This subject is a sub-topic in Computer Science and is related to how computers process tasks using Modern Processing Architectures. Modern Processor Architecture is an offshoot of computer architecture.
<h3>
What is Modern Processing Architecture?</h3>
Please note that the information is incomplete hence the general answer. The complete question should provide figures for items 1 to 3.
Modern Processor Architecture is the name given to computer processors with highly advanced capabilities.
In simple language, processors with modern architectures are those that have been built with the ability to complete many instructions or tasks at the same time.
It can also perform or execute these instructions in random order. When a processor is able to do this, it is called Asynchronous performance.
Learn more about Computer Architecture at:
brainly.com/question/18185805
Answer:
-
= 1
= 1
Explanation:
Argon atom has atomic number 18. Then, it has 18 protons and 18 electrons.
To determine the quantum numbers you must do the electron configuration.
Aufbau's principle is a mnemonic rule to remember the rank of the orbitals in increasing order of energy.
The rank of energy is:
1s < 2s < 2p < 3s < 3p < 4s < 3d < 4p < 5s < 4d < 5p < 6s < 4f < 5d < 6p < 7s < 5f < 6d < 7d
You must fill the orbitals in order until you have 18 electrons:
- 1s² 2s² 2p⁶ 3s² 3p⁶ : 2 + 2 + 6 + 2 + 6 = 18 electrons.
The last electron is in the 3p orbital.
The quantum numbers associated with the 3p orbitals are:
- = 1 (orbitals s correspond to = 0, orbitals p correspond to = 1, orbitals d, correspond to = 2 , and orbitals f correspond to = 3)
- can be -1, 0, or 1 (from - to + )
- the fourth quantum number, the spin can be +1/2 or -1/2
Thus, the six possibilities for the last six electrons are:
- (3, 1, -1 +1/2)
- (3, 1, -1, -1/2)
- (3, 1, 0, +1/2)
- (3, 1, 0, -1/2)
- (3, 1, 1, +1/2)
- (3, 1, 1, -1/2)
Hence, the correct choice is:
- = 1
- = 1
Answer:
boolean isEven = false;
if (x.length % 2 == 0)
isEven = true;
Comparable currentMax;
int currentMaxIndex;
for (int i = x.length - 1; i >= 1; i--)
{
currentMax = x[i];
currentMaxIndex = i;
for (int j = i - 1; j >= 0; j--)
{
if (((Comparable)currentMax).compareTo(x[j]) < 0)
{
currentMax = x[j];
currentMaxIndex = j;
}
}
x[currentMaxIndex] = x[i];
x[i] = currentMax;
}
Comparable a = null;
Comparable b = null;
if (isEven == true)
{
a = x[x.length/2];
b = x[(x.length/2) - 1];
if ((a).compareTo(b) > 0)
m = a;
else
m = b;
}
else
m = x[x.length/2];
