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anygoal [31]
3 years ago
15

Se ha quemado magnesio (reacción con el oxígeno) y se obtuvieron 12g de óxido de magnesio (II). ¿Cuánto magnesio se quemó? ¿Qué

volumen de oxígeno medido en condiciones normales se quemó?
Physics
1 answer:
Sedaia [141]3 years ago
7 0

Answer:

The first answer is 7,24g Mg

The second answer is 3,36L O2

Explanation:

Atomic masses: Mg= 24,3 ; O=16

Solution: MMgO = 40,3 g/mol

2 Mg + O2 => 2 MgO

1) 12 g MgO x (2x24,3 g Mg / 2x 40,3 MgO) = 7,24 g Mg

2)

At  first calculate moles from Oxygen:

12 g MgO x ( 1 mol from Oxygen/ 2 x 40,3 g MgO ) = 0,15 moles from Oxygen

then ....

0,15 moles from Oxygen x (22,4 lts / 1 mol from Oxygen ) = 3,36 lts

22.4 liters is the constant that we use in this equation, it is a value that is repeated since we are in the presence of a gas with ideal behavior, so it is considered that the normal molar volume of ideal gaseous substances is always 22.4 liters, estimated with a temperature of 0º C and a pressure of 1 atmosphere. (constant temperature and pressure)

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An electric bulb is marked 40volts ,230w another bulb is marked 40w,110v
Andrej [43]

Answer:

a. The ratio of their resistance is 2783:64

b. The ratio of their energy is 4:23

c. The charge on the first bulb is 5.75 C

The charge on the second bulb is 0.\overline {36} C

Explanation:

The voltage on one of the electric bulbs, V₁ = 40  volts

The power rating of the bulb, P₁ = 230 w

The voltage on the other electric bulbs, V₂ = 110 volts

The power rating of the bulb, P₂ = 40 w

a. The power is given by the formula, P = I·V = V²/R

Therefore, R = V²/P

For the first bulb, the resistance, R₁ = 40²/230 ≈ 6.96

The resistance of the second bulb, R₂ = 110²/40

The ratio of their resistance, R₂/R₁ = (110²/40)/(40²/230) = 2783/64

∴ The ratio of their resistance, R₂:R₁ = 2783:64

b. The energy of a bulb, E = t × P

Where;

t = The time in which the bulb is powered on

∴ The energy of the first bulb, E₁ = 230 w × t

The energy of the second bulb, E₂ = 40 w × t

The ratio of their energy, E₂/E₁ = (40 w × t)/(230 w × t) = 4/23

∴ The ratio of their energy, E₂:E₁ = 4:23

c. The charge on a bulb, 'Q', is given by the formula, Q = I × t

Where;

I = The current flowing through the bulb

From P = I·V, we get;

I = P/V

For the first bulb, the current, I = 230 w/40 V = 5.75 amperes

The charge on the first bulb per second (t = 1) is therefore;

Q₁ = 5.75 A × 1 s = 5.75 C

The charge on the first bulb, Q₁ = 5.75 C

Similarly, the charge on the second bulb, Q₂ = (40 W/110 V) × 1 s = 0.\overline {36} C

The charge on the second bulb, Q₂ = 0.\overline {36} C.

d. The question has left out parts

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3 years ago
The metric unit of force is the
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3 years ago
A film of soapy water on top of a plastic cutting board has a thickness of 255 nm. What wavelength and color is most strongly re
Alona [7]

Answer:

the reflected wavelength is lano = 4.55 10⁻⁷ m which corresponds to the blue color

Explanation:

This is a case of reflection interference, we must be careful

* There is a 180º phase change when light passes from the air to the soap film (n = 1,339), but there is no phase change when passing from the pomp to the plastic (n = 1.3)

* the wavelength within the film is modulated by the refractive index

     λₙ =  λ₀ / n

if we consider these relationships the condition for constructive interference is

     2 t = (m + ½)  λₙ

     2t = (m + ½)  λ₀ / n

      λ₀ = 2t n / (m + ½)

we substitute the values

      λ₀= 2 255 10⁻⁹ 1,339 / (m + ½)

      λ₀  = 6.829 10⁻⁷ (m + ½)

let's calculate the wavelength for various interference orders

m = 0

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       λ₀ = 13.6 10⁻⁷

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m = 1

      λ₀ = 6,829 10⁻⁷/ (1 + ½)

       λ₀ = 4.55 10⁻⁷

      color blue

m = 2

      λ₀ = 6.829 10⁻⁷ / (2 + ½)

      λ₀ = 2,7 10⁻⁷

   it is not visible

therefore the reflected wavelength is lano = 4.55 10⁻⁷ m which corresponds to the blue color

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What is the resistance ofa wire made of a material with resistivity of 3.2 x 10^-8 Ω.m if its length is 2.5 m and its diameter i
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R = 0.407Ω.

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To calculate the resistance R of a wire made of a material with resistivity of 3.2x10⁻⁸Ω.m, the length of the wire is 2.5m and its diameter is 0.50mm.

We have to use the equation R = ρL/A but first we have to calculate the cross-sectional area of the wire which is a circle. So, the area of a circle is given by A = πr², with r = d/2. The cross-sectional area of the wire is A = πd²/4.  Then:

R =[(3.2x10⁻⁸Ω.m)(2.5m)]/[π(0.5x10⁻³m)²/4]

R = 8x10⁻⁸Ω.m²/1.96x10⁻⁷m²

R = 0.407Ω

5 0
3 years ago
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