172 \; g of the first compound would contain

  • n(\text{O atoms}) = 0.016 \; g  / 16 \; g \cdot mol^{-1}= 0.001 \; mol
  • n(\text{Hg atoms}) = 0.4012 \; g  / 200.58 \; g \cdot mol^{-1}= 0.002 \; mol

n(\text{Hg}) : n(\text{O}) \approx  2:1 and therefore the empirical formula

\text{Hg}_{2} \text{O}.

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3 years ago
Why do you think different liquids have different freezing points
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I think that different liquids have different freezing points because every liquid consists of different atoms and different things that make up the atom causing them to have different freezing points.
5 0
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Which of the following pairs of elements is most likely to form an ionic bond?
balu736 [363]

A pair of elements will most likely form an ionic bond if one is a metal and one is a nonmetal. These types of ionic compounds are composed of monatomic cations and anions. ( K, Cl)...

4 0
2 years ago
Read 2 more answers
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