It is harder to remove an electron from fluorine than from carbon because the size of the nuclear charge in fluorine is larger than that of carbon.
The energy required to remove an electron from an atom is called ionization energy.
The ionization energy largely depends on the size of the nuclear charge. The larger the size of the nuclear charge, the higher the ionization energy because it will be more difficult to remove an electron from the atom owing to increased electrostatic attraction between the nucleus and orbital electrons.
Since fluorine has a higher size of the nuclear charge than carbon. More energy is required to remove an electron from fluorine than from carbon leading to the observation that; it is harder to remove an electron from fluorine than from carbon.
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Answer:
120 gram sample of a radioactive element, how many grams of that element will be left after 3 half-lives have passed? If you have a 300
Explanation:
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Answer:
molecules C6H12O6 = 2.674 E22 molecules.
Explanation:
from periodic table:
⇒ molecular mass C6H12O6 = ((6)(12.011)) + ((12)(1.008)) + ((6)(15.999))
⇒ Mw C6H12O6 = 180.156 g/mol
⇒ mol C6H12O6 = (8.00 g)(mol/180.156 g) = 0.0444 mol C6H12O6
∴ mol ≡ 6.022 E23 molecules
⇒ molecules C6H12O6 = (0.0444 mol)(6.022 E23 molecules/mol)
⇒ molecules C6H12O6 = 2.674 E22 molecules
Answer:
To find the mass percent of hydrogen in hydrogen chloride, we must divide the weight of the hydrogen atom alone by the weight of the entire molecule. Then we multiply by 100% to find the percentage. Thus, 2.77% of the mass of hydrogen chloride is hydrogen.
Explanation:
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Answer:
% = 76.75%
Explanation:
To solve this problem, we just need to use the expressions of half life and it's relation with the concentration or mass of a compound. That expression is the following:
A = A₀ e^(-kt) (1)
Where:
A and A₀: concentrations or mass of the compounds, (final and initial)
k: constant decay of the compound
t: given time
Now to get the value of k, we should use the following expression:
k = ln2 / t₁/₂ (2)
You should note that this expression is valid when the reaction is of order 1 or first order. In this kind of exercises, we can assume it's a first order because we are not using the isotope for a reaction.
Now, let's calculate k:
k = ln2 / 956.3
k = 7.25x10⁻⁴ d⁻¹
With this value, we just replace it in (1) to get the final mass of the isotope. The given time is 1 year or 365 days so:
A = 250 e^(-7.25x10⁻⁴ * 365)
A = 250 e^(-0.7675)
A = 191.87 g
However, the question is the percentage left after 1 year so:
% = (191.87 / 250) * 100
<h2>
% = 76.75%</h2><h2>
And this is the % of isotope after 1 year</h2>
