All categories

2 years ago

Why do electrons transition between energy levels within the atom, and how do we detect these transitions?

Chemistry

1 answer:

Sedaia [141]2 years ago

7 0

Answer:

See explanation

Explanation:

Electrons transition between energy levels in an atom due to gain or loss of energy. An electron may gain energy and move from its ground state to one of the accessible excited states. The electron quickly returns to ground state, emitting the energy previously absorbed as a photon of light. The wavelength of light emitted is measured using powerful spectrometers.

Atoms can be excited thermally or by irradiation with light of appropriate frequency.

You might be interested in

Copper(I) ions in aqueous solution react with NH 3 ( aq ) according to Cu + ( aq ) + 2 NH 3 ( aq ) ⟶ Cu ( NH 3 ) + 2 ( aq ) K f

Kruka [31]

Answer:

53.18 gL⁻¹

Explanation:

Given that:

$Cu^{2+}_{(aq)}$ $+$ $2NH_{3(aq)}$ $------>$ $[Cu(NH_3)_2]^+_{(aq)}$ ------equation (1)

where;

Formation Constant $(k_f) =$ $6.3*10^{10}$

However, the Dissociation of $CuBr_{(s)$ yields:

$CuBr_{(s)}$ ⇄ $Cu^{+}_{(aq)}$ $+$ $Br^-_{(aq)}$ -------------- equation (2)

where;

the Solubility Constant $(k_{sp})$ $= 6.3 *10^{-9$

From equation (1);

$(k_f) =$ $\frac{[[Cu(NH_3)_2]^+]}{[Cu^{2+}][NH_3]^{2}}$ --------- equation (3)

From equation (2)

$(k_{sp})$ $= [Cu^+][Br^-]$ --------- equation (4)

In $NH_3$ , the net reaction for $CuBr_{(s)$ can be illustrated as:

$CuBr_{(s)$ $+$ $2NH_{3(aq)}$ ⇄ $[Cu(NH_3)_2]^+_{(aq)}$ $+ Br^-_{(aq)}$

The equilibrium constant (K) can be written as :

$K=$ $\frac{[[Cu(NH_3)_2]^+][Br^-]}{[NH_3]^2}$

If we multiply both the numerator and the denominator with $[Cu^+]$ ; we have:

$K=$ $\frac{[[Cu(NH_3)_2]^+][Br^-]}{[NH_3]^2}*\frac{[Cu^+]}{[Cu^+]}$

$K=$ $\frac{[[Cu(NH_3)_2]^+}{[NH_3]^2[Cu^+]}*{[Cu^+][Br^-]}$

$K = k_f *k_{sp}$

$K= (6.3*10^{10})*(6.3*10^{-9})$

$K= 3.97*10^2$

$K$ ≅ $4.0*10^2$

Now; we can re-write our equilibrium constant again as:

$K=$ $\frac{[[Cu(NH_3)_2]^+][Br^-]}{[NH_3]^2}$

$4.0*10^2 = \frac{(x)(x)}{(0.76-2x)^2}$

$4.0*10^2 = \frac{(x)^2}{(0.76-2x)^2}$

$4.0*10^2 = (\frac{(x)}{(0.76-2x)})^2$

By finding the square of both sides, we have

$\sqrt {4.0*10^2} = \sqrt {(\frac{(x)}{(0.76-2x)})^2$

$2.0*10 = \frac{x}{(0.76-2x)}$

$20(0.76-2x) =x$

$15.2 -40x=x$

$15.2 = 40x +x$

$15.2 = 41x$

$x = \frac{15.2}{41}$

$x = 0.3707 M$

In gL⁻¹; the solubility of $CuBr_{(s)$ in 0.76 M $NH_3$ solution will be:

$= \frac{0.3707 mole of CuBr}{1L}*\frac{143.45 g}{mole of CuBr}$

= 53.18 gL⁻¹

4 0

2 years ago

How much heat energy, in kilojoules, is required to convert 72.0 gg of ice at −−18.0 ∘C∘C to water at 25.0 ∘C∘C ? Express your a

Ghella [55]

Answer: The enthalpy change is 34.3 kJ

Explanation:

The conversions involved in this process are :

$(1):H_2O(s)(-18^0C)\rightarrow H_2O(s)(0^0C)\\\\(2):H_2O(s)(0^0C)\rightarrow H_2O(l)(0^0C)\\\\(3):H_2O(l)(0^0C)\rightarrow H_2O(l)(25^0C)$

Now we have to calculate the enthalpy change.

$\Delta H=[m\times c_{s}\times (T_{final}-T_{initial})]+n\times \Delta H_{fusion}+[m\times c_{l}\times (T_{final}-T_{initial})]$

where,

$\Delta H$ = enthalpy change = ?

m = mass of water = 72.0 g

$c_{s}$ = specific heat of ice = $2.09J/g^0C$

$c_{l}$ = specific heat of liquid water = $4.184J/g^0C$

n = number of moles of water = $\frac{\text{Mass of water}}{\text{Molar mass of water}}=\frac{72.0g}{18g/mole}=4.00moles$

$\Delta H_{fusion}$ = enthalpy change for fusion = 6010 J/mole

Now put all the given values in the above expression, we get

$\Delta H=[72.0g\times 2.09J/g^0C\times (0-(-18)^0C]+4.00mole\times 6010J/mole+[72.0g\times 4.184J/g^)C\times (25-0)^0C]$ $\Delta H=34279.8J=34.3kJ$ (1 KJ = 1000 J)