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3 years ago

15

What is the energy of a light wave with a frequency of 9.8 x 10^20 Hz?

1 answer:

Wewaii [24]3 years ago

7 0

Answer:

Regions of the Electromagnetic Spectrum

Wavelength (m)Frequency (Hz)Radio> 1 x 10-1< 3 x 109Microwave1 x 10-3 - 1 x 10-13 x 109 - 3 x 1011Infrared7 x 10-7 - 1 x 10-33 x 1011 - 4 x 1014Optical4 x 10-7 - 7 x 10-74 x 1014 - 7.5 x 1014

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Looking at the Periodic Table, how can you determine if an element is a metal, metalloid or nonmetal?

Digiron [165]

Answer:

It depends on where it's located on the periodic table.

Explanation:

Metals are on left side of the periodic table except Hydrogen (H), which is a nonmetal, then there's the staircase, which has the metalloids, and on the right of the staircase are the nonmetals.

6 0

3 years ago

The standard reduction potentials for the Ag+|Ag(s) and Zn2+| Zn(s) half-cell reactions are +0.799 V and -0.762 V, respectively.

mihalych1998 [28]

<u>Answer:</u> The potential of the given cell is 1.551 V

<u>Explanation:</u>

The given chemical cell follows:

$Zn(s)|Zn^{2+}(0.125M)||Ag^{+}(0.240M)|Ag(s)$

<u>Oxidation half reaction:</u> $Zn(s)\rightarrow Zn^{2+}(0.125M)+2e^-;E^o_{Zn^{2+}/Zn}=-0.762V$

<u>Reduction half reaction:</u> $Ag^{+}(0.240M)+e^-\rightarrow Ag(s);E^o_{Ag^{+}/Ag}=0.799V$ ( × 2)

<u>Net cell reaction:</u> $Zn(s)+2Ag^{+}(0.240M)\rightarrow Zn^{2+}(0.125M)+2Ag(s)$

Oxidation reaction occurs at anode and reduction reaction occurs at cathode.

To calculate the $E^o_{cell}$ of the reaction, we use the equation:

$E^o_{cell}=E^o_{cathode}-E^o_{anode}$

Putting values in above equation, we get:

$E^o_{cell}=0.799-(-0.762)=1.561V$

To calculate the EMF of the cell, we use the Nernst equation, which is:

$E_{cell}=E^o_{cell}-\frac{0.059}{n}\log \frac{[Zn^{2+}]}{[Ag^{+}]^2}$

where,

$E_{cell}$ = electrode potential of the cell = ? V

$E^o_{cell}$ = standard electrode potential of the cell = +1.561 V

n = number of electrons exchanged = 2

$[Zn^{2+}]=0.125M$

$[Ag^{+}]=0.240M$

Putting values in above equation, we get:

$E_{cell}=1.561-\frac{0.059}{2}\times \log(\frac{(0.125)}{(0.240)^2})$

$E_{cell}=1.551V$

Hence, the potential of the given cell is 1.551 V

6 0

4 years ago

What is the molarity of 200. mL of solution containing 4.50 moles of ZnCl2?

MariettaO [177]

Answer:

Molarity of solution = 0.9 M

Explanation:

Given data:

Moles of ZnCl₂ = 4.50 mol

Volume of solution = 200 mL

Molarity of solution = ?

Solution:

Molarity is used to describe the concentration of solution. It tells how many moles are dissolve in per litter of solution.

Formula:

Molarity = number of moles of solute / L of solution

Now we will convert the volume in L.

200 mL×1 L/1000 mL = 0.2L

M = 4.50 mol / 0.2 L

M = 0.9 M

4 0

3 years ago

The nucleus of a lead atom has a charge of 82 protons what are the drection and magnitude

KATRIN_1 [288]

Answer:

left

Explanation:

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3 years ago

Describe 4 real life examples of scientific inquiry

Olegator [25]

What stress would adding a postage stamp have on a helicopters flight capabilities?

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4 years ago