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2 years ago

The equilibrium constant for the reaction is 1.1 x 106 M. HONO(aq) + CN-(aq) ⇋ HCN(aq) + ONO-(aq) This value indicates that

Chemistry

1 answer:

kakasveta [241]2 years ago

7 0

The given question is incomplete. The complete question is given here :

The equilibrium constant for the reaction is $1.1\times 10^6$ M.

$HONO(aq)+CN^- (aq)\rightleftharpoons HCN(aq)+ONO^-(aq)$

This value indicates that

A. $CN^-$ is a stronger base than $ONO^-$

B. HCN is a stronger acid than HONO

C. The conjugate base of HONO is $ONO^-$

D. The conjugate acid of CN- is HCN

Answer: A. $CN^-$ is a stronger base than $ONO^-$

Explanation:

Equilibrium constant is the ratio of product of the concentration of products to the product of concentration of reactants.

When $K_{p}>1$ ; the reaction is product favoured.

When $K_{p};$ ; the reaction is reactant favored.

$When K_{p}=1$ ; the reaction is in equilibrium.

As, $K_p>>1$ , the reaction will be product favoured and as it is a acid base reaction where $HONO$ acts as acid by donating $H^+$ ions and $CN^-$ acts as base by accepting $H^+$

Thus $HONO$ is a strong acid thus $ONO^-$ will be a weak conjugate base and $CN^-$ is a strong base which has weak $HCN$ conjugate acid.

Thus the high value of K indicates that $CN^-$ is a stronger base than $ONO^-$

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Determinación de pH Expresa las siguientes concentraciones de [H+ ] en función del pH • [H+] = 0.001 M • [H+] = 0.002 M • [H+] =

Pavel [41]

Answer:

• pH = 3.0

• pH = 2.70

• pH = 3.61

• pH = 8.28

• pH = 1.40

Explanation:

El pH es una medida en química usada para determinar el grado de acidez o basicidad en una solución.

Se define como:

pH = -log₁₀ [H⁺]

El - logaritmo de la concentración molar de H⁺

Para las concentraciones de H⁺ dadas:

• [H+] = 0.001 M

pH = -log (0.001M) = 3

pH = 3.0

• [H+] = 0.002 M

pH = -log (0.002M)

pH = 2.70

• [H+] = 2.45X10-4 M

pH = -log (2.45X10-4 M )

pH = 3.61

• [H+] = 5.2X10-9 M

pH = -log (5.2X10-9 M)

pH = 8.28

• [H+] = 0.04 M

pH = -log (0.04M)

pH = 1.40

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2 years ago

Which of the following statements about monosaccharide structure is true?

Nana76 [90]

Answer:

The only statement about monosaccharide structure which is true is b. (Monosaccharides can be classified according to the spatial arrangement of their atoms)

Explanation:

Monosaccharides are simple sugars that are classified according to the amount of carbon atoms and based on these numbers, we can call them trioses, pentoses and hexoses. They are molecules with aldehyde (aldose) or centone (ketose) groups that have more than one alcohol function, but which do not differ in their position (OH). They do not contain N, since their general formula is Cx (H2O) x. A 6-carbon monosaccharide is called hexose, since the pentose only has 5

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2 years ago

jan is holding an ice cube. what causes the ice to melt? thermal energy from the ice is transferred to the air. thermal energy f

loris [4]

Answer: Ice is melting due to the transfer of thermal energy from Jan's hand to ice.

Explanation: The melting of ice is a physical change and is happening when the thermal energy from Jan's hand is transferred to ice. Due to this energy transfer, the particles of ice starts to move faster and hence, making the ice melt.

In this, the physical state of ice is changing from solid to liquid state.

$H_2O(s)\rightleftharpoons H_2O(l)$

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2 years ago

Read 2 more answers

Gasoline is a mixture of hydrocarbons, a major component of which is octane, CH3CH2CH2CH2CH2CH2CH2CH3. Octane has a vapor pressu

Nitella [24]

Answer:

$\Delta \:H_{vap}=40383.88\ J/mol$

Explanation:

The expression for Clausius-Clapeyron Equation is shown below as:

$\ln P = \dfrac{-\Delta{H_{vap}}}{RT} + c$

Where,

P is the vapor pressure

ΔHvap is the Enthalpy of Vaporization

R is the gas constant (8.314×10⁻³ kJ /mol K)

c is the constant.

For two situations and phases, the equation becomes:

$\ln \left( \dfrac{P_1}{P_2} \right) = \dfrac{\Delta H_{vap}}{R} \left( \dfrac{1}{T_2}- \dfrac{1}{T_1} \right)$

Given:

$P_1$ = 13.95 torr

$P_2$ = 144.78 torr

$T_1$ = 25°C

The conversion of T( °C) to T(K) is shown below:

T(K) = T( °C) + 273.15

So,

T = (25 + 273.15) K = 298.15 K

$T_1$ = 298.15 K

$T_2$ = 75°C = 348.15 K

So,

$\ln \:\left(\:\frac{13.95}{144.78}\right)\:=\:\frac{\Delta \:H_{vap}}{8.314}\:\left(\:\frac{1}{348.15}-\:\frac{1}{298.15}\:\right)$

$\Delta \:H_{vap}=\ln \left(\frac{13.95}{144.78}\right)\frac{8.314}{\left(\frac{1}{348.15}-\frac{1}{298.15}\right)}$

$\Delta \:H_{vap}=\frac{8.314}{\frac{1}{348.15}-\frac{1}{298.15}}\left(\ln \left(13.95\right)-\ln \left(144.78\right)\right)$

$\Delta \:H_{vap}=\left(-\frac{863000.86966\dots }{50}\right)\left(\ln \left(13.95\right)-\ln \left(144.78\right)\right)$

$\Delta \:H_{vap}=40383.88\ J/mol$

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2 years ago

At 10°c one volume of water dissolves 3.10 volumes of chlorine gas at 1.00 atm pressure. what is the henry's law constant of cl2

s344n2d4d5 [400]

Answer is: 0,133 mol/ l· atm.
T(chlorine) = 10°C = 283K.
p(chlorine) = 1 atm.
V(chlorine) = 3,10 l.
R - gas constant, R = 0.0821 atm·l/mol·K.
Ideal gas law: p·V = n·R·T
n(chlorine) = p·V ÷ R·T.
n(chlorine) = 1atm · 3,10l ÷ 0,0821 atm·l/mol·K · 283K = 0,133mol.
Henry's law: c = p·k.
k - Henry's law constant.
c - solubility of a gas at a fixed temperature in a particular solvent.
c = 0,133 mol/l.
k = 0,133 mol/l ÷ 1 atm = 0,133 mol/ l· atm.

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2 years ago