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Reil [10]
3 years ago
5

What process forms an image in a mirror? Choices: A. absorbing light B. Reflecting light C. Refracting light D. Transmitting lig

ht
Chemistry
1 answer:
Dafna1 [17]3 years ago
4 0

Answer

Using the law of reflection—the angle of reflection equals the angle of incidence—we can see that the image and object are the same distance from the mirror. This is a virtual image, since it cannot be projected—the rays only appear to originate from a common point behind the mirror.

Explanation:

Hope this helps someone

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If you collect 1.75 L of hydrogen gas during a lab experiment, when the room temperature is 23ºC and the barometric pressure is
Marianna [84]

Answer:

n=0.0747mol

Explanation:

Hello,

In this case, since we can consider hydrogen gas as an ideal gas, we check the volume-pressure-temperature-mole relationship by using the ideal gas equation:

PV=nRT

Whereas we are asked to compute the moles given the temperature in Kelvins, thr pressure in atm and volume in L as shown below:

n=\frac{105kPa*\frac{0.009869atm}{1kPa}*1.75L}{0.082\frac{atm*L}{mol*K}*(23+273.15)K} \\\\n=0.0747mol

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6 0
3 years ago
Which of the following chemical reactions produces water via a
DiKsa [7]

Answer:

Should be A, but also I think there should be an arrow between O2 and 2H2O

6 0
3 years ago
Consider the following unbalanced redox reactions. In each case, separate the whole reactions into half-reactions, balance the h
Marina86 [1]

Answer & Explanation:

(a)

Fe^{2+} +NO_{3}^{-}  => Fe{(OH)}_3 + N_2

reducing agent = Fe²⁺

Oxidizing agent = NO₃⁻

oxidation

Fe²⁺  ⇒ Fe(OH)₃

reduction

NO₃⁻  ⇒ N₂

Oxidation Half Reaction

(<em>redox reactions are balanced by adding appropriate H⁺ and H₂O atoms)</em>

Fe²⁺ ⇒ Fe(OH)₃

Balance O atoms

Fe²⁺ + 3H₂O ⇒ Fe(OH)₃

Balance H atoms

Fe² + 3H₂0 ⇒ Fe(OH)₃ + 3H⁺

balance Charge

Fe² + 3H₂0 ⇒ Fe(OH)₃ + 3H⁺ + e⁻..............(1)

reduction Half Reaction

NO₃⁻ ⇒ N₂

Balance N atoms

2NO₃⁻ ⇒N₂

Balance O atoms by adding appropriate H₂O

2NO₃⁻ ⇒ N₂ + 6H₂O

Balance H atoms

2NO₃⁻ + 12H⁺ ⇒ N₂ + 6H₂O

Balance Charge

2NO₃⁻ + 12H⁺ + 10e⁻⇒ N₂ + 6H₂O.................(2)

Combine Equation (1) and (2)

(1) × 10: 10Fe² + 30H₂0 ⇒ 10Fe(OH)₃ + 30H⁺ + 10e⁻

(2) × 1:   2NO₃⁻ + 12H⁺ + 10e⁻⇒ N₂ + 6H₂O

(1) + (2): 10Fe² + <u><em>30H₂0</em></u> + 2NO₃⁻ + <u><em>12H⁺</em></u> + <u><em>10e⁻</em></u> ⇒10Fe(OH)₃ + <u><em>30H⁺</em></u><u><em> </em></u>+ <em><u>10e⁻</u></em> +              

            N₂ + <u><em>6H₂O</em></u>

            10Fe² + 24H₂0 + 2NO₃⁻  ⇒ 10Fe(OH)₃ + 18H⁺ + N₂

this is the balanced reaction

REDUCTION POTENTIAL

10Fe²⁺(aq) + 10e⁻ ⇒ 10Fe(OH)₃(aq)             E°ox = 10(-0.44) = -4.4V

2NO₃⁻(aq) -  2e⁻ ⇌ N₂(g) + 18H⁺       E°red = 2(+0.80) = +1.6

10Fe² + 24H₂0 + 2NO₃⁻  ⇒ 10Fe(OH)₃ + 18H⁺ + N₂    E°cell = -2.8V

E°cell = E°red + E°ox

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