The force that holds protons and neutrons together is too strong to overcome.
<h3>Explanation</h3>
Consider the location of the particles in an atom.
- Electrons are found outside the nucleus.
- Protons and neutrons are found within the nucleus.
Protons carry positive charges and repel each other. The nucleus will break apart without the strong force that holds the protons and neutrons together. This force is much stronger than the attraction between the nucleus and the electrons. X-rays are energetic enough for removing electrons from an atom. However, you'll need a collider to remove protons from a stable nucleus. You could well have ionized the atom with all that energy.
Also, changing the number of protons per nucleus will convert the halogen atom to an atom of a different element. Rather than making the halogen negative, removing a proton will convert the halogen atom to the negative ion of a different element.
The answer to your question is D
Answer:
56 L
Explanation:
We're dealing with a gas in this problem. We may, therefore, apply the ideal gas law for this problem:
![pV = nRT](https://tex.z-dn.net/?f=pV%20%3D%20nRT)
We now that we have a constant pressure. Besides, R, the ideal gas law constant, is also a constant number. Let's rearrange the equation so that we have all constant variables on the right and all changing variables on the left:
![\frac{V}{T} = \frac{nR}{p} = const](https://tex.z-dn.net/?f=%5Cfrac%7BV%7D%7BT%7D%20%3D%20%5Cfrac%7BnR%7D%7Bp%7D%20%3D%20const)
This means the ratio between volume and temperature is a constant number. For two conditions:
![\frac{V_1}{T_1} = \frac{V_2}{T_2}](https://tex.z-dn.net/?f=%5Cfrac%7BV_1%7D%7BT_1%7D%20%3D%20%5Cfrac%7BV_2%7D%7BT_2%7D)
Given initial volume of:
![V_1 = 93 L](https://tex.z-dn.net/?f=V_1%20%3D%2093%20L)
Convert the initial temperature into Kelvin:
![T_1 = 145^oC + 273.15 K = 418.15 K](https://tex.z-dn.net/?f=T_1%20%3D%20145%5EoC%20%2B%20273.15%20K%20%3D%20418.15%20K)
Convert the final temperature into Kelvin:
![T_2 = -22^oC + 273.15 K = 251.15 K](https://tex.z-dn.net/?f=T_2%20%3D%20-22%5EoC%20%2B%20273.15%20K%20%3D%20251.15%20K)
Rearrange the equation for the final volume:
![V_2 = V_1 \cdot \frac{T_2}{T_1} = 93 L\cdot \frac{251.15 K}{418.15 K} = 56 L](https://tex.z-dn.net/?f=V_2%20%3D%20V_1%20%5Ccdot%20%5Cfrac%7BT_2%7D%7BT_1%7D%20%3D%2093%20L%5Ccdot%20%5Cfrac%7B251.15%20K%7D%7B418.15%20K%7D%20%3D%2056%20L)
First calculate density of the the metal (should equal to density of iron)
- Mass=18.2g
- Volume=2.56cm^3
![\\ \bull\tt\longmapsto Density=\dfrac{Mass}{Volume}](https://tex.z-dn.net/?f=%5C%5C%20%5Cbull%5Ctt%5Clongmapsto%20Density%3D%5Cdfrac%7BMass%7D%7BVolume%7D)
![\\ \bull\tt\longmapsto Density=\dfrac{18.2}{2.56}](https://tex.z-dn.net/?f=%5C%5C%20%5Cbull%5Ctt%5Clongmapsto%20Density%3D%5Cdfrac%7B18.2%7D%7B2.56%7D)
![\\ \bull\tt\longmapsto Density=7.1g/cm^3](https://tex.z-dn.net/?f=%5C%5C%20%5Cbull%5Ctt%5Clongmapsto%20Density%3D7.1g%2Fcm%5E3)
Metal is not iron