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slavikrds [6]
3 years ago
15

A term related to frequency that increases as frequency increases

Chemistry
2 answers:
HACTEHA [7]3 years ago
8 0

Answer: Energy.

Explanation: The energy of a wave is directly proportional to its frequency, but inversely proportional to its wavelength. In other words, the greater the energy, the larger the frequency and the shorter (smaller) the wavelength. Given the relationship between wavelength and frequency described above, it follows that short wavelengths are more energetic than long wavelengths.

zvonat [6]3 years ago
7 0
As wavelength increases so does frequency
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natita [175]

Answer:

when you put the marker in the water the water gets into the marker and it will change the solid color to a plain color witch means that the marker will make the water in the cup the color from the marker.

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4 0
2 years ago
Find the no. Of electron involved in the electro deposition of 63.5g of cu from a solution of cuso4
sp2606 [1]
This is a redox reaction, meaning reduction-oxidation reaction. This represents the reaction in one side of the electrode in an electrolysis set-up. First, we find the oxidation number of Cu in CuSO4:

(ox. # of Cu)+ ox.# of S + 4(ox.# of oxygen) = 0
(ox. # of Cu) + (6) + 4(-2) = 0
ox. # of Cu = 2+

CuSO4 ---> Cu + SO42-
Cu2+ + SO42-  ---->  Cu + SO42-
Cu2+ -----> Cu + 2e-   (net ionic reaction)

The stoichiometric equation would be 2 electrons per mole Copper. Copper has a molar mass of <span>63.5 g/mol. Then, it would only need 2 electrons.


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8 0
3 years ago
Identify a reason that chemical reactions release energy during the reaction process.
Bess [88]

Answer:

option b

Explanation:

When the energy is released the process is called exothermic reaction. This happens when the bonds are broken in the reactants and the system release energy.

5 0
3 years ago
Read 2 more answers
A 50.0 mL sample containing Cd2+ and Mn2+ was treated with 64.0 mL of 0.0600 M EDTA . Titration of the excess unreacted EDTA req
tigry1 [53]

Answer:

the concentration of Cd^{2+}  in the original solution= 0.0088 M

the concentration of Mn^{2+} in the original solution = 0.058 M

Explanation:

Given that:

The volume of the sample  containing Cd2+ and Mn2+ =  50.0 mL; &

was treated with 64.0 mL of 0.0600 M EDTA

Titration of the excess unreacted EDTA required 16.1 mL of 0.0310 M Ca2+

i.e the strength of the Ca2+ = 0.0310 M

Titration of the newly freed EDTA required 14.2 mL of 0.0310 M Ca2+

To determine the concentrations of Cd2+ and Mn2+ in the original solution; we have the following :

Volume of newly freed EDTA = \frac{Volume\ of \ Ca^{2+}* Sample \ of \ strength }{Strength \ of EDTA}

= \frac{14.2*0.0310}{0.0600}

= 7.3367 mL

concentration of  Cd^{2+} = \frac{volume \ of \  newly  \ freed \ EDTA * strength \ of \ EDTA }{volume \ of \ sample}

= \frac{7.3367*0.0600}{50}

= 0.0088 M

Thus the concentration of Cd^{2+} in the original solution = 0.0088 M

Volume of excess unreacted EDTA = \frac{volume \ of \ Ca^{2+} \ * strength \ of Ca^{2+} }{Strength \ of \ EDTA}

= \frac{16.1*0.0310}{0.0600}

= 8.318 mL

Volume of EDTA required for sample containing Cd^{2+}   and  Mn^{2+}  = (64.0 - 8.318) mL

= 55.682 mL

Volume of EDTA required for Mn^{2+}  = Volume of EDTA required for

                                                                sample containing  Cd^{2+}   and  

                                                             Mn^{2+} --  Volume of newly freed EDTA

Volume of EDTA required for Mn^{2+}  = 55.682 - 7.3367

= 48.3453 mL

Concentration  of Mn^{2+} = \frac{Volume \ of EDTA \ required \ for Mn^{2+} * strength \ of \ EDTA}{volume \ of \ sample}

Concentration  of Mn^{2+} =  \frac{48.3453*0.0600}{50}

Concentration  of Mn^{2+}  in the original solution=   0.058 M

Thus the concentration of Mn^{2+} = 0.058 M

6 0
3 years ago
PLEASE HELP WITH THESE 3 QUESTIONS
Alenkinab [10]

Answer:

i think this is the answer

Explanation:

C).50 ML

D)8.5 ph

E)weak acid,strong base,basic

8 0
1 year ago
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