Answer:
All three are present
Explanation:
Addition of 6 M HCl would form precipitates of all the three cations, since the chlorides of these cations are insoluble:
.
- Firstly, the solid produced is partially soluble in hot water. Remember that out of all the three solids, lead(II) choride is the most soluble. It would easily completely dissolve in hot water. This is how we separate it from the remaining precipitate. Therefore, we know that we have lead(II) cations present, as the two remaining chlorides are insoluble even at high temperatures.
- Secondly, addition of liquid ammonia would form a precipitate with silver:
; Silver hydroxide at higher temperatures decomposes into black silver oxide:
. - Thirdly, we also know we have
in the mixture, since addition of potassium chromate produces a yellow precipitate:
. The latter precipitate is yellow.
Answer:
Energy lost is 7.63×10⁻²⁰J
Explanation:
Hello,
I think what the question is requesting is to calculate the energy difference when an excited electron drops from N = 15 to N = 5
E = hc/λ(1/n₂² - 1/n₁²)
n₁ = 15
n₂ = 5
hc/λ = 2.18×10⁻¹⁸J (according to the data)
E = 2.18×10⁻¹⁸ (1/n₂² - 1/n₁²)
E = 2.18×10⁻¹⁸ (1/15² - 1/5²)
E = 2.18×10⁻¹⁸ ×(-0.035)
E = -7.63×10⁻²⁰J
The energy lost is 7.63×10⁻²⁰J
Note : energy is lost / given off when the excited electron jumps from a higher energy level to a lower energy level
Answer:
94.61 %
Explanation:
percent yield = (actual yield / theoretical yield) X 100%
The balanced equation for the reaction is:
H2 (g ) + Cl2(g) => 2 HCl (l)
So, the theoretical yield =
7.25g of chlorine X (2mol of Cl / 35.453 g of Cl) X (2mol Cl / 2mol of HCl) X 37.469g of Hcl / 2mol of Hcl = 0.409 x 18.735 = 7.663g of Hcl
Using this theoretical yield and the provided value for actual yield, the percent yield = (actual yield / theoretical yield) X 100%
= (7.25 g / 7.663g) X 100
= 94.61 %
Answer:
343.98 nm is the longest wavelength of radiation with enough energy to break carbon–carbon bonds.
Explanation:
A typical carbon–carbon bond requires 348 kJ/mol=348000 J/mol
Energy required to breakl sigle C-C bond:E


where,
E = energy of photon
h = Planck's constant = 
c = speed of light = 
= wavelength of the radiation
Now put all the given values in the above formula, we get the energy of the photons.



343.98 nm is the longest wavelength of radiation with enough energy to break carbon–carbon bonds.