Answer:
1.09 × 10⁻⁷ m
UV region
Explanation:
Step 1: Given and required data
Energy of the photon of light (E): 1.83 × 10⁻¹⁸ J
Planck's constant (h): 6.63 × 10⁻³⁴ J.s
Speed of light (c): 3.00 × 10⁸ m/s
Step 2: Calculate the wavelength (λ) of this photon of light
We will use the Planck-Einstein's relation.
E = h × c/λ
λ = h × c/E
λ = 6.63 × 10⁻³⁴ J.s × (3.00 × 10⁸ m/s)/1.83 × 10⁻¹⁸ J = 1.09 × 10⁻⁷ m
This wavelenght falls in the UV region of the electromagnetic spectrum.
Answer:
B. CH3COOH pH > 4.7 (4.8)
Explanation:
- CH3COOH + NaOH ↔ CH3COONa + H2O
- CH3COONa + NaOH ↔ CH3COONa
∴ mol NaOH = (5 E-3 L)*(0.10 mol/L) = 5 E-4 mol
⇒ mol CH3COOH = (0.05 L)*(0.20 mol/L) = 0.01 mol
⇒ <em>C</em> CH3COOH = (0.01 mol - 5 E-4 mol) / (0.105 L)
⇒ <em>C</em> CH3COOH = 0.0905 M
∴ mol CH3COONa = (0.05 L )*(0.20 mol/L) = 0.01 mol
⇒ <em>C</em> CH3COONa = (0.01 mol + 5 E-4 mol) / (0.105 L )
⇒ <em>C</em> CH3COONa = 0.1 M
∴ Ka = ([H3O+]*(0.1 + [H3O+])) / (0.0905 - [H3O+]) = 1.75 E-5
⇒ 0.1[H3O+] + [H3O+]² = (1.75 E-5)*(0.0905 - [H3O+])
⇒ [H3O+]² 0.1[H3O+] = 1.584 E-6 - 1.75 E-5[H3O+]
⇒ [H3O+]² + 0.1000175[H3O+] - 1.584 E-6 = 0
⇒ [H3O+] = 1.5835 E-5 M
∴ pH = - Log [H3O+]
⇒ pH = - Log (1.5835 E-5)
⇒ pH = 4.8004 > 4.7
I'm not sure about this question
The Kp in this problem is expressed as Kp= [CO]^2/[CO2] and is equal to 2.25. Substituting the initial concentrations we get Kp=
![\frac{ [0.32+x]^{2} }{ 0.56-x_{2} }](https://tex.z-dn.net/?f=%20%5Cfrac%7B%20%5B0.32%2Bx%5D%5E%7B2%7D%20%7D%7B%200.56-x_%7B2%7D%20%7D%20)
where x is the amount that is produced. x s calculated is 0.255 atm. Thus, the final partial pressure of CO is 0.83 atm. Answer is A.
