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Anestetic [448]
2 years ago
15

Cobalt-60 is a radioactive isotope used to treat cancers. A gamma ray emitted by this isotope has an energy of 1.33 MeV (million

electron volts; 1 eV = 1.602 x 10¹⁹ J). What is the frequency (in Hz) and the wavelength (in m) of this gamma ray?
Chemistry
1 answer:
Natalka [10]2 years ago
5 0

Answer:

E = 1.33 MeV = 2.13 x 10^{-13} J

v = wavelength = E / h = 2.13 x 10^{-13} / 6.626 x 10^{-34} = 3.2 x 10^{20} m

f = frequency = c / 3.2 x 10^{20} m = 3 x 10^{8} / 3.2 x 10^{20} = 9.375 x 10^{-13} Hz

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Which feature of the evaporating dish makes it effective in the evaporation process
tigry1 [53]
The precise fit on the ring stand. The ability to hold more substance than a test tube. A shallow bottom to increase the surface area of a liquid, leaving the solid behind.
4 0
2 years ago
Read 2 more answers
Calculate the pH of each of following buffered solutions.?a. 0.10 M acetic acid/0.25 M sodium acetate b. 0.25 M acetic acid/0.10
Masteriza [31]

Answer:

a. 5.10.

b. 4.35.

c. 5.10.

d. 4.35.

Explanation:

<u><em>a. 0.10 M acetic acid/0.25 M sodium acetate </em></u>

For acidic buffer:

∵ pH = pKa + log [salt]/[Acid]

∴ pH = - log(Ka) + log [salt]/[Acid]

Ka for acetic acid = 1.8 x 10⁻⁵.

∴ pH = - log(1.8 x 10⁻⁵) + log(0.25)/(0.10)

∴ pH = 4.744 + 0.34 = 5.084 ≅ 5.10.

<u><em>b. 0.25 M acetic acid/0.10 M sodium acetate</em></u>

For acidic buffer:

∵ pH = pKa + log [salt]/[Acid]

∴ pH = - log(Ka) + log [salt]/[Acid]

Ka for acetic acid = 1.8 x 10⁻⁵.

∴ pH = - log(1.8 x 10⁻⁵) + log(0.10)/(0.25)

∴ pH = 4.744 - 0.34 = 4.346 ≅ 4.35.

<u><em>c. 0.080 M acetic acid/0.20 M sodium acetate</em></u>

For acidic buffer:

∵ pH = pKa + log [salt]/[Acid]

∴ pH = - log(Ka) + log [salt]/[Acid]

Ka for acetic acid = 1.8 x 10⁻⁵.

∴ pH = - log(1.8 x 10⁻⁵) + log(0.20)/(0.08)

∴ pH =  4.744 + 0.34 = 5.084 ≅ 5.10.

<u><em></em></u>

<u><em>d. 0.20 M acetic acid/0.080 M sodium acetate</em></u>

For acidic buffer:

∵ pH = pKa + log [salt]/[Acid]

∴ pH = - log(Ka) + log [salt]/[Acid]

Ka for acetic acid = 1.8 x 10⁻⁵.

∴ pH = - log(1.8 x 10⁻⁵) + log(0.08)/(0.20)

∴ pH = 4.744 - 0.34 = 4.346 ≅ 4.35.

6 0
2 years ago
Which of the reactions are spontaneous (favorable)? DHAP − ⇀ ↽ − glyceraldehyde-3-phosphate Δ G = 3.8 kJ / mol DHAP↽−−⇀glycerald
Tpy6a [65]

Answer:

Explanation:

In spontaneous reaction , there is decrease in Gibb's free energy .( Δ G is negative ). Out of given reaction , following reactions have negative Δ G so they are spontaneous.

C ₂ H ₄ + H ₂ Rh ( I ) −−−→ C ₂ H ₆ ,  Δ G = − 150.97 kJ / mol

C ₆ H₁₃O₉ P + ATP ⟶ C ₆ H₁₄ O₁₂ P₂ + ADP ,  Δ G = − 14.2 kJ / mol

7 0
2 years ago
2. A compound with the following composition by mass: 24.0% C, 7.0% H, 38.0% F, and 31.0% P. what is the empirical formula
Svetach [21]

Answer:

C₂H₇F₂P

Explanation:

Given parameters:

Composition by mass:

                C = 24%

                H = 7%

                 F  = 38%

                 P  = 31%

Unknown:

Empirical formula of compound;

Solution :

The empirical formula is the simplest formula of a compound. To solve for this, follow the process below;

                                   C                          H                         F                   P

% composition

by mass                     24                          7                        38                  31

Molar mass                 12                           1                         19                  31

Number of

moles                       24/12                          7/1                    38/19           31/31

                                     2                               7                       2                   1

Dividing

by the

smallest                      2/1                             7/1                       2/1                1/1

                                     2                                7                        2                   1

           Empirical formula        C₂H₇F₂P

5 0
2 years ago
We would like to make a golden standard kilogram in the shape of circular cylinder. The density of gold is 19.32 g/cm3. a) Find
iragen [17]

Explanation:

a)  Using the provided information about the density  of gold, the sample size, thickness, and the following  equations and comersion factors, find the area of   the gold leaf:

V=l \cdot w \cdot h=A \cdot h\\m=\rho \cdot V

Gold _{\rho}=19.32 \mathrm{g} / \mathrm{cm}^{3}

1 \mu=10^{-6} \mathrm{m}

First, find the volume of the sample and then find the area of the sample.

V=\frac{m}{\rho}=\frac{27.6 \mathrm{g}}{19.32 \mathrm{g} / \mathrm{cm}^{3}} \cdot\left(\frac{0.01 \mathrm{m}}{1 \mathrm{cm}}\right)^{3} =\frac{1.429 \times 10^{-6} \mathrm{m}^{3}}

V=A \cdot h \rightarrow A=\frac{V}{h}=\frac{1.429 \times 10^{-6} \mathrm{m}^{3}}{10^{-6} \mathrm{m}} \approx 1.429 \mathrm{m}^{2}

b.  Using the provided information from part a ), the radius of the cylinder, and the following equation for the volume of a cylinder, find the length of the fiber :

V=\pi r^{2} h \rightarrow h=\frac{V}{\pi r^{2}}

h=\frac{1.429 \times 10^{-6} \mathrm{m}^{3}}{\pi \cdot\left(2.5 \times 10^{-6} \mathrm{m}\right)^{2}} \approx 72778 \mathrm{m}

8 0
2 years ago
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