<span>The s sublevel has just one orbital, so can contain 2 electrons max. The p sublevel has 3 orbitals, so can contain 6 electrons max. The d sublevel has 5 orbitals, so can contain 10 electrons max. And the 4 sublevel has 7 orbitals, so can contain 14 electrons max.
So, having this in mind, 10 electrons in total can be contained in the 4d sublevel.
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Answer:
Gold: 1.1 x 10²² atoms/cm³
Silver: 4.8 x 10²² atoms/cm³
Explanation:
100 g of the alloy will have 29 g of Au and 71 g of Ag.
19.32 g Au ____ 1 cm³
29 g Au ______ x
x = 1.5 cm³
10.49 g Ag ____ 1 cm³
71 g Ag _______ y
y = 6.8 cm³
The total volume of 100g of the alloy is x+y = 8.3 cm³.
Gold:
196.97 g Au____ 6.022 x 10²³ atoms Au
29 g Au _______ w
w = 8.9 x 10²² atoms Au
8.9 x 10²² atoms Au ____ 8.3 cm³
A ____ 1 cm³
A = 1.1 x 10²² atoms Au
Silver:
107.87 g Ag____ 6.022 x 10²³ atoms Ag
71 g Ag _______ w
w = 4.0 x 10²³ atoms Ag
4.0 x 10²³ atoms Ag ____ 8.3 cm³
B ____ 1 cm³
B = 4.8 x 10²² atoms Ag
The absorbance reported by the defective instrument was 0.3933.
Absorbance A = - log₁₀ T
Tm = transmittance measured by spectrophotometer
Tm = 0.44
Absorbance reported in this equipment = -log₁₀ (0.44) = 0.35654
True absorbance can be calculated by true transmittance, Tm = T+S(α-T)
S = fraction of stray light = 6%= 6/100 = 0.06
α= 1, ideal case
T = true transmittance of the sample
Tm = T+S(α-T)
now, T= Tm-S/ 1-S = 0.44-0.06/ 1-0.06 = 0.404233
therefore, actual reading measured is A = -log₁₀ T = -log₁₀ (0.404233)
i.e; 0.3933
To know more about transmittance click here:
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The answer is it lost the hydrogen ;)
Answer:
The answer to your question is a) N₂ b) 3.04 g of NH₃
Explanation:
Data
mass of H₂ = 2.5 g
mass of N₂ = 2.5 g
molar mass H₂ = 2.02 g
molar mass of N₂ = 28.02 g
molar mass of NH₃ = 17.04 g
Balanced chemical reaction
3H₂ + 1 N₂ ⇒ 2NH₃
A)
Calculate the theoretical yield 3H₂ / N₂ = 3(2.02) / 28.02 = 0.22
Calculate the experimental yield H₂/N₂ = 2.5/2.5 = 1
Conclusion
The limiting reactant is N₂ (nitrogen) because the experimental proportion was higher than the theoretical proportion.
B)
28.02 g of N₂ -------------------- (2 x 17.04) g of NH₃
2.5 g of N₂ -------------------- x
x = (2.5 x 2 x 17.04) / 28.02
x = 85.2 / 28.02
x = 3.04 g of NH₃
