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3 years ago

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What is the energy efficiency of a light bulb that gives you 30 units of usable energy for every 100 units of energy put in to i

t?
a. 0.3
b. 3
c. 30

2 answers:

NeX [460]3 years ago

7 0

Answer:

Explanation:

Given parameters:

Energy output = 30units

Energy input = 100units

problem:

Energy efficiency of light bulb:

Solution:

Efficiency is given as:

Efficieny = \frac{energy_{output} }{energy_{input } x100

Efficieny = $\frac{30}{100}$ x 100 = 30%

vladimir2022 [97]3 years ago

6 0

Answer:

The correct answer is option a.

Explanation:

Efficiency is defined as the ratio of output energy to input energy.

Efficiency(e)= $\frac{E_o}{E_i}$

Energy given out by the system = $E_o$

Energy given to the system = $E_i$

Here we are given with : $E_o-30 units,E_i=100$

Efficiency of of a light bulb:

$e=\frac{30 units}{100 units}=0.3$

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3 years ago

Acetylene gas (C2H2) burns completely in the presence of oxygen gas (O2) to yield carbon dioxide

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A. The balanced equation for the reaction is

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D. The volume of water vapor produced is 0.7 L

<h3>A. Balanced equation </h3>

The balanced equation for the reaction between acetylene gas (C₂H₂) and oxygen gas (O₂) is given below

2C₂H₂ + 5O₂ —> 4CO₂ + 2H₂O

<h3>B. How to determine the volume of oxygen </h3>

2C₂H₂ + 5O₂ —> 4CO₂ + 2H₂O

Since the reaction occurred at standard temperature and pressure, we can thus say that:

From the balanced equation above,

2 L of C₂H₂ reacted with 5 L O₂.

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2 L of C₂H₂ reacted to produce 4 L of CO₂

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2 years ago

En un vaso de precipitado de un litro se coloca exactamente 500 mL de agua destilada a temperatura ambiente y se realizan dos ex

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Answer:

1) La masa del agua a temperatura ambiente es de 500 gramos, 2) La masa del agua cuando se congela es de 500 gramos, 3) La masa de agua que queda después de la evaporación es de 400 gramos, 4) Se ha evaporado 100 gramos de agua.

Explanation:

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Podemos determinar la masa inicial del agua ( $m_{o}$ ), medido en gramos, al conocer su densidad ( $\rho_{w}$ ), medida en gramos por mililitro, y volumen inicial ocupado en el vaso de precipitado ( $V_{o}$ ), medido en mililitros, a partir de la siguiente expresión:

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$m_{f} =\rho_{w}\cdot V_{f}$

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$m_{f} = \left(1\,\frac{g}{mL} \right)\cdot (400\,mL)$

$m_{f} = 400\,g$

La masa de agua que queda después de la evaporación es de 400 gramos.

4) <em>¿Qué masa de agua se evaporó? </em>

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$m_{v} =m_{o}-m_{f}$

Si $m_{o} = 500\,g$ y $m_{f} = 400\,g$ , entonces tenemos que:

$m_{v} = 500\,g -400\,g$

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Answer:

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Explanation:

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3 years ago

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elena-s [515]

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