From the following list of observations, choose the one that most clearly supports the conclusion that electrons have wave prope
1 answer:
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Answer:
Ok:
Explanation:
So, you can use the Henderson-Hasselbalch equation for this:
pH = pKa + log() where A- is the conjugate base of the acid. In other words, A- is the deprotonated form and HA is the protonated.
We can solve that
1 = log() and so 10 =
or 10HA = A-. For every 1 protonated form of adenosine (HA), there are 10 A-. So, the percent in the protonated form will be 1(1+10) or 1/11 which is close to 9 percent.
The limiting reactant is iron ore, the theoretical yield of iron metal is 701.344 kg, and the theoretical yield of carbon dioxide is 413.292 kg.
<h3>Stoichiometric problem</h3>From the equation of the reaction:
The mole ratio of iron ore to carbon is 2:3.
Mole of 1000 kg of iron ore = 1000000/159.69
= 6,262 moles
Mole of 120 kg carbon = 120000/12
= 10,000 moles
Thus, it appears that the carbon is in excess while the iron ore is limited in availability.
The mole ratio of the iron ore and the iron produced is 1:2. Thus, the equivalent number of moles of iron produced will be:
6,262 x 2 = 12,524 moles
Mass of 12,524 moles of iron = 12,524 x 56
= 701,344 g or 701.344 kg
Thus, the theoretical yield of iron is 701.344 kg.
The mole ratio of the iron ore and the carbon dioxide produced is 2:3. The equivalent mole of carbon dioxide produced will be:
6,262 x 3/2 = 9,393 moles
Mass of 9,393 moles carbon dioxide = 9,393 x 44
= 413,292 or 413.292 kg
The theoretical yield of carbon dioxide is, therefore, 413.292 kg.
More on stoichiometric problems can be found here: brainly.com/question/14465605
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Answer:
(a) AL
(b) Sc
(c)Al
Explanation:
Ionization Energy is the energy required to remove electrons from the outer most orbitals of atom.
The higher the electron is on energy level the farther its from nucleus and more loosely bonded thus need lesser energy.
By looking at electron configuration we can figure out which electron will need more energy.
<h3>(a)Na, Mg, Al</h3>1s², 2s², 2p⁶, 3s², 3p⁶,4s², 3d¹
Na₁₁ ⇒ 1s², 2s², 2p⁶, 3s¹
Mg₁₂ ⇒ 1s², 2s², 2p⁶, 3s²
Al₁₃ ⇒ 1s², 2s², 2p⁶, 3s², 2p¹
Al will have lowest IE₃ as its third electron will be in highest energy level and more loosely bonded to nucleus.
<h3>(b) K, Ca, Sc</h3>K₁₉⇒1s², 2s², 2p⁶, 3s², 3p⁶,4s²
Ca₂₀⇒1s², 2s², 2p⁶, 3s², 3p⁶,4s²
Sc₂₁⇒1s², 2s², 2p⁶, 3s², 3p⁶,4s², 3d¹
Sc will have lowest IE₃ as its third electron will be in highest energy level and more loosely bonded to nucleus.
<h3>(c) Li, Al, B</h3>Li₃ ⇒ 1s², 2s¹
Al₁₃ ⇒ 1s², 2s², 2p⁶, 3s², 2p¹
B₅ ⇒ 1s², 2s², 2p¹
Al will have lowest IE₃ as its third electron will be in highest energy level and more loosely bonded to nucleus.
Answer:
electrons should be added to one of these spheres to make it electrically neutral.
Explanation:
Total charge on each sphere = +3.0 μC =
In order to neutralize the positive charge equal magnitude of negative charge is to be added.
Total charge electrons, Q= -3.0 μC =
Number of electrons = n
Charge on a single electron, e =
Q = n × e
electrons should be added to one of these spheres to make it electrically neutral.