Answer:
helium , krypton,xenon,radon, argon are noble gasses
Answer:
0.962 atm.
97.4 kPa.
731 torr.
14.1 psi.
97,434.6 Pa.
Explanation:
Hello.
In this case, given the available factors equaling 1 atm of pressure, each required pressure turns out:
- Atmospheres: 1 atm = 760 mmHg:
![p=731mmHg*\frac{1atm}{760mmHg} =0.962atm](https://tex.z-dn.net/?f=p%3D731mmHg%2A%5Cfrac%7B1atm%7D%7B760mmHg%7D%20%3D0.962atm)
- Kilopascals:: 101.3 kPa = 760 mmHg:
![p=731mmHg*\frac{101.3kPa}{760mmHg} =97.4kPa](https://tex.z-dn.net/?f=p%3D731mmHg%2A%5Cfrac%7B101.3kPa%7D%7B760mmHg%7D%20%3D97.4kPa)
- Torrs: 760 torr = 760 mmHg:
![p=731mmHg*\frac{760 torr}{760mmHg} =731 torr](https://tex.z-dn.net/?f=p%3D731mmHg%2A%5Cfrac%7B760%20torr%7D%7B760mmHg%7D%20%3D731%20torr)
- Pounds per square inch: 14.69 psi = 760 mmHg:
![p=731mmHg*\frac{14.69}{760mmHg} =14.1psi](https://tex.z-dn.net/?f=p%3D731mmHg%2A%5Cfrac%7B14.69%7D%7B760mmHg%7D%20%3D14.1psi)
- Pascals: 101300 Pa = 760 mmHg:
![p=731mmHg*\frac{101300Pa}{760mmHg} \\\\p=97,434.6Pa](https://tex.z-dn.net/?f=p%3D731mmHg%2A%5Cfrac%7B101300Pa%7D%7B760mmHg%7D%20%5C%5C%5C%5Cp%3D97%2C434.6Pa)
Best regards.
The possible number and location of all subatomic are one of them is electrically neutral, while the other has a stable electronic configuration.
<h3>What are subatomic particles?</h3>
Subatomic particles are those particles that are present inside the atoms. They are electron, neutron, and proton. They are charged particles, protons are positively charged, electrons are negatively charged and neutrons are neutral.
The protons and electrons totally contribute to the atomic mass of the elements.
Thus, the subatomic particles are electrically neutral and stable to electronic configurations.
To learn more about subatomic particles, refer to the below link:
brainly.com/question/13303285
#SPJ1
Answer:
We need 0.375 mol of CH3OH to prepare the solution
Explanation:
For the problem they give us the following data:
Solution concentration 0,75 M
Mass of Solvent is 0,5Kg
knowing that the density of water is 1g / mL, we find the volume of water:
![d = \frac{g}{mL} \\\\ V= \frac{g}{d} = \frac{500g}{1 \frac{g}{mL} } = 500mL = 0,5 L](https://tex.z-dn.net/?f=d%20%3D%20%5Cfrac%7Bg%7D%7BmL%7D%20%5C%5C%5C%5C%20V%3D%20%5Cfrac%7Bg%7D%7Bd%7D%20%20%3D%20%5Cfrac%7B500g%7D%7B1%20%5Cfrac%7Bg%7D%7BmL%7D%20%7D%20%3D%20500mL%20%3D%200%2C5%20L)
Now, find moles of
are needed using the molarity equation:
therefore the solution is prepared using 0.5 L of H2O and 0.375 moles of CH3OH, resulting in a concentration of 0,75M