Answer:
In computer science, a literal is a notation for representing a fixed value in source code. An anonymous function is a literal for the function type. In contrast to literals, variables or constants are symbols that can take on one of a class of fixed values, the constant being constrained not to change.
Explanation:
Answer:
Linear molecule is a molecule in which atoms are deployed in a straight line (under 180° angle). Molecules with an linear electron pair geometries have sp hybridization at the central atom. An example of linear electron pair and molecular geometry are carbon dioxide (O=C=O) and beryllium hydride BeH2.
Explanation:
Answer:
λ = 6.5604 x 1016 nm
Explanation:
Given Data:
The energy of the red line in Hydrogen Spectra = 3.03 x 10-19
Formula to calculate Wave length
E= hv
Where E is Energy
h is Planks Constant = 6.626 x 10–34 J s
v is frequency
In turn
v= c/ λ
where c is speed of light = 3.00 x 108 m s–1
λ is wavelength = to find
Solution:
Formula to be Used:
E= hv………………………… (1)
Putting the value v in equation 1
E= h c/ λ…………………… (2)
Put the value in equation 2
3.03 x 10-19 J = (6.626 x 10–34 J s) x (3.00 x 108 m s–1) / λ ……………………….(3)
By rearranging equation 3
λ = (6.626 x 10–34 J s) x (3.00 x 108 m s–1) /3.03 x 10-19 J
λ = 6.5604 x 107 m
The answer is in “m”
So we have to convert it into nm
So for this to convert “m” to “nm” multiply the answer with 109
λ = 6.5604 x 107 x 109
λ = 6.5604 x 1016 nm
Answer:
[OH⁻] = 4.3 x 10⁻¹¹M in OH⁻ ions.
Explanation:
Assuming the source of the carbonate ion is from a Group IA carbonate salt (e.g.; Na₂CO₃), then 0.115M Na₂CO₃(aq) => 2(0.115)M Na⁺(aq) + 0.115M CO₃²⁻(aq). The 0.115M CO₃²⁻ then reacts with water to give 0.115M carbonic acid; H₂CO₃(aq) in equilibrium with H⁺(aq) and HCO₃⁻(aq) as the 1st ionization step.
Analysis:
H₂CO₃(aq) ⇄ H⁺(aq) + HCO₃⁻(aq); Ka(1) = 4.3 x 10⁻⁷
C(i) 0.115M 0 0
ΔC -x +x +x
C(eq) 0.115M - x x x
≅ 0.115M
Ka(1) = [H⁺(aq)][HCO₃⁻(aq)]/[H₂CO₃(aq)] = [(x)(x)/(0.115)]M = [x²/0.115]M
= 4.3 x 10⁻⁷ => x = [H⁺(aq)]₁ = SqrRt(4.3 x 10⁻⁷ · 0.115)M = 2.32 x 10⁻⁴M in H⁺ ions.
In general, it is assumed that all of the hydronium ion comes from the 1st ionization step as adding 10⁻¹¹ to 10⁻⁷ would be an insignificant change in H⁺ ion concentration. Therefore, using 2.32 x 10⁻⁴M in H⁺ ion concentration, the hydroxide ion concentration is then calculated from
[H⁺][OH⁻] = Kw => [OH⁻] = (1 x 10⁻¹⁴/2.32 x 10⁻⁴)M = 4.3 x 10⁻¹¹M in OH⁻ ions.
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NOTE: The 2.32 x 10⁻⁴M value for [H⁺] is reasonable for carbonic acid solution with pH ≅ 3.5 - 4.0.
