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Hoochie [10]
3 years ago
7

What are two important quantum-mechanical concepts associated with the Bohr model of the atom?

Chemistry
1 answer:
malfutka [58]3 years ago
4 0

Explanation:

The important quantum-mechanical concepts associated with the Bohr model of atom are :

1. Electrons are nothing but particles that revolve around the nucleus in discrete orbitals.

2. Energy is associated with each orbital is quantised. Meaning electron in each shell will have energy in multiple of a fixed quanta.

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1A: Consider these compounds:
Setler [38]

Solution :

Compound                      Ksp

$PbF_2$                               $3.3 \times 10^{-8}$

$Ni(CN)_2$                        $3 \times 10^{-23}$

FeS                                $8 \times 10^{-19}$

$CaSO_4$                           $4.93 \times 10^{-5}$

$Mg(OH)_2$                      $5.61 \times 10^{-12}$

Ksp of $Ni(CN)_2 and both compounds dissociate the same way. Hence $Mg(OH)_2$ is more soluble than $(B). \  Ni(CN)_2$

$Mg(OH)_2$  is less soluble than $(A). \ \ PbF_2 \ ()Ksp \  PbF_2 > Ksp \ \text{ of } \ Mg(OH)_2$

It is not possible to determine CD - $FeS \text{ or} \ CaSO_4$  is more or less soluble than $Mg(OH)_2$  as though they have a different Ksp values their molecular dissociation is also different and they may have a close solubility values.

$Ni(OH)_2$  can be directly compared with PbS, $AlPO_4, MnS$

$\text{For } \ Ni(OH)_2$

$AB_2(s) \rightarrow A^{2+} + 2B^{-}$

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

100

1-s s 2s

Ksp = [A2+][B-]^2 = s \times (2s)^2 = 4s^3

Hence they can be directly compared by Ksp values, smaller the Ksp, smaller the solubility.

For Silver Chloride

$AB(s) \rightarrow A^{x+}+B^{x-}$

$AgCl(s) \rightarrow Ag^+ + Cl^-$

1 0 0

1 - s s s

Ksp $=[A^{x+}][B^{x-}]=s \times s = s^2$

Hence, they can be directly compared by Ksp values, smaller the Ksp, smaller the solubility.

4 0
3 years ago
The Sun is a yellow star that's both average in brightness and temperature and is classified as
Roman55 [17]

Answer:

Classifying stars according to their spectrum is a very powerful way to begin to understand how they work.  As we said last time, the spectral sequence O, B, A, F, G, K, M is a temperature sequence, with the hottest stars being of type O (surface temperatures 30,000-40,000 K), and the coolest stars being of type M (surface temperatures around 3,000 K).  Because hot stars are blue, and cool stars are red, the temperature sequence is also a color sequence.  It is sometimes helpful, though, to classify objects according to two different properties.  Let's say we try to classify stars according to their apparent brightness, also.  We could make a plot with color on one axis, and apparent brightness on the other axis, like this:

Explanation:

7 0
3 years ago
The vapor pressure of ethanol is 30°C at 98.5 mmHg and the heat of vaporization is 39.3 kJ/mol. Determine the normal boiling poi
Gelneren [198K]

Answer : The normal boiling point of ethanol will be, 348.67K or 75.67^oC

Explanation :

The Clausius- Clapeyron equation is :

\ln (\frac{P_2}{P_1})=\frac{\Delta H_{vap}}{R}\times (\frac{1}{T_1}-\frac{1}{T_2})

where,

P_1 = vapor pressure of ethanol at 30^oC = 98.5 mmHg

P_2 = vapor pressure of ethanol at normal boiling point = 1 atm = 760 mmHg

T_1 = temperature of ethanol = 30^oC=273+30=303K

T_2 = normal boiling point of ethanol = ?

\Delta H_{vap} = heat of vaporization = 39.3 kJ/mole = 39300 J/mole

R = universal constant = 8.314 J/K.mole

Now put all the given values in the above formula, we get:

\ln (\frac{760mmHg}{98.5mmHg})=\frac{39300J/mole}{8.314J/K.mole}\times (\frac{1}{303K}-\frac{1}{T_2})

T_2=348.67K=348.67-273=75.67^oC

Hence, the normal boiling point of ethanol will be, 348.67K or 75.67^oC

3 0
3 years ago
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Vilka [71]
The period is the end of the sentence!!!
6 0
2 years ago
1. Of 100 animals on a farm, 20% are horses, 15 % are pigs, 30% are cows, and 35% are
ANEK [815]
B. a circle graph

circle graphs are the best to show percentages because they’re very easy to look at and get info from
7 0
3 years ago
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