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2 years ago

Identify the type of each reaction in the copper cycle.

Chemistry

1 answer:

ra1l [238]2 years ago

3 0

Answer:

CuSO4 (aq) + Zn ( s ) ⟶ ZnSO4 (aq) + Cu (s): single displacement.

Cu(NO3)2(aq) + 2NaOH(aq) ⟶ Cu(OH)2(s) + 2NaNO3 (aq): Double displacement.

CuO(s) + H2SO4 (aq) ⟶ CuSO4 (aq) + H2O( l ): Double displacement.

Cu(s) + 4HNO3 (aq) ⟶ Cu(NO3)2 (aq) + 2NO2(aq) + 2H2O(l): Redox

Cu(OH)2 (s) Δ −→ CuO (s) + H2O(g): Decomposition.

Explanation:

Hello!

In this case, for the given chemical reactions, we obtain:

CuSO4 (aq) + Zn ( s ) ⟶ ZnSO4 (aq) + Cu (s): single displacement as the zinc metal is displacing copper in copper (II) sulfate.

Cu(NO3)2(aq) + 2NaOH(aq) ⟶ Cu(OH)2(s) + 2NaNO3 (aq): Double displacement as sodium is getting with nitrate and copper (II) with hydroxide.

CuO(s) + H2SO4 (aq) ⟶ CuSO4 (aq) + H2O( l ): Double displacement as the oxygen in the cupper (II) oxide turns out with the hydrogen and as a result, copper (II) sulfate is formed.

Cu(s) + 4HNO3 (aq) ⟶ Cu(NO3)2 (aq) + 2NO2(aq) + 2H2O(l): Redox since the copper joins NO3 but at the same time NO2 is released (Nitrogen goes from 5+ to 4+).

Cu(OH)2 (s) Δ −→ CuO (s) + H2O(g): Decomposition as copper (II) hydroxide is separated into copper (II) oxide and water.

Regards!

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2 years ago

A certain radioactive isotope decays at a rate of 0.2% annually. Determine the half-life of this isotope, to the nearest year

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The half-life of the radioactive isotope is 346 years.

Explanation:

The decay rate of the isotope is modelled after the following first-order linear ordinary differential equation:

$\frac{dm}{dt} = -\frac{m}{\tau}$

Where:

$m$ - Current isotope mass, measured in kilograms.

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$\tau$ - Time constant, measured in years.

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Where $m_{o}$ is the initial mass of the isotope. It is known that radioactive isotope decays at a yearly rate of 0.2 % annually, then, the following relationship is obtained:

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$1 - \frac{m(t+1)}{m(t)} = 0.002$

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$1 - e^{-\frac{1}{\tau} } = 0.002$

$e^{-\frac{1}{\tau} } = 0.998$

$-\frac{1}{\tau} = \ln 0.998$

The time constant associated to the decay is:

$\tau = -\frac{1}{\ln 0.998}$

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$t_{1/2} = \tau \cdot \ln 2$

If $\tau \approx 499.500\,years$ , the half-life of the isotope is:

$t_{1/2} = (499.500\,years)\cdot \ln 2$

$t_{1/2}\approx 346.227\,years$

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