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andrezito [222]
3 years ago
14

What causes an ionic bond?a. two ions share electrons.b. two atoms share electrons.c. a positive ion is attracted to a negative

ion.d. a positive ion is attracted to a positive ion?
Chemistry
1 answer:
3241004551 [841]3 years ago
4 0

Answer:

option C= A positive ion is attracted to negative ion

Explanation:

Ionic bond:

Ionic bond is formed when oppositely charged ions attract each others i.e negative and positive ions.

we know that a neutral atom consist of equal number of proton and electron, cancel the charge of each other that is equal in magnitude and make the atom neutral or we can say that net charge is zero.

But when the atom lose or gain the electron, imbalance of neutron electron occur so charge will not remain zero and atom is no more to be said neutral.

For example:

Take the example of sodium that can lose one electron and form Na+. This Na+ ion is called cation because it carry positive charge by losing the one electron. This cation now contain eleven proton and ten electron.

Now consider the example of Chlorine atom. It consist of 17 proton and 17 electron. But in order to attain the noble gas configuration or to complete the octet it gain one electron and form anion (Cl-). The chlorine anion consist of 17 proton and 18 electron.

Now consider the sodium chloride which is an ionic compound. It is formed by the complete transfer of electron from sodium to chlorine atom and form ionic bond. In this ionic compound sodium carry positive charge and chlorine carry negative charge there is attraction between these oppositely charged atoms.

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Let suppose the Gas is acting Ideally, Then According to Ideal Gas Equation,

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Data Given;

Moles  = n = 1.20 mol

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                     P  =  (1.20 mol × 0.08206 atm.L.mol⁻¹.K⁻¹ × 303 K) ÷ 4 L

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6 0
3 years ago
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Iron and vanadium both have the BCC crystal structure and V forms a substitutional solid solution in Fe for concentrations up to
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Answer:

Explanation:

To find the concentration; let's first compute the average density and the average atomic weight.

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The average atomic weight is:

A_{avg} = \dfrac{100}{ \dfrac{C_{Fe} }{A_{Fe}} + \dfrac{C_v}{A_v} }

So; in terms of vanadium, the Concentration of iron is:

C_{Fe} = 100 - C_v

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SO; replacing V_c with a^3 ; \rho_{avg} with \dfrac{100}{ \dfrac{C_{Fe} }{\rho_{Fe}} + \dfrac{C_v}{\rho_v} } ; A_{avg} with \dfrac{100}{ \dfrac{C_{Fe} }{A_{Fe}} + \dfrac{C_v}{A_v} } and

C_{Fe} with 100-C_v

Then:

a^3 = \dfrac   { n \Big (\dfrac{100}{[(100-C_v)/A_{Fe} ] + [C_v/A_v]} \Big) }    {N_A\Big (\dfrac{100}{[(100-C_v)/\rho_{Fe} ] + [C_v/\rho_v]} \Big)  }

a^3 = \dfrac   { n \Big (\dfrac{100 \times A_{Fe} \times A_v}{[(100-C_v)A_{v} ] + [C_v/A_Fe]} \Big) }    {N_A  \Big (\dfrac{100 \times \rho_{Fe} \times  \rho_v }{[(100-C_v)/\rho_{v} ] + [C_v \rho_{Fe}]} \Big)  }

a^3 = \dfrac   { n \Big (\dfrac{100 \times A_{Fe} \times A_v}{[(100A_{v}-C_vA_{v}) ] + [C_vA_Fe]} \Big) }    {N_A  \Big (\dfrac{100 \times \rho_{Fe} \times  \rho_v }{[(100\rho_{v} - C_v \rho_{v}) ] + [C_v \rho_{Fe}]} \Big)  }

Replacing the values; we have:

(0.289 \times 10^{-7} \ cm)^3 = \dfrac{2 \ atoms/unit \ cell}{6.023 \times 10^{23}} \dfrac{ \dfrac{100 (50.94 \g/mol) (55.84(g/mol)} { 100(50.94 \ g/mol) - C_v(50.94 \ g/mol) + C_v (55.84 \ g/mol)   }   }{ \dfrac{100 (7.84 \ g/cm^3) (6.0 \ g/cm^3 } { 100(6.0 \ g/cm^3) - C_v(6.0 \ g/cm^3) + C_v (7.84 \ g/cm^3)   } }

2.41 \times 10^{-23} = \dfrac{2}{6.023 \times 10^{23} }  \dfrac{ \dfrac{100 *50*55.84}{100*50.94 -50.94 C_v +55.84 C_v} }{\dfrac{100 * 7.84 *6}{600-6C_v +7.84 C_v} }

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Answer:

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Explanation:

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We can see that all the S orbitals have been completely filled with two electrons alike each. This is also the case for the P orbital as the 3 suborbitals take in 2 each to give a total of six

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Answer:

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