Answer : (C) "Higher frequencies have larger spaces between lines".
Explanation:
In Young's experiment, the condition for constructive interference is given by :
.........(1)
n is order or number of lines observed
d is distance between slits
is the angle between the path and the line from screen to the slits.
We also know that, 
or

where,
c is the speed of light
is frequency
is wavelength
So, equation (1) turns into


So,

or
Higher frequencies have larger spaces between line.
So, correct option is (C).
Answer:
C3H6 + Br2 → C3H6Br2
Explanation:
The reaction in which C3H6Br2 (1,2-Dibromopropane) is created is:
We can see that the only difference between the product (C3H6Br2) and the known reactant (C3H6) of the reaction is two bromine atoms (Br2). Br2 is diatomic bromine - a molecule we get after combining two bromine atoms. This compound is a red-brown liquid at room temperature, which means that that is the liquid described in your question.
It is 0.5474 but you can put 0.5 hope this helps
The amount of heat required to convert H₂O to steam is : 382.62 kJ
<u>Given data :</u>
Mass of liquid water ( m ) = 150 g
Temperature of liquid water = 43.5°C
Temperature of steam = 130°C
<h3 /><h3>Determine the amount of heat required </h3>
The amount of heat required = ∑ q1 + q2 + q3 ----- ( 1 )
where ;
q1 = heat required to change Temperature of water from 43.5°C to 100°C . q2 = heat required to change liquid water at 100°C to steam at 100°C
q3 = heat required to change temperature of steam at 100°C to 130°C
M* S
*ΔT
= 150 * 4.18 * ( 100 - 43.5 )
= 35425.5 J
moles * ΔHvap
= (150 / 18 )* 40.67 * 1000
= 338916.67 J
M * S
* ΔT
= 150 * 1.84 * ( 130 -100 )
= 8280 J
Back to equation ( 1 )
Amount of heat required = 35425.5 + 338916.67 + 8280 = 382622.17 J
≈ 382.62 kJ
Hence we can conclude that The amount of heat required to convert H₂O to steam is : 382.62 kJ.
Learn more about Specific heat of water : brainly.com/question/16559442
The delta H of -484 kJ is the heat given off when 2 moles of H2 react with 1 mole of O2 to make 2 moles of H2O. You don't have anywhere near that much reactants, only 1/4 as much
<span>actual delta H = 0.34 moles H2 x (-484 kJ / 2 moles H2) = 823 kJ </span>
<span>delta E = delta H - PdeltaV = 823 kJ - 0.41 kJ = 822 kJ</span>