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exis [7]
3 years ago
13

In the reaction Zn H 2SO 4→ ZNSO4 + H2, which, if any element is oxidized

Chemistry
1 answer:
Sedbober [7]3 years ago
7 0

Answer:

I hope this answer is correct

Explanation:

This is an oxidation-reduction (redox) reaction:

Zn0 - 2 e- → ZnII (oxidation)

2 HI + 2 e- → 2 H0 (reduction)

Zn is a reducing agent, H2SO4 is an oxidizing agent.

Reactants:

Zn

Names: Zinc source: wikidata, accessed: 2019-09-07, Zinc powder (pyrophoric) source: ICSC, accessed: 2019-09-04, Zn source: wikidata, accessed: 2019-09-07

Appearance: Grey-to-blue powder source: ICSC, accessed: 2019-09-04

H2SO4 – Sulfuric acid source: wikipedia, accessed: 2019-09-27source: wikidata, accessed: 2019-09-07source: NIOSH NPG, accessed: 2019-09-02

Other names: Oil of vitriol source: wikipedia, accessed: 2019-09-27source: wikidata, accessed: 2019-09-07source: ICSC, accessed: 2019-09-04source: NIOSH NPG, accessed: 2019-09-02, Sulfuric acid, concentrated (> 51% and < 100%) source: ICSC, accessed: 2019-09-04, H2SO4 source: wikidata, accessed: 2019-09-07

Appearance: Clear, colorless liquid source: wikipedia, accessed: 2019-09-27; Odourless colourless oily hygroscopic liquid source: ICSC, accessed: 2019-09-04; Colorless to dark-brown, oily, odorless liquid. [Note: Pure compound is a solid below 51°F. Often used in an aqueous solution.] source: NIOSH NPG, accessed: 2019-09-02

Products:

ZnSO4 – Zinc sulfate source: wikipedia, accessed: 2019-09-27source: wikidata, accessed: 2019-09-02source: ICSC, accessed: 2019-09-04

Other names: White vitriol source: wikipedia, accessed: 2019-09-27, Goslarite source: wikipedia, accessed: 2019-09-27, Zinc sulfate (1:1) source: wikidata, accessed: 2019-09-02

Appearance: White powder source: wikipedia, accessed: 2019-09-27; Colourless hygroscopic crystals source: ICSC, accessed: 2019-09-04

H2

Names: Dihydrogen source: wikidata, accessed: 2019-09-07, Hydrogen source: ICSC, accessed: 2019-09-04source: wikidata, accessed: 2019-09-07, H2 source: wikidata, accessed: 2019-09-07

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A 25.0g sample of brass, which has a specific heat capacity of 0.375·J·g−1°C−1, is dropped into an insulated container containin
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Answer:

The equilibrium temperature of water is 25.6 °C

Explanation:

Step 1: Data given

Mass of the sample of brass = 25.0 grams

The specific heat capacity = 0.375 J/g°C

Mass of water = 250.0 grams

Temperature of water = 25.0 °C

The initial temperature of the brass is 96.7°C

Step 2: Calculate the equilibrium temperature

Heat lost = heat gained

Q(sample) = -Q(water)

Q = m*C* ΔT

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

⇒m(sample) = the mass of the sample of brass = 25.0 grams

⇒with c(sample) =The specific heat capacity = 0.375 J/g°C  

⇒with ΔT = the change of temperature = T2 - T1 =T2 - 96.7 °C

⇒with m(water) = the mass of the water = 250.0 grams

⇒with c(water) = the specific heat capacity = 4.184 J/g°C

⇒with ΔT(water) = the change of temperature of water = T2 - T1 = T2 - 25.0°C

25.0 * 0.375 * (T2 - 96.7) = - 250.0 * 4.184 J/g°C * (T2 - 25.0°C)

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9.375T2 + 1046T2 = 26150 + 906.56

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T2 = 25.6 °C

The equilibrium temperature of water is 25.6 °C

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