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Virty [35]
4 years ago
8

In the laboratory a student finds that it takes 44.0 Joules to increase the temperature of 10.6 grams of solid zinc from 24.9 to

36.4 degrees Celsius. The specific heat of zinc calculated from her data is J/g°C.
Chemistry
1 answer:
BaLLatris [955]4 years ago
7 0

Answer:

The specific heat of zinc is 0.361 J/g°C

Explanation:

<u>Step 1:</u> Data given

44.0 J needed

Mass of solid zinc = 10.6 grams

Initial temperature = 24.9 °C

Final temperature = 36.4 °C

<u>Step 2</u>: Calculate the specific heat of zinc

Q = m*c*ΔT

⇒ with Q = heat (in Joule) = 44.0 J

⇒ with m = the mass of the solid zinc = 10.6 grams

⇒ with c = the specific heat of the zinc = TO BE DETERMINED

⇒ with ΔT = The change in temperature = T2-T1 = 36.4 °C - 24.9 °C = 11.5 °C

44.0 J = 10.6 grams * c * 11.5°C

c = 44.0 J / (10.6g * 11.5 °C)

c = 0.361 J/g°C

The specific heat of zinc is 0.361 J/g°C

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Suppose 6.63g of zinc bromide is dissolved in 100.mL of a 0.60 M aqueous solution of potassium carbonate. Calculate the final mo
Alenkasestr [34]

Answer:

[Zn²⁺] = 4.78x10⁻¹⁰M

Explanation:

Based on the reaction:

ZnBr₂(aq) + K₂CO₃(aq) → ZnCO₃(s) + 2KBr(aq)

The zinc added produce the insoluble ZnCO₃ with Ksp = 1.46x10⁻¹⁰:

1.46x10⁻¹⁰ = [Zn²⁺] [CO₃²⁻]

We can find the moles of ZnBr₂ added = Moles of Zn²⁺ and moles of K₂CO₃ = Moles of CO₃²⁻ to find the moles of CO₃²⁻ that remains in solution, thus:

<em>Moles ZnB₂ (Molar mass: 225.2g/mol) = Moles Zn²⁺:</em>

6.63g ZnBr₂ * (1mol / 225.2g) = 0.02944moles Zn²⁺

<em>Moles K₂CO₃ = Moles CO₃²⁻:</em>

0.100L * (0.60mol/L) = 0.060 moles CO₃²⁻

Moles CO₃²⁻ in excess: 0.0600moles CO₃²⁻ - 0.02944moles =

0.03056moles CO₃²⁻ / 0.100L = 0.3056M = [CO₃²⁻]

Replacing in Ksp expression:

1.46x10⁻¹⁰ = [Zn²⁺] [0.3056M]

<h3>[Zn²⁺] = 4.78x10⁻¹⁰M</h3>

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3 years ago
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