Auroras occur when ions from the Sun strike air molecules, causing them to give off bright colors of light.-google
Let us assume that the ring is a size 7 ring, which has a circumference of 54.3 millimeters. Converting this to centimeters, the circumference of the ring is:
54.3 mm = 5.43 cm
Now, we determine the number of gold atoms that will be present in this:
5.43 / 1 x 10⁻⁹
There will be 5.43 x 10⁹ atoms
We now determine the number of moles this is by:
one mole = 6.02 x 10²³ atoms
Moles = 5.43 x 10⁹ / 6.02 x 10²³
Moles = 9.01 x 10⁻¹⁵ moles
The molar mass of gold is 197 g/mol
The mass is 9.01 x 10⁻¹⁵ * 197
The mass of the strand is 1.76 x 10⁻¹² grams
I think this is what you're after:
Cs(g) → Cs^+ + e⁻ ΔHIP = 375.7 kJ mol^-1 [1]
Convert to J and divide by the Avogadro Const to give E in J per photon
E = 375700/6.022×10^23 = 6.239×10^-19 J
Plank relationship E = h×ν E in J ν = frequency (Hz s-1)
Planck constant h = 6.626×10^-34 J s
6.239×10^-19 = (6.626×10^-34)ν
ν = 9.42×10^14 s^-1 (Hz)
IP are usually given in ev Cs 3.894 eV
<span>E = 3.894×1.60×10^-19 = 6.230×10^-19 J per photon </span>
Answer:
Explanation:
Let assume that gas behaves ideally and experiments an isobaric and isothermal processes. The following relationship is applied to determined the final volume:
Answer:
0.42%
Explanation:
<em>∵ pH = - log[H⁺].</em>
2.72 = - log[H⁺]
∴ [H⁺] = 1.905 x 10⁻³.
<em>∵ [H⁺] = √Ka.C</em>
∴ [H⁺]² = Ka.C
∴ ka = [H⁺]²/C = (1.905 x 10⁻³)²/(0.45) = 8.068 x 10⁻⁶.
<em>∵ Ka = α²C.</em>
Where, α is the degree of dissociation.
<em>∴ α = √(Ka/C) </em>= √(8.065 x 10⁻⁶/0.45) = <em>4.234 x 10⁻³.</em>
<em>∴ percentage ionization of the acid = α x 100</em> = (4.233 x 10⁻³)(100) = <em>0.4233% ≅ 0.42%.</em>
