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Nesterboy [21]
3 years ago
6

The reducing agent in a material is the one that A. gains electrons. B. increases its atomic number. C. is reduced. D. is oxidiz

ed.
Chemistry
2 answers:
Dmitry [639]3 years ago
8 0

Answer is: D. is oxidized.

Reducing agent is an element  or compound that loses an electron to another chemical species in a redox chemical reaction and they have been oxidized.

For example, chemical equation: Ca + H₃PO₄ → Ca₃(PO₄)₂ + H₂.

Oxidation reaction: Ca⁰ → Ca⁺² + 2e⁻ /×3.

3Ca⁰ → 3Ca⁺² + 6e⁻

Reduction reaction: 6e⁻ + 2H₃⁺¹ → 3H₂⁰.

Balanced chemical equation: 3Ca + 2H₃PO₄ → Ca₃(PO₄)₂ + 3H₂.

Calcium is stronger reducing agent than hydrogen, gives electrons easier.

Calcium is oxidized, it changes oxidation number from 0 to +2.

Bas_tet [7]3 years ago
4 0
D. Is Oxidized Hope I helped.
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3 years ago
A 151.5-g sample of a metal at 75.0°C is added to 151.5 g at 15.1°C. The temperature of the water rises to 18.7°C. Calculate the
Kryger [21]

Answer:

The specific heat capacity of the metal is 0.268 J/g°C

Explanation:

Step 1: Data given

Mass of the metal = 151.5 grams

The temperature of the metal = 75.0 °C

Temperature of water = 15.1 °C

The temperature of the water rises to 18.7°C.

The specific heat capacity of water is 4.18 J/°C*g

Step 2: Calculate the specific heat capacity of the metal

heat lost = heat gained

Q = m*c*ΔT

Qmetal = - Qwater

m(metal) * c(metal) * ΔT(metal) = m(water) * c(water) * ΔT(water)

⇒ mass of the metal = 151.5 grams

⇒ c(metal) = TO BE DETERMINED

⇒ΔT( metal) = T2 - T1 = 18.7 °C - 75.0 °C = -56.3 °C

⇒ mass of the water = 151.5 grams

⇒ c(water) = 4.184 J/g°C

⇒ ΔT(water) = 18.7° - 15.1 = 3.6 °C

151.5g * c(metal) * -56.3°C = 151.5g * 4.184 J/g°C * 3.6 °C

c(metal) = 0.268 J/g°C

The specific heat capacity of the metal is 0.268 J/g°C

5 0
3 years ago
Consider the half reactions below for a chemical reaction.
ladessa [460]

Answer:

Option A:

Zn(s) + Cu^(2+) (aq) → Cu(s) + Zn^(2+)(aq)

Explanation:

The half reactions given are:

Zn(s) → Zn^(2+)(aq) + 2e^(-)

Cu^(2+) (aq) + 2e^(-) → Cu(s)

From the given half reactions, we can see that in the first one, Zn undergoes oxidation to produce Zn^(2+).

While in the second half reaction, Cu^(2+) is reduced to Cu.

Thus, for the overall reaction, we will add both half reactions to get;

Zn(s) + Cu^(2+) (aq) + 2e^(-) → Cu(s) + Zn^(2+)(aq) + 2e^(-)

2e^(-) will cancel out to give us;

Zn(s) + Cu^(2+) (aq) → Cu(s) + Zn^(2+)(aq)

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