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oee [108]
3 years ago
10

One mole of copper has a mass of 63.5 grams. Approximately how many atoms of copper are present in one mole of copper?

Chemistry
1 answer:
quester [9]3 years ago
7 0

Answer: 6*10^23 atoms

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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
At equilibrium, the concentrations in this system were found to be [N2]=[O2]=0.200 M and [NO]=0.400 M.
Kipish [7]

Answer:

The concentration in equilibrium of NO is 0,550M.

Explanation:

For the reaction:

N₂(g) + O₂(g) ⇄ 2NO

The equilibrium constant is defined as:

k = [NO]² / [N₂][O₂] <em>(1)</em>

Replacing for the concentrations in equilibrium:

k = (0,400M)² / (0,200M)(0,200M)

<em>k = 4,000</em>

If you add more NO until 0,700M, the equilibrium concentrations will be:

[NO] = 0,700M-2x

[N₂] = 0,200M+x

[O₂] = 0,200M+x

Replacing in (1)

4,000 =  (0,700M-2x)² / (0,200M+x)²

4,000 =  4x²- 2,8x + 0,49 / x² + 0,4x + 0,04

4x² + 1,6x + 0,16 = 4x²- 2,8x + 0,49

4,4x = 0,33

x = 0,075M

That means that concentration in equilibrium of NO is:

[NO] = 0,700M - 2×0,075M = <em>0,550M</em>

I hope it helps!

4 0
3 years ago
HELP!!!
vredina [299]

Answer:

the answer is A

Explanation:

I dont really have an explanation but hope it helps

5 0
2 years ago
Read 2 more answers
1 mole of no2(g) has a greater entropy than 1 mole of n2o4(g). true or false
bija089 [108]
In order to answer this, you need to find the empirical data for the standard entropies. Please refer to this link: http://www.mrbigler.com/misc/energy-of-formation.PDF

For NO₂ gas, the entropy is 240 J/mol-K. For N₂O₄ gas, the entropy is 304.2 J/mol-K. Therefore, <em>the statement is false.</em>
7 0
3 years ago
What is the type of bonding in a molecule of sulfur dioxide?​
mestny [16]

Answer:

Covalent bonding

Explanation:

Sulfur Dioxide is a process of covalent bonding, since Sulfur and Oxygen are both non-metals. The Sulfur is in the center surrounded by 2 Oxygen atoms.

6 0
3 years ago
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