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aev [14]
3 years ago
15

Se disuelven 200 ml de etanol C2H6O al 70 % en volumen con 300 ml de agua calcular considerando que los volúmenes son aditivos E

l % en volumen de alcohol.
Chemistry
1 answer:
insens350 [35]3 years ago
8 0

Answer:

28%

Explanation:

Primero <u>calculamos los mililitros de etanol presentes en la primera solución</u>:

  • 200 mL * 70/100 = 140 mL etanol

Para calcular el nuevo porcentaje en volumen, dividimos los mililitros de etanol (<em>que permanecen iguales durante el proceso de dilución</em>) entre el volumen final.

  • Volumen final = 200 mL + 300 mL = 500 mL
  • % Volumen de Etanol = 140 mL / 500 mL * 100% = 28%
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Balanced force I think
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H2 + O2 ----&gt; H2O What is the mole ratio of hydrogen to water?
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A sample of a gas occupies 2.76 L at 303K. What would the volume be if the temperature was increased to 404K? Assume constant
Artist 52 [7]

Answer:

If the temperature was increased to 404 K, its volume would be 3.68 L.

Explanation:

Charles' Law gives a relationship between the volume and the temperature of the gas at constant temperature. This law states that the volume of a given amount of gas held at constant pressure is directly proportional to the temperature.

V\propto T

\dfrac{V_1}{V_2}=\dfrac{T_1}{T_2}

Let

V_1=2.76\ L\\\\T_1=303\ K\\\\T_2=404\ K

Let V_2 is new volume. Using above formula we get :

V_2=\dfrac{V_1T_2}{T_1}\\\\V_2=\dfrac{2.76\times 404}{303}\\\\V_2=3.68\ L

If the temperature was increased to 404 K, its volume would be 3.68 L.

6 0
4 years ago
For the simple decomposition reaction: AB(g) LaTeX: \longrightarrow⟶ A(g) + B(g), the rate = k{AB}2 ({ = [) and k = 0.85 1/MLaTe
Nesterboy [21]

Answer:

1.169s

Explanation:

k = 0.851 M-1s-1

The unit of the rate constant, k tells us this is a second order reaction.

From the question;

Initial Concentration  [A]o = 2.01M

Final Concentration  [A] = One third of 2.10 = (1/3) * 2.10 = 0.67M

Time = ?

The integrated rate law for second order reactions is given as;

1 / [A] = (1 / [A]o) + kt

Making t subject of interest, we have;

kt = (1 / [A] ) - (1 / [A]o )

t = (1 / [A] ) - (1 / [A]o ) / k

Inserting the values;

t = [ (1 / 0.67 ) - (1 /  2.10) ] / 0.851

t = ( 1.4925 - 0.4975 ) / 0.851

t = 0.995 / 0.851

t = 1.169s

[A] = 0.13073 M ≈ 0.13 M ( 2 s.f)

7 0
4 years ago
In vacuum, the speed of light is c= 2.998 x 108m/s. However, the speed generally decreases when light travels through media othe
Kaylis [27]

Answer:

Light of wavelength 200 nm will have lowest frequency in while traveling through diamond.

Explanation:

Speed of the light in vacuum = c

Relation between speed of light , wavelength (λ) and frequency (ν);

\nu =\frac{c}{\lambda }

Speed of light in a medium = c' = c/n

\nu =\frac{c'}{\lambda }=\frac{c}{n\times \lambda }...[1]

Where : n = refractive index of a medium

So, the medium with greater value of refractive index lower the speed of light to greater extent.

From [1] , we can see that frequency of light is inversely proportional to the refractive index of the medium :

\nu \propto \frac{1}{n}

This means that higher the value of refractive index lower will be the value of frequency of light in that medium or vice-versa.

According to question, light of wavelength 200 nm will have lowest frequency in while traveling through diamond because refractive index of diamond out of the given mediums is greatest.

Increasing order of refractive indices:

H_2

Decreasing order of frequency of light 210 nm in these medium :

H_2>H_2O>CCL_4>\text{Silicon oil}> Diamond

4 0
3 years ago
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