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densk [106]
3 years ago
5

What are the correct half reactions for the following reaction:

Chemistry
2 answers:
d1i1m1o1n [39]3 years ago
8 0

Answer:

D. Zn → Zn²⁺ + 2e⁻, 2H⁺ + 2e⁻ → H₂.

Explanation:

  • It is a redox reaction that is consisted of two half-reactions:

Oxidation reaction:

Zn losses 2 electrons and is oxidized to Zn²⁺:

<em>Zn → Zn²⁺ + 2e⁻.</em>

<em></em>

Reduction reaction:

H⁺ gains 1 electron and is reduced to H:

<em>2H⁺ + 2e⁻ → H₂.</em>

<em></em>

<em>So, the right choice is: D. Zn → Zn²⁺ + 2e⁻, 2H⁺ + 2e⁻ → H₂.</em>

<em></em>

Valentin [98]3 years ago
7 0

Answer:

D. Zn → Zn²⁺ + 2e⁻, 2H⁺ + 2e⁻ → H₂.

Explanation:

Answer via Educere/ Founder's Education

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Nucleic acids are made of which of the following?
abruzzese [7]

Answer: I believe the correct answer would be A.

5 0
2 years ago
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A mixture of 117.9 g of P and 121.8 g of O, reacts completely to form P4O6 and P4O10. Find the masses of P4O6 and P4O10
Alex17521 [72]

Answer:

Mass of P4O6=103.4

            P4O10=133.48

Explanation:

Balanced reaction is:

8P +8O_{2}  ⇒P_{4} O_{6} +P_{4} O_{10}

Both reactant completely vanishes as equivalent of bot are equal.

Moles of P=\frac{117.9}{31} =3.80

Moles of O_{2} =\frac{117.9}{31} =3.80

No. of moles of formed product are equal and is \frac{1}{8}th of mole of any of reactant.

Thus weight of  P_{4} O_{6} =\frac{3.80}{8}×220 ≈103.41

        weight of  P_{4} O_{10} =\frac{3.80}{8}×284 ≈133.48

4 0
2 years ago
In a reaction, 25 grams of reactant
Vsevolod [243]

Answer:

D. 15g

Explanation:

The law of conservation of mass states that, in a chemical reaction, mass can neither be created nor destroyed. This means that the amount of matter in the elements of the reactants must be equal to the amount in the resulting products.

In this question, 25 grams of a reactant AB, was broken down in a reaction to produce 10 grams of products A and X grams of product B. According to the law of conservation of mass, the mass of the reactant must be equal to the total mass of the products. This means that 25 grams must also be the total mass of both products in this reaction. Hence, if product A is 10 grams, product B will be 25 grams - 10 grams = 15 grams.

Therefore, product B must be 15 grams in order to form a total of 25 grams when added to the mass of product A. This will equate the mass of the reactant AB and fulfill the law of conservation of mass.

8 0
3 years ago
Is it a physical or chemical change when a candle is lit
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Answer:

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Explanation:

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2 years ago
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A 85.2 g copper bar was heated to 221.32 degrees Celsius and placed in a coffee cup calorimeter containing 4250 mL of water at 2
Assoli18 [71]

Answer:

The specific heat of copper is 0.385 J/g°C

Explanation:

A 85.2 g copper bar was heated to 221.32 degrees Celsius and placed in a coffee cup calorimeter containing 425 mL of water at 22.55 degrees Celsius. The final temperature of the water was recorded to be 26.15 degrees Celsius. What is the specific heat of the copper?

Step 1: Data given

Mass of copper = 85.2 grams

Temperature of copper = 221.32 °C

Volume of water = 425 mL

Temperature of water = 22.55 °C

Final temperature = 26.15 °C

Specific heat of water = 4.184 J/g°C

Step 2: Calculat the specific heat of copper

Heat lost = heat gained

Q = m*c*ΔT

Qcopper = -Qwater

m(copper)*c(copper)*ΔT(copper) = - m(water) * c(water) * ΔT(water)

⇒ m(copper) = 85.2 grams

⇒ c(copper) = TO BE DETERMINED

⇒ ΔT(copper) = the change in temeprature = T2 -T1 = 26.15 -221.32 = -195.17 °C

⇒ m(water) = The mass of water = 425 mL * 1g/mL = 425 grams

⇒ c(water) = The specific heat of water = 4.184 J/g°C

⇒ ΔT(water) = The change of temperature of water = 26.15 - 22.55 = 3.6

85.2 * c(copper) * (-195.17) = -425 * 4.184 * 3.6

c(copper) = 0.385 J/g°C

The specific heat of copper is 0.385 J/g°C

(Note, The original question says the volume of the water is 4250 mL. IF this is not an error, the specific heat of copper is 3.85 J/g°C (10x higher than the normal value).

8 0
2 years ago
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