Answer:
11.9g remains after 48.2 days
Explanation:
All isotope decay follows the equation:
ln [A] = -kt + ln [A]₀
<em>Where [A] is actual amount of the isotope after time t, k is decay constant and [A]₀ the initial amount of the isotope</em>
We can find k from half-life as follows:
k = ln 2 / Half-Life
k = ln2 / 27.7 days
k = 0.025 days⁻¹
t = 48.2 days
[A] = ?
[A]₀ = 39.7mg
ln [A] = -0.025 days⁻¹*48.2 days + ln [39.7mg]
ln[A] = 2.476
[A] = 11.9g remains after 48.2 days
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I need more characters but it is b
Answer:
Dipole, less electronegativity, higher electronegativity
Explanation:
!!! Incomplete Question
When two atoms with different electronegativities are bonded together, a bond dipole exists.
A bond dipole is a partial charge assigned to bonded atoms due to difference in electron density, difference in electronegativity is a factor to this.
These are displayed with an arrow originating at the atom with the less electronegativity and pointing toward the atom with the higher electronegativity value
Answer:
Ratio is 4:1
Explanation:
from graham's law of gaseous diffusion:

but Ra = 2Rb:

Answer:
To determine the enthalpy and entropy of dissolving a compound, you need to measure the Ksp at multiple temperatures. Then, plot ln(Ksp) vs. 1/T. The slope of the plotted line relates to the enthalpy (ΔH) of dissolving and the intercept of the plotted line relates to the entropy (ΔS) of dissolving.
Explanation:
Hello there!
In this case, according to the given information, it turns out possible for us use the thermodynamic definition of the Gibbs free energy and its relationship with Ksp as follows:

Thus, by combining them, we obtain:

Which is related to the general line equation:

Whereas:

It means that we answer to the blanks as follows:
To determine the enthalpy and entropy of dissolving a compound, you need to measure the Ksp at multiple temperatures. Then, plot ln(Ksp) vs. 1/T. The slope of the plotted line relates to the enthalpy (ΔH) of dissolving and the intercept of the plotted line relates to the entropy (ΔS) of dissolving.
Regards!