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Delvig [45]
3 years ago
5

Which produces wave particles that move and displace one another? A) a radio B) a guitar C) a light bulb D) a microwave oven

Chemistry
2 answers:
Semmy [17]3 years ago
8 0
The answer is B. A guitar generally produces sound waves that propagate when the strings are strummed. The strings are displaced through the vibrations caused by contact of the hand and the guitar. You will also notice the vibrations by looking closely to the string. Wave particles continuously collide with each other to make a sustaining or prolonging sound.
Lesechka [4]3 years ago
4 0

Answer:

B: A guitar

Explanation:

A guitar produces sound wave particles, which move and displace one another. A microwave oven, a light bulb, and a radio have electromagnetic waves, which do not displace one another.

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Liquid nitrogen is used for: electrosurgery cryosurgery laser surgery
Otrada [13]
Cryosurgery. You automatically know this because 'cryo' means ice or something cold, thus you can assume that it is that
4 0
3 years ago
The cell potential of a redox reaction occurring in an electrochemical cell under any set of temperature and concentration condi
avanturin [10]

Answer : The actual cell potential of the cell is 0.47 V

Explanation:

Reaction quotient (Q) : It is defined as the measurement of the relative amounts of products and reactants present during a reaction at a particular time.

The given redox reaction is :

Ni^{2+}(aq)+Zn(s)\rightarrow Ni(s)+Zn^{2+}(aq)

The balanced two-half reactions will be,

Oxidation half reaction : Zn\rightarrow Zn^{2+}+2e^-

Reduction half reaction : Ni^{2+}+2e^-\rightarrow Ni

The expression for reaction quotient will be :

Q=\frac{[Zn^{2+}]}{[Ni^{2+}]}

In this expression, only gaseous or aqueous states are includes and pure liquid or solid states are omitted.

Now put all the given values in this expression, we get

Q=\frac{(0.0141)}{(0.00104)}=13.6

The value of the reaction quotient, Q, for the cell is, 13.6

Now we have to calculate the actual cell potential of the cell.

Using Nernst equation :

E_{cell}=E^o_{cell}-\frac{RT}{nF}\ln Q

where,

F = Faraday constant = 96500 C

R = gas constant = 8.314 J/mol.K

T = room temperature = 316 K

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

E^o_{cell} = standard electrode potential of the cell = 0.51 V

E_{cell} = actual cell potential of the cell = ?

Q = reaction quotient = 13.6

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

E_{cell}=0.51-\frac{(8.314)\times (316)}{2\times 96500}\ln (13.6)

E_{cell}=0.47V

Therefore, the actual cell potential of the cell is 0.47 V

4 0
3 years ago
A. If the mass of Hydrogen is 1 amu, what is the mass of Hydrogen in the reactant side of the equation above?
dimulka [17.4K]

Answer:

there are 4 hydrogen so

A.the mass of Hydrogen in the reactant side of the equation above is 1×4=4 amu.

B.the mass of Hydrogen on the product side of the equation above =1×4=4 amu.

<u>Note</u><u>:</u><u> </u><u>mass</u><u> </u><u>of</u><u> </u><u>reactant</u><u> </u><u>=</u><u>mass</u><u> </u><u>of</u><u> </u><u>product</u><u>.</u>

7 0
2 years ago
What is the percent yield if 248.7 g of Cu is produced when 87 g of Al reacts with an excess
valkas [14]

Answer:

2.09 g Cu

Explanation:

First, we have to balance the equation:

3Al(s) + 3CuSO4(aq) = Al2(SO4)3(aq) + 3Cu(s)

Then we have to change grams of Al to mol:

1 mol Al = 26.98 g

1.37 g Al x (1 mol/26.98 g) = 0.051 mol

Then, we use the balanced equation to find the mol of Cu produced:

According to the equation:

3 mol Al produces 3 mol of Cu

3 mol Al = 3 mol Cu

So:

0.051 mol Al x (3 mol Cu/3 mol Al) = 0.051 mol Cu

This mol value is the theoric yield of the reaction.

Remember the equation for yield percent:

%Y = (Y real / Y theoric) x 100

Y theoric = 0.051 mol Cu

% Y = 67.4%

Let's replace these values on the equation and then fin Y real.

67.4 % = (Y real / 0.051) x 100

Y real = 0.033 mol Cu

So the real produced mol is 0.033 mol Cu.

Let's change these moles to grams:

1 mol Cu = 63.55 g

0.033 mol Cu x ( 63.55 g Cu/ 1 mol Cu) = 2.09 g Cu

If you want, watch the image attached is more clear the

mathematical procedure

Explanation:

6 0
2 years ago
Please help :) hope you have a good day
sineoko [7]

Have a wonderful day :) thanks for the points

3 0
2 years ago
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