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2 years ago

Electron configuration for Be2+

Chemistry

1 answer:

rusak2 [61]2 years ago

7 0

Answer:

Beryllium has an electronic configuration of 1s2 2s2 in its natural state. When in an ion state of Be2+ it loses the electrons in a 2s shell and has a configuration of 1s2. This means that Be2+ has 2 electrons.

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c) The complete equation for the reaction that occurs when copper sulfate and potassium hydroxide are mixed:

$CuSO_4(aq)+2KOH (aq)\rightarrow Cu(OH)_2(s)+K_2SO_4 (aq)$

d) $CuSO_4(aq)\rightarrow Cu^{2+}(aq) +SO_{4}^{2-}(aq)$ ..[1]

$KOH (aq)\rightarrow 2K^+(aq) +OH^-(aq)$ ..[2]

$Cu^{2+}(aq) +SO_{4}^{2-}(aq)+2K^+(aq) +2OH^-(aq)\rightarrow Cu(OH)_2(s)+SO_{4}^{2-}(aq)+2K^+(aq)$

Common ion both sides are removed. The net ionic equation is given as:

$Cu^{2+}(aq) +2OH^-(aq)\rightarrow Cu(OH)_2(s)(aq)$

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Mass of final solution ,m= 1 mL

Density of solution ,d= 1 g/mL (same as pure water)

$m=d\time V=1 g/ml\times 100 mL = 100 g$

Heat capacity of the solution = c = 4.186 J/g°C (same as pure water)

Change in temperature of the solution,ΔT = 27.7 °C- 21.5 °C=6.2°C

$Q=mc\Delta T$

$Q=100 g\times 4.186 J/g ^oC\times 6.2^oC=2595.32 J=2.595 kJ$

Enthalpy of the reaction = ΔH = $\frac{Q}{\text{Moles of copper}}$

ΔH = $\frac{2.595 kJ}{0.050 mol}=51.90 kJ/mol$

The ΔH for the reaction that occurs on mixing is 51.90 kJ/mol.

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