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Nataly [62]
3 years ago
14

Most radiation exposure comes from A. consumer products. B. nuclear medicine. C. medical X-rays. D. natural sources.

Chemistry
2 answers:
Verizon [17]3 years ago
5 0
The answer is B) nuclear medicine
Lady_Fox [76]3 years ago
5 0

The answer is D. natural sources.

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11. Calculate the number of atoms in 2.00g of platinum.
Serggg [28]

Answer:

Explanation:

All you need to know is the atomic mass of platinum, and Avogadro's number.

2.00g Pt divided by atomic mass gives you the moles of platinum, and multiplying by avogadro's number (6.022 x 10^23) gives you the number of atoms.

Atomic mass of platinum can be found on any periodic table.

Hope this helped.

7 0
3 years ago
Read 2 more answers
How is the empirical formula of copper chloride hydrate found?
s2008m [1.1K]
CuCl2
IUPAC ID: Copper dichloride, Copper(II) chloride.

Start with the number of grams of each element, given in the problem.
Convert the mass of each element to moles using the molar mass from the periodic table.
Divide each mole value by the smallest number of moles calculated.
Round to the nearest whole number.
7 0
3 years ago
When 155 mL of water at 26 C is mixed with 75 mL of water at 85 C, what is the final temperature? (assume that no heat is releas
Ket [755]
Depends if it desolved with the water or not

6 0
4 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
3 years ago
Read the sentence from the introduction [paragraphs 1-3].
ss7ja [257]

Answer: the answer is B

Explanation:

6 0
3 years ago
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