c. A full s subshell is able to shield a newly filled p subshell from the nucleus, making the first electron in a p subshell easy to remove.
Explanation:
From the given options, a full s-sublevel is able to shield a newly filled p-subshell from the nucleus thereby making the first electron in a p-subshell easy to remove is correct.
What is ionization energy?
Ionization energy is a measure of the readiness of an atom to lose an electron.
First ionization energy is the energy required to remove the most loosely held electron in the gas phase.
The size of an atom/element depends on the number of electrons it contains. The more the electrons, the larger its size.
- The larger an atom becomes the lesser the ionization energy needed to remove the first electron from its outermost shell.
Electron - electron repulsion occurs when two electrons in the same sub-level repels one another.
Shielding effect is the ability of the inner electrons to protect the outer electrons from the pull of the nuclear charge.
In option C, a s-subshell has a greater shielding effect than the p,d and f sub-shell in that order.
A newly introduced electron in the p-sublevel will be loosely held and easier to remove.
Learn more:
First ionization energy brainly.com/question/2153804
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Answer:
Explanation:
Glucose + ATP → glucose 6-phosphate + ADP The equilibrium constant, Keq, is 7.8 x 102.
In the living E. coli cells,
[ATP] = 7.9 mM;
[ADP] = 1.04 mM,
[glucose] = 2 mM,
[glucose 6-phosphate] = 1 mM.
Determine if the reaction is at equilibrium. If the reaction is not at equilibrium, determine which side the reaction favors in living E. coli cells.
The reaction is given as
Glucose + ATP → glucose 6-phosphate + ADP
Now reaction quotient for given equation above is
![q=\frac{[\text {glucose 6-phosphate}][ADP]}{[Glucose][ATP]}](https://tex.z-dn.net/?f=q%3D%5Cfrac%7B%5B%5Ctext%20%7Bglucose%206-phosphate%7D%5D%5BADP%5D%7D%7B%5BGlucose%5D%5BATP%5D%7D)
so,
⇒ following this criteria the reaction will go towards the right direction ( that is forward reaction is favorable until q = Keq
The answer is 200 g.
If the molar mass of CaCl2 is 110.98 g/mol, this means there are 110.98 g in 1 L of 1 M solution.
Let's find how many g of CaCl2 are present in 0.720 M.
110.98 g : 1 M = x : 0.720 M
x = 110.98 g * 0.720 M : 1 M
x = 79.90 g
So there are 79.90 g in 0.720 M. In other words, in 1 L of 0.720 M solution there will be 79.90 g.
Now, we need to prepare ten beakers with 250 mL of solutions:
10 * 250 mL = 2500 mL = 2.5 L
79.90 g : 1 L = x : 2.5 L
x = 79.90 g * 2.5 L : 1 L
x = 199.75 g ≈ 200 g
Answer:
Explanation:
The half-life of K-40 (1.3 billion years) is the time it takes for half of it to decay.
After one half-life, half (50 %) of the original amount will remain.
After a second half-life, half of that amount (25 %) will remain, and so on.
We can construct a table as follows:
No. of Fraction
<u>half-lives</u> <u> t/yr </u> <u>Remaining</u>
0 0 1
1 1.3 billion ½
2 2.6 ¼
3 3.9 ⅛
We see that after 2 half-lives, ¼ of the original mass remains.
Conversely, if two half-lives have passed, the original mass must have been four times the mass we have now.
Original mass = 4 × 2.10 g =
Answer:
This is due the different charges of fluoride and oxide ions.
Explanation:
When calcium reacts it is oxidized to Ca²⁺. In the same way, fluoride ion is reduced to F⁻ and oxide ion to O²⁻.
When these ions are combined, the molecule must be neutral. That means 2 ions of F⁻ are necessaries and just 1 O²⁻ ion will reacts producing:
CaF₂ and CaO.
The different charges of these ions is the reason why calcium will combine in different ratios.
