Answer:
λ = 0.38 ×10⁻⁹ m
Explanation:
Given data:
Wavelength of xray = ?
Frequency of xray = 7.8 ×10¹⁷ Hz
Solution:
Formula:
Speed of light = wavelength × frequency
speed of light = 3×10⁸ m/s
Now we will put the values in formula.
3×10⁸ m/s = λ × 7.8 ×10¹⁷ Hz
λ = 3×10⁸ m/s / 7.8 ×10¹⁷ Hz
Hz = s⁻¹
λ = 3×10⁸ m/s / 7.8 ×10¹⁷s⁻¹
λ = 0.38 ×10⁻⁹ m
Answer:
See explanation
Explanation:
In this case, we have to remember that if we want to remove water from the reaction vessel we have to heat the vessel. So, we can convert the liquid water into <u>gas water</u> and we can remove it from the vessel. In this case, the products of dehydration for both molecules are <u>(E)-4-methylpent-2-ene</u> and <u>cyclohexene</u> with boiling points of <u>59.2 ºC</u> and <u>89 ºC</u> respectively. The boiling point of water is <u>100 ºC</u>, therefore if we heat the vessel the products and water would leave the system, and the products would be lost.
See figure 1
I hope it helps!
Answer:
Sodium - malleable, soft, and shiny
Silicon - has properties of both metals and nonmetals
Bromine - highly reactive gas
Argon - non-reactive gas
Explanation:
Sodium is an alkaline metal. Just like other alkaline metals, it's malleable, soft, and shiny.
Silicon is a metalloid. Metalloids are elements that have properties of both metals and nonmetals.
Bromine a highly reactive chemical element. It is a fuming red-brown liquid at room temperature that evaporates to form a similarly coloured gas.
Argon is a noble gas. Just like other noble gases, it's non-reactive.
<span>the balanced chemical equation for the reaction is as follows;
C</span>₃H₈ + 5O₂ ---> 3CO₂ + 4H₂<span>O
stoichiometry of </span> C₃H₈ to O₂ is 1:5
number of moles of C₃H₈ reacted - 0.025 g / 44.1 g/mol = 0.000567 mol according to molar ratio of 1:5
number of O₂ moles required are 5 times the amount of C₃H₈ moles reacted therefore number of O₂ moles required - 0.000567 x 5 = 0.00284 mol .
mass of O₂ required - 0.00284 mol x 32.00 g/mol = 0.091 mol .
answer is 0.091 mol
Answer:
1. 
was the 
value calculated by the student.
2. 
was the 
of ethylamine value calculated by the student.
Explanation:
1.
The 
value of Aspirin solution = 2.62
![pH=-\log[H^+]](https://tex.z-dn.net/?f=pH%3D-%5Clog%5BH%5E%2B%5D)
![[H^+]=10^{-2.62}=0.00240 M](https://tex.z-dn.net/?f=%5BH%5E%2B%5D%3D10%5E%7B-2.62%7D%3D0.00240%20M)
Moles of s asprin = 
Volume of the solution = 0.600 L
The initial concentration of Aspirin = c = 
initially
c 0 0
At equilibrium
(c-x) x x
The expression of dissociation constant :
![K_a=\frac{[As^-][H^+]}{[HAs]}](https://tex.z-dn.net/?f=K_a%3D%5Cfrac%7B%5BAs%5E-%5D%5BH%5E%2B%5D%7D%7B%5BHAs%5D%7D)
:
was the value calculated by the student.
2.
The value of ethylamine = 11.87
![pOH=-\log[OH^-]](https://tex.z-dn.net/?f=pOH%3D-%5Clog%5BOH%5E-%5D)
![[OH^-]=10^{-2.13}=0.00741 M](https://tex.z-dn.net/?f=%5BOH%5E-%5D%3D10%5E%7B-2.13%7D%3D0.00741%20M)
The initial concentration of ethylamine = c = 0.100 M
initially
c 0 0
At equilibrium
(c-x) x x
The expression of dissociation constant :
![K_b=\frac{[C_2H_5NH_3^{+}][OH^-]}{[C_2H_5NH_2]}](https://tex.z-dn.net/?f=K_b%3D%5Cfrac%7B%5BC_2H_5NH_3%5E%7B%2B%7D%5D%5BOH%5E-%5D%7D%7B%5BC_2H_5NH_2%5D%7D)
:
was the of ethylamine value calculated by the student.
