Answer:
PART A: 412.98 nm
PART B: 524.92 nm
Explanation:
The equation below can be used for a diffraction grating of nth order image:
n*λ = d*sinθ
Therefore, for first order images, n = 1 and:
λ = d*sinθ
.
The angle θ can be calculated as follow:
tan θ = 9.95 cm/15.0 cm = 0.663 and
θ = 
(0.663) = 33.56°
Thus: d =λ/sin θ = 461/sin 33.56° = 833.97 nm
PART A:
For a position of 8.55 cm:
tan θ = 8.55 cm/15.0 cm = 0.57 and
θ = (0.57) = 29.68°
Therefore:
λ =d*sin θ = 833.97*sin 29.68° = 412.98 nm
PART B:
For a position of 12.15 cm:
tan θ = 12.15 cm/15.0 cm = 0.81 and
θ = (0.81) = 39.01°
Therefore:
λ =d*sin θ = 833.97*sin 39.01° = 524.92 nm
Answer:
oxygen and other non-metals
Answer:
9% solution by mass
Explanation:
If there is 500 gm of solution and 45 g of it is glucose then:
45/500 * 100% = 9 % solution
The equilibrium will shift to the left or the backward reaction since addition of <span>CH3COONa will add more CH3COO- ions to the solution. The formation of reactants are promoted.</span>
Answer:
Explanation:
Ag₂CO₃(s) ⇌2Ag⁺(aq) + CO₃²⁻(aq); Ksp = 8.10 × 10⁻¹²
2x 0.007 50 + x
![K_{sp} =\text{[Ag$^{+}$]$^{2}$[CO$_{3}^{2-}$]} = (2x)^{2}\times 0.00750 = 8.10 \times 10^{-12}\\0.0300x^{2} = 8.10 \times 10^{-12}\\x^{2} = 2.70 \times 10^{-10}\\x = \sqrt{2.70 \times 10^{-10}} = \mathbf{1.64\times 10^{5}} \textbf{ mol/L}\\\text{The maximum concentration of Ag$^{+}$ is $\large \boxed{\mathbf{1.64\times 10^{-5}}\textbf{ mol/L }}$}](https://tex.z-dn.net/?f=K_%7Bsp%7D%20%3D%5Ctext%7B%5BAg%24%5E%7B%2B%7D%24%5D%24%5E%7B2%7D%24%5BCO%24_%7B3%7D%5E%7B2-%7D%24%5D%7D%20%3D%20%282x%29%5E%7B2%7D%5Ctimes%200.00750%20%3D%208.10%20%5Ctimes%2010%5E%7B-12%7D%5C%5C0.0300x%5E%7B2%7D%20%3D%208.10%20%5Ctimes%2010%5E%7B-12%7D%5C%5Cx%5E%7B2%7D%20%3D%202.70%20%5Ctimes%2010%5E%7B-10%7D%5C%5Cx%20%3D%20%5Csqrt%7B2.70%20%5Ctimes%2010%5E%7B-10%7D%7D%20%3D%20%5Cmathbf%7B1.64%5Ctimes%2010%5E%7B5%7D%7D%20%5Ctextbf%7B%20mol%2FL%7D%5C%5C%5Ctext%7BThe%20maximum%20concentration%20of%20Ag%24%5E%7B%2B%7D%24%20is%20%24%5Clarge%20%5Cboxed%7B%5Cmathbf%7B1.64%5Ctimes%2010%5E%7B-5%7D%7D%5Ctextbf%7B%20mol%2FL%20%7D%7D%24%7D)
