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3 years ago

A phosphorus atom and a chlorine atom are likely to form a covalent bond because both elements:

Chemistry

2 answers:

quester [9]3 years ago

5 0

Answer: Option (a) is the correct answer.

Explanation:

When there occurs sharing of electrons between the two combining atoms then the bond formed is known as a covalent bond. Whereas when there occurs transfer of electrons between the combining atoms then the bond formed is known as ionic bond.

An ionic bond is formed between a metal and a non-metal.

Both phosphorous and chlorine belongs to third period and both of them are non-metals.

Atomic number of P is 15 and its electronic distribution is 2, 8, 5. Atomic number of Cl is 17 and its electronic distribution is 2, 8, 7.

Therefore, in order to attain stability both P and Cl will combine with each other by sharing electrons and hence, they will form a covalent bond.

Thus, we can conclude that a phosphorus atom and a chlorine atom are likely to form a covalent bond because both elements are non-metals.

Jobisdone [24]3 years ago

4 0

A phosphorus atom and a chlorine atom are likely to form a covalent bond because both elements are nonmetals.

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3 years ago

Which statement is correct about molarity and volume?

Oksana_A [137]

Answer:

Explanation:

If the choices are:

A. Molarity is defined as moles of solute per liter of solvent.

B. % by mass is defined as grams of solute per 100 g of solvent.

C. % by volume is defined as grams of solute per 100 L of solution.

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3 years ago

Calculate the electric double layer thickness of a alumina colloid in a dilute (0.1 mol/dm3) CsCI electrolyte solution at 30 °C.

Ad libitum [116K]

Explanation:

The given data is as follows.

Concentration = 0.1 $mol/dm^{3}$

= 0.1 \frac{mol dm^{3}}{dm^{3}} \frac{10^{3}}{dm^{3}} \times \frac{6.022 \times 10^{23}}{1 mol} ions

= $6.022 \times 10^{25} ions/m^{3}$

T = $30^{o}C$ = (30 + 273) K = 303 K

Formula for electric double layer thickness ( $\lambda_{D}$ ) is as follows.

$\lambda_{D}$ = $\frac{1}{k} = \sqrt \frac{\varepsilon \varepsilon_{o} K_{g}T}{2 n^{o} z^{2} \varepsilon^{2}}$

where, $n^{o}$ = concentration = $6.022 \times 10^{25} ions/m^{3}$

Hence, putting the given values into the above equation as follows.

$\lambda_{D}$ = $\sqrt \frac{\varepsilon \varepsilon_{o} K_{g}T}{2 n^{o} z^{2} \varepsilon^{2}}$

= $\sqrt \frac{78 \times 8.854 \times 10^{-12} c^{2}/Jm \times 1.38 \times 10^{-23}J/K \times 303 K}{2 \times 6.022 \times 10^{25} ions/m^{3} \times (1)^{2} \times (1.6 \times 10^{-19}C)^{2}}$

= $9.669 \times 10^{-10}$ m

or, = $9.7 A^{o}$

= 1 nm (approx)

Also, it is known that $\lambda_{D}$ = $\sqrt \frac{1}{n^{o}}$

Hence, we can conclude that addition of 0.1 $mol/dm^{3}$ of KCl in 0.1 $mol/dm^{3}$ of NaBr " $\lambda_{D}$ " will decrease but not significantly.

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3 years ago

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Fofino [41]

A physical property does not change the substance.

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3 years ago

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Triss [41]

Answer:

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2 years ago