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1 year ago

Chlorine and bromine have very similar chemical properties. This is best explained by the fact that both elements:

Chemistry

1 answer:

34kurt1 year ago

7 0

What is the similarity between chlorine and bromine?

Chlorine and bromine have very similar chemical properties and are best explained by the fact that they both belong to the category of halogens.

Similarities shared by halogens:

Chemical elements are represented as periods and groups on a periodic table. Chemical elements with comparable properties make up each category. Halogens are group 17 elements that are primarily non-metallic.

In the case of halogens, each element's atom possesses seven valence electrons, $ns^{2} np^{5}$ , in which n stands for the energy level. The electron configuration of chlorine is $[Ne]3s^{2} 3p^{5}$ , and the electron configuration of bromine is $[Ar]3d^{10}4s^{2} 4p^{5}$ .

Because of comparable electron configurations, elements of the same group have similar properties. The number of valence electrons, which are located in the outermost (highest energy) s and p orbitals, is particularly similar.

Learn more about halogens here;

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Now the mass of the 10 mL of water is 10 g as density of water is 10 g/mL. Thus after addition the total mass of the solution is (39 + 10) = 49g and the volume is (30 + 10) = 40 mL. Thus the density of the mixture will be $\frac{49}{40}=1.225$ g/mL. Thus the specific gravity of the mixed sample will be 1.225 g/mL.

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3 years ago

Read 2 more answers

The molar solubility of pbi2 is 1.5 103 m.

Vsevolod [243]

Answer: -

Concentration of PbI₂ = 1.5 x 10⁻³ M

PbI₂ dissociates in water as

PbI₂ ⇄ Pb²⁺ + 2 I⁻

So PbI₂ releases two times the amount of I⁻ as it's own concentration when saturated.

Thus the molar concentration of iodide ion in a saturated PbI₂ solution = [ I⁻] =

= 1.5 x 10⁻³ x 2 M

= 3 x 10⁻³ M

PbI₂ releases the same amount of Pb²⁺ as it's own concentration when saturated.

[Pb²⁺] = 1.5 x 10⁻³ M

So solubility product for PbI₂

Ksp = [Pb²⁺] x [ I⁻]²

=1.5 x 10⁻³ x (3 x 10⁻³)²

= 4.5 x 10⁻⁹

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2 years ago

Nitrogen dioxide decomposes to nitric oxide and oxygen via the reaction: 2NO2 → 2NO + O2 In a particular experiment at 300 °C, [

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Answer:

$rate=-1.75x10^{-5}\frac{M}{s}$

Explanation:

Hello,

In this case, for the given information, we can compute the rate of disappearance of NO₂ by using the following rate relationship:

$rate=\frac{1}{2}*\frac{C_f-C_0}{t_f-t_0}$

Whereas it is multiplied by the the inverse of the stoichiometric coefficient of NO₂ in the reaction that is 2. Moreover, the subscript <em>f</em> is referred to the final condition and the subscript <em>0</em> to the initial condition, thus, we obtain:

$rate=\frac{1}{2}*\frac{0.00650M-0.0100M}{100s-0s}\\\\rate=-1.75x10^{-5}\frac{M}{s}$

Clearly, it turns out negative since the concentration is diminishing due to its consumption.

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