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almond37 [142]
2 years ago
7

The rate law of the reaction NH3 + HOCl → NH2Cl + H2O is rate = k[NH3][HOCl] with k = 5.1 × 106 L/mol·s at 25°C. The reaction is

made pseudo-first-order in NH3 by using a large excess of HOCl. How long will it take for 40% of the NH3 to react if the initial concentration of HOCl is 2 × 10−3 M?
Chemistry
1 answer:
SIZIF [17.4K]2 years ago
6 0

Answer:

40% of the ammonia will take 4.97x10^-5 s to react.

Explanation:

The rate is equal to:

R = k*[NH3]*[HOCl] = 5.1x10^6 * [NH3] * 2x10^-3 = 10200 s^-1 * [NH3]

R = k´ * [NH3]

k´ = 10200 s^-1

Because k´ is the psuedo first-order rate constant, we have the following:

b/(b-x) = 100/(100-40) ; 40% ammonia reacts

b/(b-x) = 1.67

log(b/(b-x)) = log(1.67)

log(b/(b-x)) = 0.22

the time will equal to:

t = (2.303/k) * log(b/(b-x)) = (2.303/10200) * (0.22) = 4.97x10^-5 s

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The Principle of Le Chatelier states that if a system in equilibrium is subjected to a disturbance, the system will react in such a way that it will diminish the effect of that disturbance. Thus, when the concentration of one of the substances in an equilibrium system is changed, the equilibrium varies in such a way that it can compensate for this change.

For example, if the concentration of one of the reactants is increased, the equilibrium shifts to the right or to the side of the products. Also, if you add more reagents, the reaction will move even more to the right until the balance is re-established again, increasing the quantity of products.

In this way, adding HCl to a solution of CuSO4 will produce the following reaction:

CuSO4 (aq) + 2HCl (aq) ⇔ CuCl2 (aq) + H2SO4 (aq)

Initially the solution of CuSO4 in water will be blue, but when adding HCl the solution will change color to green, since the aqueous solutions of CuCl2 are green. By adding more HCl this color will intensify as the balance shifts to the right, producing more CuCl2 and H2SO4.

On the other hand, adding NH3 to a solution of CuSO4 will produce the following reaction

CuSO4 (aq) + 4NH3 (aq) ⇔ [Cu(NH3)4] SO4 (s)

Thus, by adding NH3 to the CuSO4 solution we will observe the formation of a precipitate corresponding to [Cu(NH3) 4] SO4. <u>When adding more NH3, the formation of more precipitate will be observed as the equilibrium moves to the right, producing a greater quantity of [Cu (NH3) 4] SO4.</u>

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A radio technician measures the frequency of an AM radio transmitter. The frequency is 11979kHz. What is the frequency in megahe
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Answer:

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Due to the presence of mobile or moving electrons in an atom they are good conductor of heat and electricity. Thus, the heat conduction and current conduction properties of metals are explained by its mobile electrons.

The other mentioned properties of metal are strength which can be explained by type of bonding within the metals, malleability explains the tendency of metals to be flattened into thin sheets, ductility explains the tendency to be stretched into wires, luster means the surface of metal is shiny and opacity is measure of impermeability that is to what extent they can pass light through them, metals are opaque, can not pass light through them or they are not transparent . All these properties are not because of mobile electrons in metals.

Therefore, correct properties are heat conduction and current conduction.


8 0
2 years ago
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A common way of initiating certain chemical reactions with light involves the generation of free halogen atoms in solution. if δ
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Answer: The longest wavelength of light that will produce free chlorine atoms in solution is 493 nm.

Explanation:

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1 mole = 6.022\times 10^{23} molecules

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For one molecule of chlorine gas =  \frac{242800 Joules/mol}{6.022\times 10^{23} mol^{-1}}=40,318.83\times 10^{-23}Joules

According to photoelectric equation:

E=h\nu=\frac{hc}{\Lambda }

E = Energy of the photon of light used to produce free chlorine atoms

\nu= frequency of the light used to produce free chlorine atoms

h = Planck's constant =6.626\times 10^{-34}J.s, c = speed of light=3\times 10^8 m/s

\lambda = wavelength of the light used to produce free chlorine atoms

40,318.83\times 10^{-23}J=\frac{hc}{\Lambda }=\frac{6.626\times 10^{-34} J.s\times 3\times 10^8 m/s}{\lambda }

\lambda=0.0004930203\times 10^{-3} m=493.0203\times 10^{-9} m=493 nm

The longest wavelength of light that will produce free chlorine atoms in solution is 493 nm.

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The type of gasoline with the highest percentage of octane among the options is premium.

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