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3 years ago

Valence electron of the first 20 elements

Physics

2 answers:

Dima020 [189]3 years ago

4 0

Answer:

Hydrogen

1 valence electron

Helium

2 valence electrons

lithium

1 valence electrons

beryllium

2 valence electrons

boron

3 valence electrons

carbon

4 valence electrons

nitrogen

5 valence electrons

oxygen

6 valence electrons

flourine

7 valence electrons

neon

8 valence electrons

sodium

1 valence electron

magnesium

2 valence electrons

aluminum

3 valence electrons

silicon

4 valence electrons

phosphorus

5 valence electrons

bixtya [17]3 years ago

4 0

Answer: 17 Chlorine -1, +1, (+2), +3, (+4), +5, +7

18 Argon 0

19 Potassium +1

20 Calcium +2

Explanation:

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Answer and Explanation:

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Use the information and diagram to answer the following question.

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ome metal oxides can be decomposed to the metal and oxygen under reasonable conditions. 2 Ag2O(s) → 4 Ag(s) + O2(g) Thermodynami

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Answer : The values of $\Delta H^o,\Delta S^o\text{ and }\Delta G^o$ are $62.2kJ,132.67J/K\text{ and }22.66kJ$ respectively.

Explanation :

The given balanced chemical reaction is,

$2Ag_2O(s)\rightarrow 4Ag(s)+O_2(g)$

First we have to calculate the enthalpy of reaction $(\Delta H^o)$ .

$\Delta H^o=H_f_{product}-H_f_{product}$

$\Delta H^o=[n_{Ag}\times \Delta H_f^0_{(Ag)}+n_{O_2}\times \Delta H_f^0_{(O_2)}]-[n_{Ag_2O}\times \Delta H_f^0_{(Ag_2O)}]$

where,

$\Delta H^o$ = enthalpy of reaction = ?

n = number of moles

$\Delta H_f^0$ = standard enthalpy of formation

Now put all the given values in this expression, we get:

$\Delta H^o=[4mole\times (0kJ/mol)+1mole\times (0kJ/mol)}]-[2mole\times (-31.1kJ/mol)]$

$\Delta H^o=62.2kJ=62200J$

conversion used : (1 kJ = 1000 J)

Now we have to calculate the entropy of reaction $(\Delta S^o)$ .

$\Delta S^o=S_f_{product}-S_f_{product}$

$\Delta S^o=[n_{Ag}\times \Delta S_f^0_{(Ag)}+n_{O_2}\times \Delta S_f^0_{(O_2)}]-[n_{Ag_2O}\times \Delta S_f^0_{(Ag_2O)}]$

where,

$\Delta S^o$ = entropy of reaction = ?

n = number of moles

$\Delta S_f^0$ = standard entropy of formation

Now put all the given values in this expression, we get:

$\Delta S^o=[4mole\times (42.55J/K.mole)+1mole\times (205.07J/K.mole)}]-[2mole\times (121.3J/K.mole)]$

$\Delta S^o=132.67J/K$

Now we have to calculate the Gibbs free energy of reaction $(\Delta G^o)$ .

As we know that,

$\Delta G^o=\Delta H^o-T\Delta S^o$

At room temperature, the temperature is 298 K.

$\Delta G^o=(62200J)-(298K\times 132.67J/K)$

$\Delta G^o=22664.34J=22.66kJ$

Therefore, the values of $\Delta H^o,\Delta S^o\text{ and }\Delta G^o$ are $62.2kJ,132.67J/K\text{ and }22.66kJ$ respectively.

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3 years ago

What may have been the first step in the formation of the solar system?

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Milky way galaxy / stars

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3 years ago

Your friend, who is in a field 100 meters away from you, kicks a ball towards you with an initial velocity of 16 m/s. Assuming t

LekaFEV [45]

Answer:

Time, t = 5.355 seconds

Explanation:

Given the following data;

Distance = 100 m

Initial velocity = 16 m/s

Deceleration = 1 m/s²

To find the time, we would use the second equation of motion;

But since the ball is decelerating, it's acceleration would be negative.

S = ut + ½at²

Where;

S represents the displacement or height measured in meters.

u represents the initial velocity measured in meters per seconds.

t represents the time measured in seconds.

a represents acceleration measured in meters per seconds square.

Substituting into the equation, we have;

100 = 16t - 0.5t²

200 = 32t - t²

t² + 32t - 200 = 0

Solving the quadratic equation using the quadratic formula;

The quadratic equation formula is;

$x = \frac {-b \; \pm \sqrt {b^{2} - 4ac}}{2a}$

Substituting into the equation, we have;

$x = \frac {-32 \; \pm \sqrt {32^{2} - 4*1*(-200)}}{2*1}$

$x = \frac {-32\pm \sqrt {1024 - (-800)}}{2}$

$x = \frac {-32 \pm \sqrt {1024 + 800}}{2}$

$x = \frac {-32 \pm \sqrt {1824}}{2}$

$x = \frac {-32 \pm 42.71}{2}$

$x_{1} = \frac {-32 + 42.71}{2}$

$x_{1} = \frac {10.71}{2}$

x1 = 5.355

We do not need the negative value of x, so we proceed.

Therefore, time = 5.355 seconds

3 0

3 years ago

Valence electron of the first 20 elements​

Valence electron of the first 20 elements