First we find for the wavelength of the photon released due
to change in energy level. We use the Rydberg equation:
1/ʎ = R [1/n1^2 – 1/n2^2]
where,
ʎ is the wavelength
R is the rydbergs constant = 1.097×10^7 m^-1
n1 is the 1st energy level = 1
n2 is the higher energy level = infinity, so 1/n2 = 0
Calculating for ʎ:
1/ʎ = 1.097×10^7 m^-1 * [1/1^2 – 0]
ʎ = 9.1158 x 10^-8 m
Then calculate the energy using Plancks equation:
E = hc/ʎ
where,
h is plancks constant = 6.626×10^−34 J s
c is speed of light = 3x10^8 m/s
E = (6.626×10^−34 J s * 3x10^8 m/s) / 9.1158 x 10^-8 m
E = 2.18 x 10^-18 J = 2.18 x 10^-21 kJ
This is still per atom, so multiply by Avogadros number =
6.022 x 10^23 atoms / mol:
E = (2.18 x 10^-21 kJ / atom) * (6.022 x 10^23 atoms /
mol)
E = 1312 kJ/mol
A or B depends on what you mean by lit or glowing but when you place a wooden split in the sample the gas must reignite but there can be some confusion between hydrogen and oxygen mainly because a splint can cause a slight popping sound while it reignites but hydrogen pops are more violent and can most time extinguish the splint.
Answer:
Molar mass = 94972.745 g/mol
Explanation:
Given data:
Density = 2.25 g/ml
Pressure = 700 mmHg
Temperature = 200°C
Molar mass = ?
Solution:
Density = 2.25 g/ml (2.25×1000 = 2250 g/L)
Pressure = 700 mmHg (700/760 = 0.92 atm)
Temperature = 200°C (200+273 = 473K)
Formula:
d = PM/RT
M = dRT/P
M = 2250 g/L × 0.0821 atm.L /mol.K × 473K / 0.92 atm
M = 87374.93 g/mol / 0.92
M = 94972.745 g/mol
Answer:
Caesium and Barium
Explanation:
Once again, if you look at the periodic table, you can see that Caesium and Barium are in the same period (Period 6).
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pH of the buffer solution is 1.76.
Chemical dissociation of formic acid in the water:
HCOOH(aq) ⇄ HCOO⁻(aq) + H⁺(aq)
The solution of formic acid and formate ions is a buffer.
[HCOO⁻] = 0.015 M; equilibrium concentration of formate ions
[HCOOH] + [HCOO⁻] = 1.45 M; sum of concentration of formic acid and formate
[HCOOH] = 1.45 M - 0.015 M
[HCOOH] = 1.435 M; equilibrium concentration of formic acid
pKa = -logKa
pKa = -log 1.8×10⁻⁴ M
pKa = 3.74
Henderson–Hasselbalch equation: pH = pKa + log(cs/ck)
pH = 3.74 + log (0.015 M/1.435 M)
pH = 3.74 - 1.98
pH = 1.76
More about buffer: brainly.com/question/4177791
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