Who kept scientific experiments from being performed in the middle

<em>A chemist must prepare 550.0 mL of hydrochloric acid solution with a pH of 1.60 at 25 °C. He will do this in three steps: Fill a 550.0 mL volumetric flask about halfway with distilled water. Measure out a small volume of concentrated (8.0 M) stock hydrochloric acid solution and add it to the flask. Fill the flask to the mark with distilled water. Calculate the volume of concentrated hydrochloric acid that the chemist must measure out in the second step. Round your answer to 2 significant digits.</em>

Step 1: Calculate [H⁺] in the dilute solution

We will use the following expresion.

pH = -log [H⁺]

[H⁺] = antilog - pH = antilog -1.60 = 0.0251 M

Since HCl is a strong monoprotic acid, the concentration of HCl in the dilute solution is 0.0251 M.

Step 2: Calculate the volume of the concentrated HCl solution

We want to prepare 550.0 mL of a 0.0251 M HCl solution. We can calculate the volume of the 8.0 M solution using the dilution rule.

C₁ × V₁ = C₂ × V₂

V₁ = C₂ × V₂/C₁

V₁ = 0.0251 M × 550.0 mL/8.0 M = 1.7 mL

3 0

3 years ago

A light source of wavelength, l, illuminates a metal and ejects photoelectrons with a maximum kinetic energy of 1 eV. A second l

madreJ [45]

Explanation:

From first source, kinetic energy ( $K.E_{1}$ ) ejected is 1 eV and wavelength of light is $\lambda$ .

From second source, kinetic energy ( $K.E_{2}$ ) ejected is 4 eV and wavelength of light is $\frac{\lambda}{2}$ .

Relation between work function, wavelength, and kinetic energy is as follows.

K.E = $\frac{hc}{\lambda} - \phi$

where, h = Plank's constant = $6.63 \times 10^{-34}$ J.s

c = speed of light = $3 \times 10^{8}$ m/s

Also, it is known that 1 eV = $1.6 \times 10^{-19}$ J

Therefore, substituting the values in the above formula as follows.

From first source,

$K.E_{1}$ = $\frac{hc}{\lambda} - \phi$

1 eV = $1.6 \times 10^{-19} J = \frac{6.63 \times 10^{-34} \times 3 \times 10^{8}}{\lambda} - \phi$

$1.6 \times 10^{-19} J = \frac{1.98 \times 10^{-25} J.m}{\lambda} - \phi$ ........... (1)

From second source,

$K.E_{2}$ = $\frac{hc}{\lambda} - \phi$

$4 \times 1.6 \times 10^{-19} J = \frac{6.63 \times 10^{-34} \times 3 \times 10^{8}}{\frac{\lambda}{2}} - \phi$

$6.4 \times 10^{-19} J = \frac{2 \times 1.98 \times 10^{-25} J.m}{\lambda} - \phi$ ........... (2)

Now, divide equation (2) by 2. Therefore, it will become

${6.4 \times 10^{-19}J}{2} = \frac{2 \times 1.98 \times 10^{-25} J.m}{2\lambda} - \frac{\phi}{2}$

$3.2 \times 10^{-19}J = \frac{1.98 \times 10^{-25} J.m}{\lambda} - \frac{\phi}{2}$ ......... (3)

Now, subtract equation (3) from equation (1), we get the following.

$1.6 \times 10^{-19} = \frac{\phi}{2}$

$\phi$ = $3.2 \times 10^{-19}$

= 2 eV

Thus, we can conclude that work function of the metal is 2 eV.

3 0

4 years ago