The answer is most likely C
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:
2.01 moles of P → 1.21×10²⁴ atoms
2.01 moles of N → 1.21×10²⁴ atoms
4.02 moles of Br → 2.42×10²⁴ atoms
Explanation:
We begin from this relation:
1 mol of PNBr₂ has 1 mol of P, 1 mol of N and 2 moles of Br
Then 2.01 moles of PNBr₂ will have:
2.01 moles of P
2.01 moles of N
4.02 moles of Br
To determine the number of atoms, we use the relation:
1 mol has NA (6.02×10²³) atoms
Then: 2.01 moles of P will have (2.01 . NA) = 1.21×10²⁴ atoms
2.01 moles of N (2.01 . NA) = 1.21×10²⁴ atoms
4.02 moles of Br (4.02 . NA) = 2.42×10²⁴ atoms
According to the balanced equation of the reaction:
2C2H2 + 5O2 → 4CO2 + 2H2O
So we can mention all as liters,
A) as we see that 2 liters of C2H2 react with 5 liters of oxygen to produce 4 liters of CO4 and 2 liters of H2O
So, when we have 75L of CO2
and when we have 2 L of C2H2 reacts and gives 4 L of CO2
2C2H2 → 4CO2
∴ The volume of C2H2 required is:
= 75L / 2
= 37.5 L
B) and, when we have 75 L of CO2
and 4CO2 → 2H2O
∴ the volume of H2O required is:
= 75 L /2
= 37.5 L
C) and from the balanced equation and by the same way:
when 5 liters O2 reacts to give 4 liters of CO2
and we have 75 L of CO2:
5 O2 → 4 CO2
?? ← 75 L
∴ the volume of O2 required is:
= 75 *(5/4)
= 93.75 L
D) about the using of the number of moles the answer is:
no, there is no need to find the number of moles as we called everything in the balanced equation by liters and use it as a liter unit to get the volume, without the need to get the number of moles.
The hockey puck will continue in the same speed in a continues straight line
