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3 years ago

What is the standard unit of frequency Amplitude Hertz Period Second

1 answer:

7 0

Hertz. And since I need 20 characters I’ll give fascinating facts. To find the velocity of a wave you multiply lambda by frequency. Frequency can be found by dividing energy by planks constant ( 6.626x10e-34) or speed divided by wavelength.

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Elements that typically give up electrons

Naddika [18.5K]

Same answer as the first one above

8 0

3 years ago

Consider the following reaction proceeding at 298.15 K: Cu(s)+2Ag+(aq,0.15 M)⟶Cu2+(aq, 1.14 M)+2Ag(s) If the standard reduction

lutik1710 [3]

Answer : The cell potential for this cell 0.434 V

Solution :

The balanced cell reaction will be,

$Cu(s)+2Ag^{+}(aq)\rightarrow Cu^{2+}(aq)+2Ag(s)$

Here copper (Cu) undergoes oxidation by loss of electrons, thus act as anode. silver (Ag) undergoes reduction by gain of electrons and thus act as cathode.

First we have to calculate the standard electrode potential of the cell.

$E^o_{[Cu^{2+}/Cu]}=0.34V$

$E^o_{[Ag^{+}/Ag]}=0.80V$

$E^o=E^o_{[Ag^{+}/Ag]}-E^o_{[Cu^{2+}/Cu]}$

$E^o=0.80V-(0.34V)=0.46V$

Now we have to calculate the concentration of cell potential for this cell.

Using Nernest equation :

$E_{cell}=E^o_{cell}-\frac{0.0592}{n}\log \frac{[Cu^{2+}][Ag]^2}{[Cu][Ag^+]^2}$

where,

n = number of electrons in oxidation-reduction reaction = 2

$E_{cell}$ = ?

Now put all the given values in the above equation, we get:

$E_{cell}=0.46-\frac{0.0592}{2}\log \frac{(1.14)\times (1)^2}{(1)\times (0.15)}$

$E_{cell}=0.434V$

Therefore, the cell potential for this cell 0.434 V

8 0

4 years ago

A student is pushing a box across the room. To push the box three times farther, the student needs to do how much work?

Travka [436]

Answer:

Removing some of the books reduced the mass of the box, and less force was needed to push it across the floor.

8 0

3 years ago

Select the correct answer.

Stels [109]

Answer:

An electric bell is placed inside a transparent glass jar. The bell can be turned on and off using a switch on the outside of the jar. A vacuum is created inside the jar by sucking out the air. Then the bell is rung using the switch. What will we see and hear?

We’ll see the bell move, but we won’t hear it ring.

We won’t see the bell move, but we’ll hear it ring.

We’ll see the bell move and hear it ring.

We won’t see the bell move or hear it ring.

We’ll see the sound waves exit the vacuum pump.

Explanation:

so, the answer to the question is

We'll see the bell move, but we won’t hear it ring.

5 0

3 years ago

Read 2 more answers

I need help with this

fredd [130]

We have here what is known as parallel combination of resistors.

Using the relation:

$\frac{1}{ r_{eff} } = \frac{1}{ r_{1} } + \frac{1}{ r_{2} } + \frac{1}{ r_{3} }.. . + \frac{1}{ r_{n} } \\$
And then we can turn take the inverse to get the effective resistance.

Where r is the magnitude of the resistance offered by each resistor.

In this case we have,
(every term has an mho in the end)
$\frac{1}{10000} + \frac{1}{2000} + \frac{1}{1000} \\ \\ = \frac{1}{1000} ( \frac{1}{10} + \frac{1}{2} + \frac{1}{1} ) \\ \\ = \frac{1}{1000} ( \frac{31}{20}) \\ \\ = \frac{31}{20000}$

To ger effective resistance take the inverse:
we get,
$\frac{20000}{31} \: ohm \\ = 645 .16 \: ohm$

The potential difference is of 9V.

So the current flowing using ohm's law,

V = IR

will be, 0.0139 Amperes.

7 0

3 years ago