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tankabanditka [31]
2 years ago
8

How to tell how many valence electrons an atom has?

Chemistry
1 answer:
lbvjy [14]2 years ago
6 0

Answer:  Count only the electrons in the highest shell s and p orbitals when determining valence.

Explanation:  I'm glad you asked this question.  It is often not well explained.   The term valence electrons is assigned to only the electrons in an element's highest energy level.  These reside only in the s and p orbitals, and not the d or f, as I'll explain later.  The s orbital can hold 2 electrons and the p can hold 6.  Potassium, K, has an s orbital in its highest energy shell, 4.  It contains only 1 electron, so it has a valency of 1.  

Calcium, Ca, has 2 in its highest energy level: 4s^2, so it has a valency of 2.   Moving to the right, the element scandium, Sc, add another electron, but it goes into the 3d orbital.  3d is in the 3rd energy shell, so it is not counted as a valence electron.  Only after we move further right, to gallium, Ga, do we start adding electrons to the 4th energy level again - the 4p orbitals can accept up to 6 electrons.  Ga has 3 valence electrons - 2 in the 4s and 1 in the 4p.  Oxygen has 2 in the 2s and 4 in the 4p orbitals, for a total of 6.  It is close to having a comple outer shell (2 in the 2s and 6 in the 2p).  Just 2 more electrons would fill both the 2s and 2p orbitals for a total of 8 valence electrons, a stable configuration (the same configuration as thje stable gas Neon).

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An electrode reaction is reduction. This means that the electrode is a(n) a. cathode. c. electrochemical cell. b. anode. d. elec
Romashka [77]
Oxidation happens at the anode and reduction happens at the cathode.<span />
5 0
2 years ago
In an electrically heated boiler, water is boiled at 140°C by a 90 cm long, 8 mm diameter horizontal heating element immersed in
RideAnS [48]

Explanation:

The given data is as follows.

Volume of water = 0.25 m^{3}

Density of water = 1000 kg/m^{3}

Therefore,  mass of water = Density × Volume

                       = 1000 kg/m^{3} \times 0.25 m^{3}

                       = 250 kg  

Initial Temperature of water (T_{1}) = 20^{o}C

Final temperature of water = 140^{o}C

Heat of vaporization of water (dH_{v}) at 140^{o}C  is 2133 kJ/kg

Specific heat capacity of water = 4.184 kJ/kg/K

As 25% of water got evaporated at its boiling point (140^{o}C) in 60 min.

Therefore, amount of water evaporated = 0.25 × 250 (kg) = 62.5 kg

Heat required to evaporate = Amount of water evapotaed × Heat of vaporization

                           = 62.5 (kg) × 2133 (kJ/kg)

                           = 133.3 \times 10^{3} kJ

All this heat was supplied in 60 min = 60(min)  × 60(sec/min) = 3600 sec

Therefore, heat supplied per unit time = Heat required/time = \frac{133.3 \times 10^{3}kJ}{3600 s} = 37 kJ/s or kW

The power rating of electric heating element is 37 kW.

Hence, heat required to raise the temperature from 20^{o}C to 140^{o}C of 250 kg of water = Mass of water × specific heat capacity × (140 - 20)

                      = 250 (kg) × 40184 (kJ/kg/K) × (140 - 20) (K)

                     = 125520 kJ  

Time required = Heat required / Power rating

                       = \frac{125520}{37}

                       = 3392 sec

Time required to raise the temperature from 20^{o}C to 140^{o}C of 0.25 m^{3} water is calculated as follows.

                    \frac{3392 sec}{60 sec/min}

                     = 56 min

Thus, we can conclude that the time required to raise the temperature is 56 min.

4 0
3 years ago
Chem quiz please help
GalinKa [24]

The theoretical and percentage yield for the reaction are:

  • The theoretical yield is 21 g
  • The percentage yield is 119%

<h3>Balanced equation </h3>

CH₄ + 2O₂ —> CO₂ + 2H₂O

Molar mass of CH₄ = 16 g/mol

Mass of CH₄ from the balanced equation = 1 × 16 = 16 g

Molar mass of O₂ = 32 g/mol

Mass of O₂ from the balanced equation = 2 × 32 = 64 g

Molar mass of CO₂ = 44 g/mol

Mass of CO₂ from the balanced equation = 1 × 44 = 44 g

SUMMARY

From the balanced equation above,

16 g of CH₄ reacted with 64 g of O₂ to produce 44 g of CO₂

<h3>How to determine the limiting reactant</h3>

From the balanced equation above,

16 g of CH₄ reacted with 64 g of O₂

Therefore,

20 g of CH₄ will react with = (20 × 64 ) / 16 = 80 g of O₂

From the above calculation, a higher mass (i.e 80 g) of O₂ than what was given (i.e 30 g) is needed to react completely with 20 g of CH₄.

Therefore, O₂ is the limiting reactant

<h3>How to determine the theoretical yield of CO₂</h3>

From the balanced equation above,

64 g of O₂ reacted to produce 44 g of CO₂

Therefore,

30g of O₂ will react to produce = (30 × 44) / 64 = 21 g of CO₂

<h3>How to determine the percentage yield </h3>
  • Actual yield of CO₂ = 25 g
  • Theoretical yield of CO₂ = 21 g
  • Percentage yield =?

Percentage yield = (Actual / Theoretical) × 100

Percentage yield = (25 / 21) ×100

Percentage yield = 119%

Learn more about stoichiometry:

brainly.com/question/14735801

#SPJ1

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Answer:

Explanation:

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On what two days would the UK, Chile, South Africa, and Japan have an equal amount of day and night? O Winter Solstice and Summe
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I say both equinox because winter solstice has the least amount of sunlight and the summer solstice has the most ampunt of daylight so i said both equinox
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2 years ago
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