Absorbed photon energy
Ea = hc/λ.. (Planck's equation)
Ea = hc / 92.05^-9m
<span>Energy emitted
Ee = hc/ 1736^-9m </span>
Energy retained ..
∆E = Ea - Ee = hc(1/92.05<span>^-9 - 1/1736^-9) </span>
<span>∆E = (6.625^-34)(3.0^8) (1.028^7)
∆E = 2.04^-18 J </span>
<span>Converting J to eV (1.60^-19 J/eV)
∆E = 2.04^-18 / 1.60^-19
∆E = 12.70 eV </span>
<span>Ground state (n=1) energy for Hydrogen = - 13.60eV </span>
<span>New energy state = (-13.60 + 12.70)eV = -0.85 eV </span>
<span>Energy states for Hydrogen
En = - (13.60 / n²) </span>
n² = -13.60 / -0.85 = 16
n = 4
Answer:
The concentration of hydroxide ions is 3.02*10⁻³ M
Explanation:
The pOH (or OH potential) is a measure of the basicity or alkalinity of a contamination and is defined as the negative logarithm of the activity of the hydroxide ions. That is, the concentration of OH- ions:
pOH= -log [OH-]
The pOH has a value between 0 and 14 in aqueous solution, the solutions with pOH being greater than 7 being acidic, and those with pOH less than 7 being basic.
If pOH= 2.52 then
2.52= -log [OH-]
[OH-]= 3.02*10⁻³ M
<u><em>The concentration of hydroxide ions is 3.02*10⁻³ M</em></u>
<u><em></em></u>
Answer: The volume of gas is 3020 ml
Explanation:
According to ideal gas equation:
P = pressure of gas = 821.4 torr = 1.08 atm (760 torr = 1atm)
V = Volume of gas in L = ?
n = number of moles = 
R = gas constant =
T =temperature =
Thus volume of gas is 3020 ml
Answers:
- a.) 10.0 mL of 0.0500 M HCl: 5.00 ml
- b.) 25.0 mL of 0.126 M HNO₃: 31.5 ml
- c.) 50.0 mL of 0.215 M H₂SO4: 215. ml
Explanation:
All the reactions are the neutralization of strong acids with the same strong base.
At the neutralization point you have:
- number of equivalents of the base = number of equivalent of the acid
And the number of equivalents (#EQ) may be calculated using the normality (N) concentration and the volume (V)
Then, at the neutralization point:
- # EQ acid = N acid × V acid
- # EQ base = N base × V base
- N acid × V acid = N base × V base
Also, you can use the formula that relates normality with molarity
- N = M × number of hydrogen or hydroxide ions
<u>a.) 10.0 mL of 0.0500 M HCl</u>
- The number of hydrogen ions for HCl is 1 and the number of hydroxide ions for NaOH is 1.
- 10.0 ml × 0.0500 M × 1 = V base × 0.100 M × 1
⇒ V base = 10.0 ml ×0.0500 M / 0.100 M = 5.00 ml
<u>b.) 25.0 mL of 0.126 M HNO₃</u>
- The number of hydrogen ions for HNO₃ is 1 and the number of hydroxide ions for NaOH is 1.
- 25.0 ml × 0.126 M × 1 = V base × 0.100 M × 1
⇒ V base = 25.0 ml ×0.126 M / 0.100 M = 31.5 ml
<u>c.) 50.0 mL of 0.215 M H₂SO4</u>
- The number of hydrogen ions for H₂SO4 is 2 and the number of hydroxide ions for NaOH is 1.
- 50.0 ml × 0.215 M × 2 = V base × 0.100 M × 1
⇒ V base = 50.0 ml ×0.215 M × 2 / 0.100 M = 215. ml
