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fomenos
2 years ago
8

What is amplitude?

Chemistry
2 answers:
NISA [10]2 years ago
6 0

What is amplitude?

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<u>A: amplitude is the height of a wave  </u>

B: amplitude is the number of complete waves that pass a point in a second

C: amplitude is the time it takes for one complete waves to a given point  

D: amplitude is the distance between two crest or two troughs

Ede4ka [16]2 years ago
4 0

A. The height of a sound wave

Happy to help. PLEASE MARK AS BRAINLIEST!

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Circle the correct one from given alternatives.
mash [69]

Answer:

24 is the correct anwer

this the anwer text this u no

3 0
1 year ago
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4. How many milligrams are in 5.25 x 10-13 kg?<br><br> the “-13” is an exponent
rusak2 [61]

5. 25 x 10⁻⁷mg

Explanation:

This is mass conversion from mg to kg;

The kg is a quantity of mass used to measure the amount of matter in a substance.

   Given mass = 5.25 x 10⁻¹³kg

The kilo-  is a prefix that denotes 10³

  therefore;

         1000g = 1kilogram

 the milli-  is a prefix that denotes 10⁻⁻³

       1000mg = 1g

Now that we know this, we can convert:

   5.25 x 10⁻¹³kg  x \frac{1000g}{1kg}  x \frac{1000mg}{1g}   =  5. 25 x 10⁻¹³ x 10⁶mg

      =  5. 25 x 10⁻⁷mg

learn more:

Conversion brainly.com/question/1548911

#learnwithBrainly

8 0
3 years ago
Circulacion de electrones en un material conductor en determinado tiempo.
barxatty [35]

Answer:

falso

Explanation:

tiempo no determinado El circulation de elctrones

4 0
1 year ago
Conduction is when heat is transferred through
Nastasia [14]
The correct answer is a the sun
3 0
3 years ago
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At 2000°C, the equilibrium constant for the reaction below is Kc = 4.10 ´ 10–4 . If 0.600 moles of NO is placed in a 1.0-L react
erastova [34]

Answer:

At equilibrium, the concentration of N_{2 (g)} is going to be 0.30M

Explanation:

We first need the reaction.

With the information given we can assume that is:

N_{2 (g)} + O_{2 (g)} ⇄ 2NO_{(g)}

If there is placed 0.600 moles of NO in a 1.0-L vessel, we have a initial concentration of 0.60 M NO; and no N_{2 (g)} nor  O_{2 (g)} present. Immediately, N_{2 (g)} andO_{2 (g)} are going to be produced until equilibrium is reached.

By the ICE (initial, change, equilibrium) analysis:

I: [N_{2 (g)}]=0   ;     [O_{2 (g)} ]= 0    ; [NO_{(g)}]=0.60M

C: [N_{2 (g)}]=+x   ;     [O_{2 (g)} ]= +x    ; [NO_{(g)}]=-2x

E: [N_{2 (g)}]=0+x   ;     [O_{2 (g)} ]= 0+x   ; [NO_{(g)}]=0.60-2x

Now we can use the constant information:

K_{c}=\frac{[products]^{stoichiometric coefficient} }{[reactants]^{stoichiometric coefficient} }

4.10* 10^{-4} =\frac{(0.60-2x)^{2}}{(x)*(x)}

4.10* 10^{-4}= \frac{(0.60-2x)^{2}}{x^{2} }

4.10* 10^{-4} * x^{2}= (0.60-2x)^{2}}

\sqrt{4.10* 10^{-4} * x^{2}}= \sqrt{(0.60-2x)^{2}}}

0.0202 x =0.60 - 2x

2x+0.0202x=0.60

x=\frac{0.60}{2.0202}= 0.30

At equilibrium, the concentration of N_{2 (g)} is going to be 0.30M

3 0
2 years ago
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