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GuDViN [60]
3 years ago
9

For a grating, how many lines per millimeter would be required for the first order diffraction line for λ = 400 nm to be observe

d at a reflection angle of 5o when the angle of incidence is 45o?
Chemistry
1 answer:
MA_775_DIABLO [31]3 years ago
6 0

Answer:

For a grating, how many lines per millimeter would be required for the first order diffraction line for λ = 400 nm to be observed at a reflection angle of 5^{o} when the angle of incidence is  45^{o}?

<em>The number of lines per mm is 1985 lines </em>

Explanation:

A diffraction grating is used to separate a light source with multiple wavelengths into single wavelengths of different colors using the principle of diffraction.

Calculating for the spacing between the reflected surfaces

The spacing between the reflecting surfaces can be obtained using the relationship with the wavelength as shown below;

nA = d (sin i + sin r) ....................................1

where n is the order of diffraction = 1;

A is the wavelength = 400 nm

i is the angle of incidence = 45^{o}

r is the angle of refraction =  5^{o}

making d the subject formula from equation 1 we have;

d = \frac{nA}{sin i + sin r}...........................2

Now we substitute the given parameters into equation 2;

d = \frac{1(400 nm)}{(sin 45^{o}  ) + (sin 5^{o} )}

d = \frac{400 nm}{0.707 +0.087}

d = 503.7 nm

The spacing between reflecting surfaces is 503.7 nm

Calculating for the number of lines per millimeter

To calculate the number of spacing per millimeter we convert nanometer to millimeters, 1 mm =10^{6} and 1 line = 503.7 nm;

1 mm *\frac{1 line}{503.7 nm} *\frac{10^{6} }{mm}

= 1985 lines

Therefore the number of lines per millimeter would be 1985 lines

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Answer:

                     Mass = 11.78 g of P₄

Explanation:

                     The balance chemical equation is as follow:

                                          6 Sr + P4 → 2 Sr₃P₂

Step 1: Calculate moles of Sr as;

Moles = Mass / M/Mass

Moles = 50.0 g / 87.62 g/mol

Moles = 0.570 moles

Step 2: Find moles of P₄ as;

According to equation,

6 moles of Sr reacted with  =  1 mole of P₄

So,

0.570 moles of Sr will react with  =  X moles of P₄

Solving for X,

X = 1 mol × 0.570 mol / 6 mol

X = 0.0952 mol of P₄

Step 3: Calculate mass of P₄ as,

Mass = Moles × M.Mass

Mass = 0.0952 mol × 123.89 g/mol

Mass = 11.78 g of P₄

8 0
3 years ago
Electron affinity is the measure of the attraction of an electron toward an isolated gaseous atom. When an electron is added to
zhuklara [117]

False

Explanation:

Electron affinity is negative when energy is absorbed and it is positive when energy is released.

Electron affinity is defined as the energy released in adding an electron to a neutral atom in the gas phase.

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Endothermic reactions in which energy is absorbed have negative electron affinity values.

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What is the relationship between elements ,atoms ,and compounds
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What volume (in L) will a 32 g sample of butane gas, C4H10(g), occupy at a temperature of 45.0 oC and a pressure of 728 mm Hg?
larisa86 [58]

Answer:

15.0 L

Explanation:

To find the volume, you need to use the Ideal Gas Law:

PV = nRT

In this equation,

-----> P = pressure (mmHg)

-----> V = volume (L)

-----> n = moles

-----> R = Ideal Gas constant (62.36 L*mmHg/mol*K)

-----> T = temperature (K)

To calculate the volume, you need to (1) convert grams C₄H₁₀ to moles (via the molar mass), then (2) convert the temperature from Celsius to Kelvin, and then (3) calculate the volume (via the Ideal Gas Law).

Molar Mass (C₄H₁₀): 4(12.011 g/mol) + 10(1.008 g/mol)

Molar Mass (C₄H₁₀): 58.124 g/mol

32 grams C₄H₁₀              1 moles
-------------------------  x  -----------------------  = 0.551 moles C₄H₁₀
                                    58.124 grams

P = 728 mmHg                      R = 62.36 L*mmHg/mol*K

V = ? L                                    T = 45.0 °C + 273.15 = 318.15 K

n = 0.551 moles

PV = nRT

(728 mmHg)V = (0.551 moles)(62.36 L*mmHg/mol*K)(318.15 K)

(728 mmHg)V = 10922.7632

V = 15.0 L

6 0
1 year ago
EXTRA POINTSSS 1. A solution at 25 degrees Celsius is 1.0 × 10–5 M H3O+. What is the concentration of OH– in this solution?
AlekseyPX

Answer:

Concentration of OH⁻:

1.0 × 10⁻⁹ M.

Explanation:

The following equilibrium goes on in aqueous solutions:

\text{H}_2\text{O}\;(l)\rightleftharpoons \text{H}^{+}\;(aq) + \text{OH}^{-}\;(aq).

The equilibrium constant for this reaction is called the self-ionization constant of water:

K_w = [\text{H}^{+}]\cdot[\text{OH}^{-}].

Note that water isn't part of this constant.

The value of K_w at 25 °C is 10^{-14}. How to memorize this value?

  • The pH of pure water at 25 °C is 7.
  • [\text{H}^{+}] = 10^{-\text{pH}} = 10^{-7}\;\text{mol}\cdot\text{dm}^{-3}
  • However, [\text{OH}^{-}] = [\text{H}^{+}]=10^{-7}\;\text{mol}\cdot\text{dm}^{-3} for pure water.
  • As a result, K_w = [\text{H}^{+}] \cdot[\text{OH}^{-}] = (10^{-7})^{2} = 10^{-14} at 25 °C.

Back to this question. [\text{H}^{+}] is given. 25 °C implies that K_w = 10^{-14}. As a result,

\displaystyle [\text{OH}^{-}] = \frac{K_w}{[\text{H}^{+}]} = \frac{10^{-14}}{1.0\times 10^{-5}} = 10^{-9} \;\text{mol}\cdot\text{dm}^{-3}.

8 0
3 years ago
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