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kolbaska11 [484]
1 year ago
7

Zinc metal (Zn) will react with aqueous hydrochloric acid (HCI aq) to produce aqueous zinc chloride (ZnCl2 aq) and hydrogen gas

(H2). Which of the following isthe complete, balanced equation for this reaction?1. 2 HCI (aq) + 2 Zn (s) > 2 H2 (g) + ZnCl2 (aq)2. HCI (aq) + Zn (s) > H2 (g) + ZnCl2 (ag)3. 2 HCI (aq) + Zn (s) › 2 H2 (g) + ZnCl2 (ag)4. 2 HCI (aq) + Zn (s) › H2 (g) + 2 ZnCl2 (ag)5. 2 HCI (aq) + Zn (s) › H2 (g) + ZnCl2 (ad)
Chemistry
1 answer:
Debora [2.8K]1 year ago
3 0

Answer:

5

Explanation:

Here, we want to get the equation of the reaction between Hydrochloric acid and Zinc metal

Zinc metal displaces the hydrogen from hydrochloric acid to form zinc chloride

We have the equation of reaction as:

2HCl_{(aq)}\text{ + Zn}_{(s)}\text{ }\rightarrow\text{ ZnCl}_{2(aq)}\text{ + H}_{2(g)}

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Since in every mole of NaCl, there is only one mole of sodium then, we conclude that there are also 108.7 moles of NaCl. Multiplying the number of moles to the molar mass of sodium chloride (58.44), the answer would be 6,352.17 grams.
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When two ions form an ionic bond, what type of compound is formed?
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Name two elements, compounds and mixtures that are essential for living organisms. Also write why are they essential?​
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Read 2 more answers
Find the amount of heat energy needed to convert 400 grams of ice at -38°C to steam at 160°C.
Marianna [84]

The amount of heat energy needed to convert 400 g of ice at -38 °C to steam at 160 °C is 1.28×10⁶ J (Option D)

<h3>How to determine the heat required change the temperature from –38 °C to 0 °C </h3>
  • Mass (M) = 400 g = 400 / 1000 = 0.4 Kg
  • Initial temperature (T₁) = –25 °C
  • Final temperature (T₂) = 0 °
  • Change in temperature (ΔT) = 0 – (–38) = 38 °C
  • Specific heat capacity (C) = 2050 J/(kg·°C)
  • Heat (Q₁) =?

Q = MCΔT

Q₁ = 0.4 × 2050 × 38

Q₁ = 31160 J

<h3>How to determine the heat required to melt the ice at 0 °C</h3>
  • Mass (m) = 0.4 Kg
  • Latent heat of fusion (L) = 334 KJ/Kg = 334 × 1000 = 334000 J/Kg
  • Heat (Q₂) =?

Q = mL

Q₂ = 0.4 × 334000

Q₂ = 133600 J

<h3>How to determine the heat required to change the temperature from 0 °C to 100 °C </h3>
  • Mass (M) = 0.4 Kg
  • Initial temperature (T₁) = 0 °C
  • Final temperature (T₂) = 100 °C
  • Change in temperature (ΔT) = 100 – 0 = 100 °C
  • Specific heat capacity (C) = 4180 J/(kg·°C)
  • Heat (Q₃) =?

Q = MCΔT

Q₃ = 0.4 × 4180 × 100

Q₃ = 167200 J

<h3>How to determine the heat required to vaporize the water at 100 °C</h3>
  • Mass (m) = 0.4 Kg
  • Latent heat of vaporisation (Hv) = 2260 KJ/Kg = 2260 × 1000 = 2260000 J/Kg
  • Heat (Q₄) =?

Q = mHv

Q₄ = 0.4 × 2260000

Q₄ = 904000 J

<h3>How to determine the heat required to change the temperature from 100 °C to 160 °C </h3>
  • Mass (M) = 0.4 Kg
  • Initial temperature (T₁) = 100 °C
  • Final temperature (T₂) = 160 °C
  • Change in temperature (ΔT) = 160 – 100 = 60 °C
  • Specific heat capacity (C) = 1996 J/(kg·°C)
  • Heat (Q₅) =?

Q = MCΔT

Q₅ = 0.4 × 1996 × 60

Q₅ = 47904 J

<h3>How to determine the heat required to change the temperature from –38 °C to 160 °C</h3>
  • Heat for –38 °C to 0°C (Q₁) = 31160 J
  • Heat for melting (Q₂) = 133600 J
  • Heat for 0 °C to 100 °C (Q₃) = 167200 J
  • Heat for vaporization (Q₄) = 904000 J
  • Heat for 100 °C to 160 °C (Q₅) = 47904 J
  • Heat for –38 °C to 160 °C (Qₜ) =?

Qₜ = Q₁ + Q₂ + Q₃ + Q₄ + Q₅

Qₜ = 31160 + 133600 + 167200 + 904000 + 47904

Qₜ = 1.28×10⁶ J

Learn more about heat transfer:

brainly.com/question/10286596

#SPJ1

7 0
2 years ago
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