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mars1129
11 days ago
9

What message would you have gotten if your computer become infected with the elk cloner virus

Chemistry
2 answers:
Anarel [852]11 days ago
7 0

Elk Cloner: A program that personifies itself. It’ll attach to all your disks. It infiltrates your chips. Yes, it’s Cloner! It adheres to you like glue. It adjusts RAM to engage the Cloner.

VMariaS [1K]11 days ago
4 0

Response:

Elk Cloner is a computer virus made for the Apple II platform. This virus propagates by infecting the disks used by computer operating systems. Although it wasn't primarily coded to cause harm, it could damage discs by overwriting critical tracks irrespective of their content. When a computer is infected with this virus, upon starting up, it displays the following message:

Elk Cloner: The program with personality

You will receive all your records

It will enter your chips

Yes, it is Cloner!

It will adhere to you as glue

Also changes your RAM

Pass it along, Elk Cloner!

You might be interested in
Consider the following system at equilibrium:
VMariaS [1037]

Answer:

A - Increase (R), Decrease (P), Decrease(q), Triple both (Q) and (R)

B - Increase(P), Increase(q), Decrease (R)

C - Triple (P) and cut (q) down to a third

Explanation:

According to the principle of Le Chatelier, when a system reaches equilibrium and a change is introduced, the system will respond to counteract that change.

Since P and Q are reactants, raising the amount of either one or both without a proportional rise in R (which is a product) will cause the equilibrium to move towards the right. Similarly, if R decreases while P and Q remain constant, this too will push the equilibrium to the right. Thus, Increase(P), Increase(q), and Decrease(R) will lead to a rightward shift in the equilibrium.

Conversely, raising R without increasing P and Q will draw the equilibrium to the left. Likewise, cutting down P and/or Q without a similar reduction in R will shift the equilibrium leftward. Therefore, Increase(R), Decrease(P), Decrease(q), and triple both (Q) and (R) will shift the equilibrium to the left.

If there are equivalent changes in P and Q, with R remaining unchanged, then the equilibrium remains stationary. So, tripling (P) while reducing (q) to one third will not alter the equilibrium.

6 0
3 days ago
How much volume (in cm3) is gained by a person who gains 11.1 lbs of pure fat?
Anarel [852]

The density of human fat is 0.918 g/cm^{3}. The mass of the pure fat is 11.1 lbs.

First, convert the mass from pounds to grams as follows:

1 lb=453.592 g

Thus,

11.1 lb\rightarrow 11.1\times 453.592 g=5034.875 g

Density is defined as mass per unit volume, meaning volume can be calculated as:

V=\frac{m}{d}

By substituting the values,

V=\frac{5034.875 g}{0.918 g/cm^{3}}=5484.61 cm^{3}

Consequently, the volume gained by the individual will be 5484.61 cm^{3}.

6 0
8 days ago
"The compound K2O2 also exists. A chemist can determine the mass of K in a sample of known mass that consists of either pure K2O
lorasvet [956]

Answer:

Indeed, the chemist is capable of identifying the compound present in the sample.

Explanation:

In one mole of K₂O, potassium has a mass of 2 × 39.1 g = 78.2 g, while the total mass of K₂O is 94.2 g. The mass ratio of K compared to K₂O is calculated as 78.2 g / 94.2 g = 0.830.

For 1 mole of K₂O₂, potassium's mass remains the same at 78.2 g, but the total mass of K₂O₂ is 110.2 g. The mass ratio of K to K₂O₂ then equates to 78.2 g / 110.2 g = 0.710.

When the chemist measures the mass of K in relation to the overall sample, the mass ratio can be computed.

  • If the mass ratio is 0.830, then it indicates a pure K₂O compound.
  • If the mass ratio is 0.710, it indicates a pure K₂O₂ compound.
  • If the mass ratio falls outside of 0.830 or 0.710, the sample is assessed to be a mixture.
6 0
13 days ago
When drawing the Lewis structure for a molecule, after drawing the skeletal structure and distributing all of the electrons arou
Anarel [852]

Answer: Rearrange the lone pairs of electrons from the outer atom(s) to create double or triple bonds with the central atom.

Explanation:

7 0
2 days ago
1. For which of these elements would the first ionization energy of the atom be higher than that of the diatomic molecule?
alisha [964]

Answer: The correct selection is (b).

Explanation:

The energy required to detach an electron from an atom or ion in its gaseous state is termed ionization energy.

This indicates that a smaller atom necessitates a greater amount of energy to remove its valence electron. The reason for this is that there exists a strong attraction between the nucleus and the electrons in smaller atoms or elements.

Therefore, a significant amount of energy is needed to dislodge the valence electrons.

The electronic configuration for helium is 1s^{2}. Hence, due to its fully occupied valence shell, it exhibits greater stability.

Consequently, a large amount of energy is needed to remove an electron from a helium atom.

In conclusion, from the choices provided, the ionization energy of helium will be greater than that of the diatomic molecule.

7 0
9 days ago
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