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julsineya [31]
2 years ago
10

Carbon dioxide readily absorbs radiation with an energy of 4.67 x 10-20 J. What is the wavelength and frequency of this radiatio

n? Does this radiation fall in the ultraviolet, visible, or infared range?
Chemistry
1 answer:
Tanya [424]2 years ago
7 0

Answer:

ν = 7.04 × 10¹³ s⁻¹

λ = 426 nm

It falls in the visible range

Explanation:

The relation between the energy of the radiation and its frequency is given by Planck-Einstein equation:

E = h × ν

where,

E is the energy

h is the Planck constant (6.63 × 10⁻³⁴ J.s)

ν is the frequency

Then, we can find frequency,

\nu = \frac{E}{h}=  \frac{4.67 \times 10^{-20}J  }{6.63 \times 10^{-34}J.s} = 7.04 \times 10^{13} s^{-1}

Frequency and wavelength are related through the following equation:

c = λ × ν

where,

c is the speed of light (3.00 × 10⁸ m/s)

λ is the wavelength

\lambda = \frac{c}{\nu } =\frac{3.00 \times 10^{8} m/s }{7.04 \times 10^{13} s^{-1} } =4.26 \times 10^{-6}m.\frac{10^{9}nm }{1m} = 426 nm

A 426 nm wavelength falls in the visible range (≈380-740 nm)

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Two species, A and B, are separated on a 2.00 m long column which has 5.000 x 103 plates when the flow rate is 15.0 mL/min. A sp
sergiy2304 [10]

Explanation:

The given data is as follows.

         t_{m} = 30.0 sec,     t_{r1} = 5 min = 5 \times 60 sec = 300 sec

         t_{r2} = 12.0 min = 12 \times 60 sec = 720 sec

Formula for adjusted retention time is as follows.

      t'_{r} = t_{r} - t_{m}

                 = 300 sec - 30.0 sec

                 = 270 sec

   t'_{r2} = 720 sec - 30 sec

                            = 690 sec

Formula for relative retention (\alpha) is as follows.

          \alpha = \frac{t'_{r2}}{t'_{r1}}

                     = \frac{690 sec}{270 sec}

                     = 2.56

Thus, we can conclude that the relative retention is 2.56.

4 0
2 years ago
A(n) _______________ can be formed by linking together several monosaccharides via glycosidic bonds.
mario62 [17]

Answer:

A polysaccharide (n) can be formed by linking several monosaccharides through glycosidic linkages.

Explanation:

Polysaccharides are carbohydrates or complex carbohydrates, where monosaccharides join with glucosidic bonds to form a more complex structure that would be the polysaccharide.

An example of a polysaccharide is starch, or glycogen.

Starch is found in many foods such as potatoes or rice, and glycogen is a form of energy reserve of our organism housed in muscles and liver to fulfill locomotion, physical activity, and other activities that consist of glycolysis.

Polysaccharides are degraded in our body by different stages, and several enzymes unlike monosoccharides or disaccharides, since they have more unions and a more complex structure to disarm in our body and thus assimilate it.

Polysaccharides are also part of animal structures, such as insect shells or nutritional sources, among others.

7 0
2 years ago
Assuming equal concentrations of conjugate base and acid, which one of the following mixtures is suitable for making a buffer so
BartSMP [9]

Answer:

NH₃/NH₄Cl

Explanation:

We can calculate the pH of a buffer using the Henderson-Hasselbalch's equation.

pH=pKa+log\frac{[base]}{[acid]}

If the concentration of the acid is equal to that of the base, the pH will be equal to the pKa of the buffer. The optimum range of work of pH is pKa ± 1.

Let's consider the following buffers and their pKa.

  • CH₃COONa/CH3COOH (pKa = 4.74)
  • NH₃/NH₄Cl (pKa = 9.25)
  • NaOCl/HOCl (pKa = 7.49)
  • NaNO₂/HNO₂ (pKa = 3.35)
  • NaCl/HCl Not a buffer

The optimum buffer is NH₃/NH₄Cl.

4 0
1 year ago
Please help i need to do good in this class
ioda

Answer:

Explanation:

di) number of protons is 12 for all, number of neutrons is 13 for mg- 25 and 14 for mg-26

8 0
1 year ago
BH+ClO4- is a salt formed from the base B (Kb = 1.00e-4) and perchloric acid. It dissociates into BH+, a weak acid, and ClO4-, w
Len [333]

Answer:

The pH of 0.1 M BH⁺ClO₄⁻ solution is <u>5.44</u>

Explanation:

Given: The base dissociation constant: K_{b} = 1 × 10⁻⁴, Concentration of salt: BH⁺ClO₄⁻ = 0.1 M

Also, water dissociation constant: K_{w} = 1 × 10⁻¹⁴

<em><u>The acid dissociation constant </u></em>(K_{a})<em><u> for the weak acid (BH⁺) can be calculated by the equation:</u></em>

K_{a}. K_{b} = K_{w}    

\Rightarrow K_{a} = \frac{K_{w}}{K_{b}}

\Rightarrow K_{a} = \frac{1\times 10^{-14}}{1\times 10^{-4}} = 1\times 10^{-10}

<em><u>Now, the acid dissociation reaction for the weak acid (BH⁺) and the initial concentration and concentration at equilibrium is given as:</u></em>

Reaction involved: BH⁺  +  H₂O  ⇌  B  +  H₃O+

Initial:                     0.1 M                    x         x            

Change:                   -x                      +x       +x

Equilibrium:           0.1 - x                    x         x

<u>The acid dissociation constant: </u>K_{a} = \frac{\left [B \right ] \left [H_{3}O^{+}\right ]}{\left [BH^{+} \right ]} = \frac{(x)(x)}{(0.1 - x)} = \frac{x^{2}}{0.1 - x}

\Rightarrow K_{a} = \frac{x^{2}}{0.1 - x}

\Rightarrow 1\times 10^{-10} = \frac{x^{2}}{0.1 - x}

As, x

\Rightarrow 0.1 - x = 0.1

\therefore 1\times 10^{-10} = \frac{x^{2}}{0.1 }

\Rightarrow x^{2} = (1\times 10^{-10})\times 0.1 = 1\times 10^{-11}

\Rightarrow x = \sqrt{1\times 10^{-11}} = 3.16 \times 10^{-6}

<u>Therefore, the concentration of hydrogen ion: x = 3.6 × 10⁻⁶ M</u>

Now, pH = - ㏒ [H⁺] = - ㏒ (3.6 × 10⁻⁶ M) = 5.44

<u>Therefore, the pH of 0.1 M BH⁺ClO₄⁻ solution is 5.44</u>

5 0
1 year ago
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