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2 years ago

8

You are given 10.00 mL of a solution of an unknown acid. The pH of this solution is exactly 2.95. You determine that the concent

ration of the unknown acid was 0.1224 M. You also determined that the acid was monoprotic (HA). What is the K_a and pK_a of your unknown acid

1 answer:

Naily [24]2 years ago

8 0

Answer:

$1.038\times 10^{-5}$ is dissociation constant and the value of $pK_a$ is 4.98.

Explanation:

The pH of the solution = 2.95 M

$pH=-\log[H^+]$

$2.95=-\log[H^+]$

$[H^+]=10^{-2.95}=0.001122 M$ ..[1]

Concentration of unknown monoprotic acid = C = 0.1224 M

$HA\rightleftharpoons A^-+H^+$

Initially

C 0 0

At equilibrium

(C-x) x x

The expression of a dissociation reaction can be written as;

$K_a=\frac{[A^-][H^+]}{[HA]}$

$K_a=\frac{x^2}{(C-x)}$

$[H^+]=x =0.001122 M$ ( from [1])

$K_a=\frac{(0.001122 M)^2}{(0.1224 M-0.001122 M)}$

$K_a=1.038\times 10^{-5}$

The value of $pK_a$ :

$pK_a=-\log[K_a]$

$=-\log[1.038\times 10^{-5}]=4.98$

$1.038\times 10^{-5}$ is dissociation constant and the value of $pK_a$ is 4.98.

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Ammonia gas is compressed from 21°C and 200 kPa to 1000 kPa in an adiabatic compressor with an efficiency of 0.82. Estimate the

Evgen [1.6K]

Explanation:

It is known that efficiency is denoted by $\eta$ .

The given data is as follows.

$\eta$ = 0.82, $T_{1}$ = (21 + 273) K = 294 K

$P_{1}$ = 200 kPa, $P_{2}$ = 1000 kPa

Therefore, calculate the final temperature as follows.

$\eta = \frac{T_{2} - T_{1}}{T_{2}}$

0.82 = $\frac{T_{2} - 294 K}{T_{2}}$

$T_{2}$ = 1633 K

Final temperature in degree celsius = $(1633 - 273)^{o}C$

= $1360^{o}C$

Now, we will calculate the entropy as follows.

$\Delta S = nC_{v} ln \frac{T_{2}}{T_{1}} + nR ln \frac{P_{1}}{P_{2}}$

For 1 mole, $\Delta S = C_{v} ln \frac{T_{2}}{T_{1}} + R ln \frac{P_{1}}{P_{2}}$

It is known that for $NH_{3}$ the value of $C_{v}$ = 0.028 kJ/mol.

Therefore, putting the given values into the above formula as follows.

$\Delta S = C_{v} ln \frac{T_{2}}{T_{1}} + R ln \frac{P_{1}}{P_{2}}$

= $0.028 kJ/mol \times ln \frac{1633}{294} + 8.314 \times 10^{-3} kJ \times ln \frac{200}{1000}$

= 0.0346 kJ/mol

or, = 34.6 J/mol (as 1 kJ = 1000 J)

Therefore, entropy change of ammonia is 34.6 J/mol.

3 0

2 years ago

In a hexagonal-close-pack (hcp) unit cell, the ratio of lattice points to octahedral holes to tetrahedral holes is 1:1:2. What i

Alex777 [14]

Explanation:

In a hexagonal-close-pack (HCP) unit cell, the ratio of lattice points to octahedral holes to tetrahedral holes = 1 : 1 : 2

let the :

Number of lattice point = 1x.

Number of octahedral points = 1x

Number of tetrahedral points = 2x

If anions occupy the HCP lattice points and cations occupy half of the tetrahedral holes.

Number of anions occupying the HCP lattice points, A= 1x

Number of cations occupying the tetrahedral points, B = $\frac{2x}{2}=x$

The formula of the compound will be = $A_{1x}B_{1x}=AB$

If anions occupy the HCP lattice points and cations occupy all of the tetrahedral holes.

Number of anions occupying the HCP lattice points, A= 1x

Number of cations occupying the tetrahedral points, B = 2x

The formula of the compound will be = $A_{1x}B_{2x}=AB_2$

6 0

2 years ago

You will be using the observations given in the scenarios i - vi below to determine which of the following is true for G (a - c)

WITCHER [35]

Answer:

A spontaneous reaction is taking place in a flask. When you touch the flask, it feels cold and gets colder as the reaction progresses........(c,f,g)

A spontaneous reaction is taking place in a flask as evidenced by the formation of gas bubbles. When you touch the flask, it feels hot and gets hotter as the reaction progresses.... (c,d,h)

A spontaneous reaction is taking place in a flask. When you touch the flask, it feels hot and gets hotter as the reaction progresses....(c,f,h)

A spontaneous reaction is taking place in a flask. You notice the formation of gas bubbles intensifies as the reaction progresses....(c,d,I)

A solid in a closed flask begins to melt. (You can consider this a phase change.)....(b,d,g)

A liquid begins to appear on the inside of a closed flask. At first glance before the liquid began to appear, you might have thought the flask was empty. (You can consider this a phase change.)....(a,e,h)

Explanation:

Now it is necessary here to state the equation for ∆G

∆G= ∆H-T∆S

The values of ∆G depends on the relative values of ∆H, T and ∆S.

When heat is evolved by a system, ∆H is negative, when heat is absorbed by a system, ∆H is positive.

When gases are evolved, ∆S is positive, when gases turn to liquids or solids ∆S is negative.

Let us also recall that, for the melting process ∆S positive ( solid changes to liquid), ∆H is positive(energy is absorbed) and T increases( the substance is heated). Hence ∆G is negative

For the condensation process; ∆H is negative (heat is evolved), ∆S is negative (gas changes to liquid) and T is decreased hence ∆G is positive.

5 0

2 years ago

It takes 839./kJmol to break a carbon-carbon triple bond. Calculate the maximum wavelength of light for which a carbon-carbon tr

tresset_1 [31]

Answer:

The maximum wavelength of light for which a carbon-carbon triple bond could be broken by absorbing a single photon is 143 nm.

Explanation:

It takes 839 kJ/mol to break a carbon-carbon triple bond.

Energy required to break 1 mole of carbon-carbon triple bond = E = 839 kJ

E = 839 kJ/mol = 839,000 J/mol

Energy required to break 1 carbon-carbon triple bond = E'

$E'=\frac{ 839,000 J/mol}{N_A}=\frac{839,000 J}{6.022\times 10^{23} mol^{-1}}=1.393\times 10^{-18} J$

The energy require to single carbon-carbon triple bond will corresponds to wavelength which is required to break the bond.

$E'=\frac{hc}{\lambda }$ (Using planks equation)

$\lambda =\frac{6.626\times 10^{-34} Js\times 3\times 10^8 m/s}{1.393\times 10^{-18} J}$

$\lambda =1.427\times 10^{-7} m =142.7 nm = 143 nm$

$(1 m = 10^9 nm)$

The maximum wavelength of light for which a carbon-carbon triple bond could be broken by absorbing a single photon is 143 nm.

6 0

2 years ago

A 108 49in source emits a 633-kev gamma photon and a 606-kev internal-conversion electron from the k shell. what is the binding

____ [38]

Binding energy is the energy needed to emit the electron from the shell. Using the formula below to compute for BE. Binding Energy BE = Energy of photon - Kinetic energy electron
where
Energy proton= 633 keV
KE electron = 606 keV
Binding energy BE = 27 keVThe binding energy of the k subshell is equal to 27 keV.

6 0

2 years ago