Let suppose the Gas is acting Ideally, Then According to Ideal Gas Equation,
P V = n R T
Solving for P,
P = n R T / V ----- (1)
Data Given;
Moles = n = 1.20 mol
Volume = V = 4 L
Temperature = T = 30 + 273 = 303 K
Gas Constant = R = 0.08206 atm.L.mol⁻¹.K⁻¹
Putting Values in Eq.1,
P = (1.20 mol × 0.08206 atm.L.mol⁻¹.K⁻¹ × 303 K) ÷ 4 L
P = 7.45 atm
According to Bohr's model of the atom, the higher the orbital in which the electrons are found, the higher their energy or excitation state. Therefore, the electrons with the least amount of energy are those at the lowest orbitals, which are closer to the nucleus.
These orbitals are characterized by 4 quantum numbers, namely the principal quantum number (n), orbital angular momentum quantum number (l), the magnetic quantum number (ml), and the electron spin quantum number (ms). The principal quantum number reflects the distance of the electrons from the nucleus with n=1 as the orbital closest to the nucleus. Thus, according to Bohr's model, electrons in the orbital with n=1 have the lowest energy.
Answer:
25 gm would be left
Explanation:
45 hours is 3 half lives
200 gm * 1/2 * 1/2 * 1/2 = 25 gm
.100 mol CO2 x

using the values of the periodic you first add the masses of C (12.01g) and O (there are two so it'll be 32.00g). That value will give the mass of 1mole of CO2.
I hate to do this, but
https://youtu.be/Pft2CASl0M0 is a link to a mr andersen video. I dislike watching these cause this is what my teacher uses instead of actually having to teach herself.
Answer:
9.430 * 10¹⁷ protons per second whill shine on the book from a 62 W bulb
Explanation:
To answer this question, first let's calculate the energy of a single photon with a wavelength (λ) of 504 nm:
E = hc/λ
Where h is Planck's constant (6.626*10⁻³⁴ J·s) and c is the speed of light (3*10⁸ m/s).
E = 6.626*10⁻³⁴ J·s * 3*10⁸ m/s ÷ (504*10⁻⁹m) = 3.944 * 10⁻¹⁹ J.
So now we can make the equivalency for this problem, that
<u>1 proton = 3.944 * 10⁻¹⁹ J</u>
Now we convert watts from J/s to proton/s:
1
= 1 W
Solving the problem, a 62 W bulb converts 5% of its output into light, so:
3.1 watts are equal to [ 2.535*10¹⁸ proton/s * 3.1 ] = 7.858 * 10¹⁸ proton/s
Of those protons per second, 12% will shine on the chemistry textbook, thus:
7.858 * 10¹⁸ proton/s * 12/100 = 9.430 * 10¹⁷ protons/s