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2 years ago

In KCI how are the valence electrons distributed

Chemistry

1 answer:

-BARSIC- [3]2 years ago

3 0

Answer:

Explanation:

In KCL, the elements to combine to form KCl are K and Cl.

Potassium (K) is an element of group 1 and they're known to have 1 valence electron in its outer most orbital. This electron is donated and used for bonding. It is every element desirability to attain octet or duet electronic configuration I.e by having 2 electrons or 8 electrons in its outer orbital. Potassium has a lower electro negativity and higher ionization energy, hence it is easier to donate electron than to recieve it. Meanwhile, Chlorine (Cl) is an element of group 17 which are characterized for having 7 electrons in its outermost orbital requiring just one electron to form an octet configuration, chlorine has a higher electro negativity and lower ionization energy hence it is easier to accept electron than to donate it.

The nature of bond formed from K and Cl to give KCl is called electrovalent bonds.

Equation of reaction

K + Cl → KCl

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Why does graphite and diamond sublime?

GREYUIT [131]

Answer:

The intermolecular bonding in the solid is van der Waals dispersion forces. That means that it will have a lower melting point than diamond or graphite where you have giant covalent bonding. In fact C60 sublimes at 800 K (527°C). Diamond and graphite sublime at around 4000 K.

Explanation:

8 0

2 years ago

On a cool morning (12"C), a balloon is filled with 1.5 L of helium. By mid afternoon, the temperature has soared to 32°C. What i

ddd [48]

Answer:

1.6 L

Explanation:

Using Charle's law

$\frac {V_1}{T_1}=\frac {V_2}{T_2}$

Given ,

V₁ = 1.5 L

V₂ = ?

T₁ = 12 °C

T₂ = 32 °C

The conversion of T( °C) to T(K) is shown below:

T(K) = T( °C) + 273.15

So,

T₁ = (12 + 273.15) K = 285.15 K

T₂ = (32 + 273.15) K = 305.15 K

Using above equation as:

$\frac{1.5}{285.15}=\frac{V_2}{305.15}$

$V_2=\frac{1.5\cdot \:305.15}{285.15}$

New volume = 1.6 L

8 0

2 years ago

Which of the following is not a nanoscale object?

KiRa [710]

If I am correct I would believe that it would be a muscle cell.

6 0

2 years ago

Read 2 more answers

Phosphorous acid, h3po3(aq), is a diprotic oxyacid that is an important compound in industry and agriculture. calculate the ph f

Varvara68 [4.7K]

Answer:

Explanation:

(a)

Before the addition of KOH :-

Given pKa1 of H3PO3 = 1.30

we know , pKa1 = - log10Ka1

Ka1 = 10-pKa1

Ka1 = 10-1.30

Ka1 = 0.0501

similarly pKa2 = 6.70 ,therefore Ka2 = 1.99 x 10-7

because Ka1 >> Ka2 , therefore pH of diprotic acid i.e H3PO3 can be calculated from first dissociation only .

ICE table is :-

H3PO3 (aq) <-------------> H+ (aq) + H2PO3-(aq)

I 2.4 M 0 M 0 M

C - x + x + x

E (2.4 - x )M x M x M

x = degree of dissociation

Now expression of Ka1 is :

Ka1 = [ H+ ] [ H2PO3-] / [ H3PO3]

0.0501 = x2 / 2.4 - x

on solving for x by using quadratic formula , we have

x = 0.32

Now [ H+ ] = [ H2PO3-] = 0.32 M

pH = - log [H+]

pH = - log 0.32

pH = - ( - 0.495)

pH = 0.495

Hence pH before the addition of KOH = 0.495

(b)

After the addition of 25.0 mL of 2.4 M KOH :-

Number of moles of KOH = 2.4 M x 0.025 L = 0.06 mol

Number of moles of H3PO3 = 2.4 M x 0.050 L = 0.12 mol

Now 0.06 moles of KOH is equal to the half of the moles required for the first equivalent point . therefore pH at this point is equal to pKa1 .

Hence pH = 1.30 M

(c)

After the addition of 50.0 mL of 2.4 M KOH :-

Number of moles of KOH = 2.4 M x 0.050 L = 0.12 mol

Number of moles of H3PO3 = 2.4 M x 0.050 L = 0.12 mol

because Number of moles of H3PO4 = Number of moles of KOH

therefore , this point is the first equivalence point

and pH = pKa1 + pKa2 / 2

pH = 1.30 + 6.70 / 2

pH = 4.00

Hence pH = 4.00

(d)

After the addition of 75.0 mL of 2.4 M KOH :-

Number of moles of KOH = 2.4 M x 0.075 L = 0.18 mol

Number of moles of H3PO3 = 2.4 M x 0.050 L = 0.12 mol

This is the half way of the second equivalence point , therefore pH is equal to pKa2 .

Hence pH = 6.70

5 0

2 years ago

In the PhET simulation, make sure that the checkbox Stable/Unstable in the bottom right is checked. Using the PhET simulation as

lord [1]

Answer:

kindly check the EXPLANATION SECTION

Explanation:

In order to be able to answer this question one has to consider the neutron proton ratio. Considering this ratio will allow us to determine the stability of a nuclei. The most important rule that helps us in determination of stability is that when the Neutron- Proton ratio of any nuclei ranges from to 1 to 1.5, then we say the nuclei is STABLE.

Also, we need to understand that when the Neutron- Proton ratio is LESS THAN 1 or GREATER THYAN 1.5, then we say the nuclei is UNSTABLE.

So, let us check which is stable and which is unstable:

a. 4 protons and 5 neutrons = Neutron- proton ratio = N/P = 5/4= stable.

b. 7 protons and 7 neutrons = Neutron- proton ratio = N/P = 7/7= 1 = stable.

c. 2 protons and 3 neutrons = Neutron- proton ratio = N/P = 3/5 =0.6 =unstable.

d. 3 protons and 0 neutrons = Neutron- proton ratio = N/P = 0/3= 0= unstable.

e. 6 protons and 5 neutrons = Neutron- proton ratio = N/P = 5/6= 0.83 = unstable.

f. 9 protons and 9 neutrons = Neutron- proton ratio = N/P = 9/9 = 1 = stable.

g. 8 protons and 7 neutrons = Neutron- proton ratio = N/P = 7/8 =0.875 = unstable.

h. 1 proton and 0 neutrons = Neutron- proton ratio = N/P = 0/1 =0 = unstable

6 0

2 years ago