6.07 moles
Explanation:
Given parameters:
Number of moles of reacting hydrogen gas = 9.1 moles
Unknown:
Number of moles of ammonia produced = ?
Solution:
Balanced chemical equation:
N₂ + 3H₂ → 2NH₃
We should work from the known to the unknown specie. We known the number of moles of hydrogen gas reacting. We simply relate this to that of the ammonia.
From the reaction:
3 mole of hydrogen gas produced 2 mole of ammonia
9.1 moles of the hydrogen gas will produce: 
= 6.07 moles
Learn more:
Number of moles brainly.com/question/1841136
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Answer:
1.696 nm
Explanation:
For a diffraction grating, dsinθ = mλ where d = number of lines per metre of grating = 5510 lines per cm = 551000 lines per metre and λ = wavelength of light = 467 nm = 467 × 10⁻⁹ m. For a principal maximum, m = 1. So,
dsinθ = mλ = (1)λ = λ
dsinθ = λ
sinθ = λ/d.
Also tanθ = w/D where w = distance of center of screen to principal maximum and D = distance of grating to screen = 1.03 m
From trig ratios 1 + cot²θ = cosec²θ
1 + (1/tan²θ) = 1/(sin²θ)
substituting the values of sinθ and tanθ we have
1 + (D/w)² = (d/λ)²
(D/w)² = (d/λ)² - 1
(w/D)² = 1/[(d/λ)² - 1]
(w/D) = 1/√[(d/λ)² - 1]
w = D/√[(d/λ)² - 1] = 1.03 m/√[(551000/467 × 10⁻⁹ )² - 1] = 1.03 m/√[(1179.87 × 10⁹ )² - 1] = 1.03 m/1179.87 × 10⁹ = 0.000848 × 10⁻⁹ = 0.848 × 10⁻¹² m = 0.848 nm.
w is also the distance from the center to the other principal maximum on the other side.
So for both principal maxima to be on the screen, its minimum width must be 2w = 2 × 0.848 nm = 1.696 nm
So, the minimum width of the screen must be 1.696 nm
Factors that affect heat transfer are:
1) Difference in temperature,
2) Mass of the object
3) Specific heat of the object
Hope this helps!
First we will find the speed of the ball just before it will hit the floor
so in order to find the speed of the cart we will first use energy conservation
So by solving above equation we will have
now in order to find the momentum we can use
Answer:
c. 60 feet is the correct answer
Explanation:
what is the contour interval of this map? a.20 b.-20 c. 60 feet 11
