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3 years ago

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If I leave 750 mL of 0.75 M sodium chloride solution uncovered on a windowsill and 150 mL of the solvent evaporates, what will t

he new concentration of the sodium chloride solution be? M1V1 = M2V2

1 answer:

MAVERICK [17]3 years ago

8 0

M1V1=M2V2

V2=750-150=600 ml

0.75M*750 ml = M2*600

M2=0.75*750/600 ≈ 9.38 M

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Answer:

1.64x10⁻¹⁸ J

Explanation:

By the Bohr model, the electrons surround the nucleus of the atom in shells or levels of energy. Each one has it's energy, and the electron doesn't fall to the nucleus because it can reach another level of energy, and then return to its level.

When the electrons go to another level, it absorbs energy, and then, when return, this energy is released, as a photon (generally as luminous energy). The value of the energy can be calculated by:

E = hc/λ

Where h is the Planck constant (6.626x10⁻³⁴ J.s), c is the light speed (3.00x10⁸ m/s), and λ is the wavelength of the photon.

The wavelength can be calculated by:

1/λ = R*(1/nf² - 1/ni²)

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1/λ = 1.097x10⁷ *(1/1² - 1/2²)

1/λ = 8.227x10⁶

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E = 1.64x10⁻¹⁸ J

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An ionic bond forms when atoms blank electrons

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Answer:

An ionic bond forms when atoms transfer electrons.

Explanation:

Ionic bonds are formed when atoms transfer electrons. (In contrast, covalent bonds are formed when atoms share electrons.)

There's a distinction between the two: when two atoms react to form an ionic bond, one atom would completely lose one electron, while the other would completely gain that electron. The atom that loses the electron becomes a positively-charged ion called a cation, whereas the atom that gains the electron becomes a negatively-charged ion called an anion.

For example, consider the reaction between a sodium $\rm Na$ atom and a chlorine $\rm Cl$ atom: $\rm Na + Cl \to NaCl$ .

When the sodium atom and the chlorine atom encounter, the sodium atom would lose one electron to form a positively-charged sodium ion, $\rm Na^{+}$ . The chlorine atom would gain that electron to form a negatively-charged chlorine ion $\rm Cl^{-}$ .

These two ions will readily attract each other because of the opposite electrostatic charges on them. This electrostatic attraction (between two ions of opposite charges) is an ionic bond.

Overall, it would appear as if the sodium $\rm Na$ atom transferred an electron to the chlorine $\rm Cl$ atom to form an ionic bond.

In contrast, when two atoms react to form a covalent bond, they share electrons without giving any away completely. Therefore, it is possible to break certain covalent bonds apart (using a beam of laser, for example) and obtain neutral atoms.

On the other hand, when an ionic bond was broken, the result would be two charged ions- not necessarily two neutral atoms. The electron transfer could not be reversed by simply breaking the bond.

For example, when table salt $\rm NaCl$ is melted (at a very high temperature,) the ionic bond between the sodium ions and chloride ions would (mostly) be broken. However, doing so would only generate a mixture of $\rm Na^{+}$ and $\rm Cl^{-}$ ions- not sodium and chlorine atoms.

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2 years ago

Consider the following mechanism for the oxidation of bromide ions by hydrogen peroxide in aqueous acid solution. H+ + H2O2 ? H3

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<u>Answer:</u> The rate law for the reaction is $\text{Rate}=k'[H+][H_2O_2][Br^-]$

<u>Explanation:</u>

Rate law is the expression which is used to express the rate of the reaction in terms of the molar concentration of reactants where each term is raised to the power their stoichiometric coefficient respectively from a balanced chemical equation.

In a mechanism of the reaction, the slow step in the mechanism determines the rate of the reaction.

The chemical equation for the oxidation of bromide ions by hydrogen peroxide in aqueous acid solution follows:

$2H^++2Br^-+H_2O_2\rightarrow Br_2+2H_2O$

The intermediate reaction of the mechanism follows:

<u>Step 1:</u> $H^++H_2O_2\rightleftharpoons H_3O_2^+;\text{ (fast)}$

<u>Step 2:</u> $H_3O_2^++Br^-\rightarrow HOBr+H_2O;\text{(slow)}$

<u>Step 3:</u> $HOBr+H^++Br^-\rightarrow Br_2+H_2O;\text{(fast)}$

As, step 2 is the slow step. It is the rate determining step

Rate law for the reaction follows:

$\text{Rate}=k[H_3O_2^+][Br^-]$ ......(1)

As, $[H_3O_2^+]$ is not appearing as a reactant in the overall reaction. So, we apply steady state approximation in it.

Applying steady state approximation for $[H_3O_2^+]$ from step 1, we get:

$K=\frac{[H_3O_2^+]}{[H^+][H_2O_2]}$

$[H_3O_2^+]=K[H^+][H_2O_2]$

Putting the value of $[H_3O_2^+]$ in equation 1, we get:

$\text{Rate}=k.K[H^+][H_2O_2][Br^-]\\\\\text{Rate}=k'[H+][H_2O_2][Br^-]$

Hence, the rate law for the reaction is $\text{Rate}=k'[H+][H_2O_2][Br^-]$

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