Answer:
3.1% is the fraction of the sample after 28650 years
Explanation:
The isotope decay follows the equation:
Ln[A] = -kt + Ln[A]₀
<em>Where [A] could be taken as fraction of isotope after time t, k is decay constant and [A]₀ is initial fraction of the isotope = 1</em>
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k could be obtained from Half-Life as follows:
K = Ln 2 / Half-life
K = ln 2 / 5730 years
K = 1.2097x10⁻⁴ years⁻¹
Replacing in isotope decay equation:
Ln[A] = -1.2097x10⁻⁴ years⁻¹*28650 years + Ln[1]
Ln[A] = -3.4657
[A] = 0.0313 =
<h3>3.1% is the fraction of the sample after 28650 years</h3>
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The answer that your looking for is b
Answer:
a. 5.36x10⁻⁴ g/mL
b. 4.29x10⁻⁵ g/mL
Explanation:
As the units for concentration are not specified, I'll respond using g/mL.
a. We <em>divide the sample mass by the final volume</em> in order to <u>calculate the concentration</u>:
- 0.268 g / 500 mL = 5.36x10⁻⁴ g/mL
b. We can use C₁V₁=C₂V₂ for this question:
- 8.00 mL * 5.36x10⁻⁴ g/mL = C₂ * 100.00 mL
Answer:
Explanation:
Hello!
In this case, since the relationship between entropy and enthalpy for any process is defined below:
For the vaporization of ammonia or any liquid, we can write:
In such a way, solving the temperature of vaporization, or boiling point, we have:
Plugging in the given enthalpy and entropy of vaporization we obtain:
Best regards!