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amid [387]
3 years ago
6

What is the wavelength of the line corresponding to n=4 in the balmer series? express your answer in nanometers to three signifi

cant figures?
Chemistry
1 answer:
liq [111]3 years ago
8 0
Expression for the Balmer series to find the wavelength of the spectral line is as follows:
1 / λ = R (\frac{1}{2^{2} } -  \frac{1}{ n^{2}} )
Where, λ is wavelength, R is Rydberg constant, and n is integral value (4 here →  Fourth level) 
Substitute 1.097 x 10⁷ m⁻¹ for R and 4 for n in the above equation 
1 / λ = (1.097 x 10⁷ m⁻¹) (\frac{1}{2^{2} } -  \frac{1}{4^{2} }  ) 
   = 0.20568 x 10⁷ m⁻¹
λ = 4.86 x 10⁻⁷ m 
since 1 m = 10⁹ nm
λ = 486 nm 
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Making methanol the element hydrogen is not abundant in nature, but it is a useful reagent in, for example, the potential synthe
stealth61 [152]
ΔG⁰ = ΔH⁰ - TΔS
ΔH⁰ = Hf,(CH₃OH) - Hf,(CO) = -238.7 + 110.5 = -128.2 kJ/mol
ΔS = S(CH₃OH) - S(CO) - 2S(H₂) = 126.8 - 197.7 - 2 x 130.6 = -332.1 J/mol.K
So 
ΔG⁰ = - 128200 + 332.1 T
For the reaction to be spontaneous:
ΔG⁰ < 0
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332.1 T < 128200
T < 386.028 K
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3 years ago
At what point in the cooking process is food typically seared?
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Answer: when you look at the meat, and it looks beautiful like a crispy donut made by Gordan Ramsay

Explanation:

Hells Kitchen is kewl

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balu736 [363]

Answer:

Explanation:

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2 years ago
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Which is the most abundant isotope of carbon??
Gwar [14]
Carbon-12, my friend
3 0
3 years ago
The standard heats of formation for CO2(g), C2H6(g), and H2O(l) are -394.0 kJ/mol, -84.00 kJ/mol, and -286.0 kJ, respectively. W
Ivanshal [37]

Answer:

ΔH°r = -1562 kJ

Explanation:

Let's consider the following combustion.

C₂H₆(g) + 7/2 O₂(g) ⇒ 2 CO₂(g) + 3 H₂O(l)

We can calculate the standard heat of reaction (ΔH°r) using the following expression:

ΔH°r = ∑np × ΔH°f(p) - ∑nr × ΔH°f(r)

where,

ni are the moles of reactants and products

ΔH°f(i) are the standard heats of formation of reactants and products

The standard heat of formation of simple substances in their most stable state is zero. That means that ΔH°f(O₂(g)) = 0

ΔH°r = ∑np × ΔH°f(p) - ∑nr × ΔH°f(r)

ΔH°r = [2 mol × ΔH°f(CO₂) + 3 mol × ΔH°f(H₂O)] - [1 mol × ΔH°f(C₂H₆) + 7/2 mol × ΔH°f(O₂)]

ΔH°r = [2 mol × (-394.0 kJ/mol) + 3 mol × (-286.0 kJ/mol)] - [1 mol × (-84.00 kJ/mol) + 7/2 mol × 0]

ΔH°r = -1562 kJ

6 0
3 years ago
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