Answer:
n = 4678.13 mol
Explanation:
Given data:
Temperature = 21°C
Pressure = 746 mmHg
Volume = 115 m³
Number of moles = ?
Solution:
21+273 = 294 k
746 /760 = 0.982 atm
115×1000 = 115000 L
PV = nRT
n = PV/RT
n = 0.982 atm × 115000 L /0.0821 atm. L. K⁻¹. mol ⁻¹× 294 k
n = 112930 atm. L/24.14 atm. L. mol ⁻¹
n = 4678.13 mol
<span> this represents the relative overall energy of each orbital, and the energy of each orbital increases as the distance from the nucleus increases</span>
the force between the electron and the proton.
a) Use F = k * q1 * q2 / d²
where k = 8.99e9 N·m²/C²
and q1 = -1.602e-19 C (electron)
and q2 = 1.602e-19 C (proton)
and d = distance between point charges = 0.53e-10 m
The negative result indicates "attraction".
the radial acceleration of the electron.
b) Here, just use F = ma
where F was found above, and
m = mass of electron = 9.11e-31kg, if memory serves
a = radial acceleration
the speed of the electron.
c) Now use a = v² / r
where a was found above
and r was given
<span> the period of the circular motion.</span>
d) period T = 2π / ω = 2πr / v
where v was found above
and r was given
Answer:
Here is the ANSWER KEY, it may also has other of the questions you don't know :)
Explanation:
https://sciencewithhorne.weebly.com/uploads/5/7/3/5/57358947/unit_11_hw_key.pdf
<u>Answer:</u> The concentration of 
ions is 0.10 M and that of 
ions is 0.30 M
<u>Explanation:</u>
We are given:
Concentration of 
= 0.10 M
The chemical equation for the ionization of aluminium bromide follows:
1 mole of aluminium bromide produces 1 mole of aluminium ions and 3 moles of bromide ions
So, concentration of aluminium ions = 0.10 M
Concentration of bromide ions = ![[(3\times 0.1)]=0.3M](https://tex.z-dn.net/?f=%5B%283%5Ctimes%200.1%29%5D%3D0.3M)
Hence, the concentration of ions is 0.10 M and that of ions is 0.30 M
