Answer:
A jump occurs when a core electron is removed.
Explanation:
A jump in ionization energy occurs when a core electron is removed. A large jump in the ionization energy easily be seen from the electronic configuration of an element.
For Beryllium, the electronic configuration of is 1s2 2s2.
There are two valence electrons in the outermost shell hence the ionization energy data for beryllium will show a sudden jump or increase in going from the second to the third ionization energy owing to the removal of a core electron
The electronic configuration for Nitrogen is 1s2 2s2 2p3. Five valence electrons are found in the outermost shell so the ionization energy data for nitrogen will show a sudden jump or increase in going from the fifth to sixth ionization energy because of the removal of a core electron
The electronic configuration of oxygen is 1s2 2s2 2p4. There are six valence electrons hence ionization energy for oxygen atom will show a sudden jump or increase in going from the sixth to the seventh ionization energy because of the removal of a core electron
The electronic configuration of Lithium is 1s2 2s1
There is one valence electron in its outermost shell so its ionization energy data will show a sudden jump or increase in going from the first to the second ionization energy because of the removal of a core electron.
Answer:
23.8g
Explanation :
Convert 2.0M into mol using mol= concentration x volume
2.0M x 0.1L (convert 100mL to L since the units for M is mol/L)
= 0.2 mol
We can now find grams by using the molar mass of KBr
=119.023 g/mol (Found online) webqc.org
but can be be calculated by using the molecular weight of K and Br found on the periodic table
We can now calculate the grams by using grams=mol x molar mass
119.023g/mol x 0.2mol
= 23.8046 g
=23.8g (rounded to 1decimal place)
The theoretical yield of NaBr given that 2.36 moles of FeBr₃ reacts is 7.08 moles
<h3>Balanced equation </h3>
2FeBr₃ + 3Na₂S → Fе₂S₃ + 6NaBr
From the balanced equation above,
2 moles FeBr₃ reacted to produce 6 moles of NaBr
<h3>How to determine the theoretical yield of NaBr</h3>
From the balanced equation above,
2 moles FeBr₃ reacted to produce 6 moles of NaBr
Therefore,
2.36 moles FeBr₃ will react to produce = (2.36 × 6) / 2 = 7.08 moles of NaBr
Therefore,
Thus, the theoretical yield of NaBr is 7.08 moles
Learn more about stoichiometry:
brainly.com/question/14735801
#SPJ1
MgCl2
Mg=24
Cl=35.5×2 = 71
rmm= 71+24=95g
1 mole = molar mass
1 mole = 95g
1g = 1/95 moles
345g= 1/95 × 345
3.6 moles
Answer: 75%
Explanation:
The following information can be gotten from the question:
Waste = 70kg
Theoretical yield = 280kg
Therefore, the actual yield will be the difference between the theoretical yield and the waste which will be:
= 280kg - 70kg = 210kg
The percent yield will now be:
= Actual yield / Theoretical yield × 100
= 210/280 × 100
= 3/4 × 100
= 75%
