Three complete orders on each side of the m=0 order can be produced in addition to the m = 0 order.
The ruling separation is
d=1 / (470mm −1) = 2.1×10⁻³ mm
Diffraction lines occur at angles θ such that dsinθ=mλ, where λ is the wavelength and m is an integer.
Notice that for a given order, the line associated with a long wavelength is produced at a greater angle than the line associated with a shorter wavelength.
We take λ to be the longest wavelength in the visible spectrum (538nm) and find the greatest integer value of m such that θ is less than 90°.
That is, find the greatest integer value of m for which mλ<d.
since d / λ = 538×10⁻⁹m / 2.1×10 −6 m ≈ 3
that value is m=3.
There are three complete orders on each side of the m=0 order.
The second and third orders overlap.
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Answer: If I have 17 moles of gas at a temperature of 67°C, and a pressure of 5.34 atmospheres, what is the volume of the gas? write down all the givens of the problem
Answer : The enthalpy change for the solution is 166.34 kJ/mol
Explanation :
First we have to calculate the enthalpy change of the reaction.
Formula used :
where,
= change in enthalpy = ?
C = heat capacity of water = 
m = total mass of sample = 2.174 + 127.4 = 129.6 g
= initial temperature = 
= final temperature = 
Now put all the given values in the above expression, we get:
Now we have to calculate the moles of AX added to water.
Now we have to calculate the enthalpy change for the solution.
As, 0.04592 moles releases heat = 7638.36 J
So, 1 moles releases heat = 
Therefore, the enthalpy change for the solution is 166.34 kJ/mol
Answer is: mass of water is 432 grams.
Chemical reaction: 2H₂ + O₂ → 2H₂O.
m(O₂) = 384 g.
M(O₂) = 2 · 16 g/mol = 32 g/mol, molar mass.
n(O₂) = m(O₂) ÷ M(O₂).
n(O₂) = 384 g ÷ 32 g/mol.
n(O₂) = 12 mol, amount of substance.
From chemical reaction: n(O₂) : n(H₂O) = 1 : 2.
n(H₂O) = 12 mol · 2 = 24 mol.
m(H₂O) = n(H₂O) · M(H₂O).
m(H₂O) = 24 mol · 18 g/mol.
m(H₂O) = 432 g.
Answer:
Explanation:
Let final temperature be T .
vapor is at 102⁰C
loss of heat by vapor in turning into water at 100⁰C
= 2.6 x 2 x 1.996 + 2.6 x 2260 = 5886.37 J
loss of heat to lower temperature to T
2.6 x 4.186 x ( 100 - T )
1088.36 - 10.88 T
Total heat loss = 5886.37 + 1088.36 - 10.88 T
= 6974.73 - 10.88 T
heat gain by silver to gain temperature from 1⁰C to T⁰C
= 1250 x ( T - 1 ) x .235 = 293.75 T - 293.75
heat gain = heat loss
293.75 T - 293.75 = 6974.73 - 10.88 T
304.63 T = 7268.48
T = 23.86°C .
