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Leona [35]
4 years ago
7

Imagine a moving boat making waves in the middle of a pool of water. Also, imagine two observers, one behind the boat and one in

front of the boat. Describe how each observer will see the waves generated by the boat and explain their observations.
Chemistry
1 answer:
Simora [160]4 years ago
3 0
The person who is behind the boat will see much powerful and bigger waves than the person who is in front of the boat. This phenomenon is caused by the Doppler effect which describes the changes in the observed frequency of a wave whenever there is relative motion between the wave source and the observer.
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A 170.0 g sample of an unidentified compound contains 29.84 g sodium, 67.49
GREYUIT [131]

The empirical formula : Na₂Cr₂O₇

<h3>Further explanation</h3>

Given

170 g sample contains :

29.84 g sodium, 67.49  g chromium, and 72.67 g oxygen

Required

The compound's empirical formula

Solution

mol ratio of elements :

Na : 29.84 : 23 g/mol = 1.297

Cr : 67.49 : 51,9961 g/mol = 1.297

O : 72.67 : 16 g/mol = 4.54

Divide by 1.297

Na : Cr : O = 1 : 1 : 3.5 = 2 : 2 : 7

6 0
3 years ago
When atoms combine they form<br> A. molecules<br> B.periodic table<br> C.particles<br> D.protons
IceJOKER [234]

Answer

A. molecules

Explanation

When atoms or elements combine, they are called molecules. When molecules combine they form compounds.

6 0
4 years ago
Calculate E ° for the half‑reaction, AgCl ( s ) + e − − ⇀ ↽ − Ag ( s ) + Cl − ( aq ) given that the solubility product constant
antoniya [11.8K]

Answer: The value of E^{o} for the half-cell reaction is 0.222 V.

Explanation:

Equation for solubility equilibrium is as follows.

          AgCl(s) \rightleftharpoons Ag^{+}(aq) + Cl^{-}(aq)

Its solubility product will be as follows.

       K_{sp} = [Ag^{+}][Cl^{-}]

Cell reaction for this equation is as follows.

     Ag(s)| AgCl(s)|Cl^{-}(0.1 M)|| Ag^{+}(1.0 M)| Ag(s)

Reduction half-reaction: Ag^{+} + 1e^{-} \rightarrow Ag(s),  E^{o}_{Ag^{+}/Ag} = 0.799 V

Oxidation half-reaction: Ag(s) + Cl^{-}(aq) \rightarrow AgCl(s) + 1e^{-},   E^{o}_{AgCl/Ag} = ?

Cell reaction: Ag^{+}(aq) + Cl^{-}(aq) \rightarrow AgCl(s)

So, for this cell reaction the number of moles of electrons transferred are n = 1.

    Solubility product, K_{sp} = [Ag^{+}][Cl^{-}]

                                               = 1.77 \times 10^{-10}

Therefore, according to the Nernst equation

           E_{cell} = E^{o}_{cell} - \frac{0.0592 V}{n} log \frac{[AgCl]}{[Ag^{+}][Cl^{-}]}

At equilibrium, E_{cell} = 0.00 V

Putting the given values into the above formula as follows.

         E_{cell} = E^{o}_{cell} - \frac{0.0592 V}{n} log \frac{[AgCl]}{[Ag^{+}][Cl^{-}]}

        0.00 = E^{o}_{cell} - \frac{0.0592 V}{1} log \frac{1}{[Ag^{+}][Cl^{-}]}    

       E^{o}_{cell} = \frac{0.0592}{1} log \frac{1}{K_{sp}}

                  = 0.0591 V \times log \frac{1}{1.77 \times 10^{-10}}

                  = 0.577 V

Hence, we will calculate the standard cell potential as follows.

           E^{o}_{cell} = E^{o}_{cathode} - E^{o}_{anode}

       0.577 V = E^{o}_{Ag^{+}/Ag} - E^{o}_{AgCl/Ag}

       0.577 V = 0.799 V - E^{o}_{AgCl/Ag}

       E^{o}_{AgCl/Ag} = 0.222 V

Thus, we can conclude that value of E^{o} for the half-cell reaction is 0.222 V.

3 0
3 years ago
Atoms have _______ different subatomic particles.
valkas [14]

Answer:

C. 3

Explanation:

I hope my answer will be useful

6 0
3 years ago
What determines whether an atom becomes a positive ion or a negative ion?
Ugo [173]
<span> Whether the atom has extra electrons or missing electrons</span>
6 0
3 years ago
Read 2 more answers
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