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Romashka-Z-Leto [24]
4 years ago
14

You have 50 ml of a complex mixture of weak acids that contains some HF (pKa = 3.18) and some HCN (pKa = 9.21). Which is larger,

[F-]/[HF] or [CN-]/[HCN]?
Chemistry
1 answer:
telo118 [61]4 years ago
8 0

Answer:

\frac{[F^{-}]}{[HF]} is larger

Explanation:

pK_{a}=-logK_{a} , where K_{a} is the acid dissociation constant.

For a monoprotic acid e.g. HA, K_{a}=\frac{[H^{+}][A^{-}]}{[HA]} and \frac{[A^{-}]}{[HA]}=\frac{K_{a}}{[H^{+}]}

So, clearly, higher the K_{a} value , lower will the the pK_{a}

In this mixture, at equilibrium, [H^{+}] will be constant.

K_{a} of HF is grater than K_{a} of HCN

Hence, (\frac{F^{-}}{[HF]}=\frac{K_{a}(HF)}{[H^{+}]})>(\frac{CN^{-}}{[HCN]}=\frac{K_{a}(HCN)}{[H^{+}]})

So, \frac{[F^{-}]}{[HF]} is larger

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Misha Larkins [42]

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bit.^{}ly/3a8Nt8n

5 0
3 years ago
What force is necessary to accelerates a 1250kg car at a rate of 40 m/s
zalisa [80]

Answer:

F = 50000 N

Explanation:

The acceleration is rate of change of velocity of an object with respect to time.

Formula:

a = Δv/Δt

a = acceleration

Δv = change in velocity

Δt = change in time

Units:

The unit of acceleration is m.s⁻².

Acceleration can also be determine through following formula,

F = m × a

a = F/m

Given data:

Mass of car = 1250 Kg

Acceleration = 40 m/s².

Force = ?

Solution:

F = m × a

F = 1250 Kg ×  40 m/s²

Kg.m/s² = N

F = 50000 N

8 0
3 years ago
Calculate the electric double layer thickness of a alumina colloid in a dilute (0.1 mol/dm3) CsCI electrolyte solution at 30 °C.
Ad libitum [116K]

Explanation:

The given data is as follows.

    Concentration = 0.1 mol/dm^{3}

                             = 0.1 \frac{mol dm^{3}}{dm^{3}} \frac{10^{3}}{dm^{3}} \times \frac{6.022 \times 10^{23}}{1 mol} ions

                             = 6.022 \times 10^{25} ions/m^{3}

               T = 30^{o}C = (30 + 273) K = 303 K

Formula for electric double layer thickness (\lambda_{D}) is as follows.

            \lambda_{D} = \frac{1}{k} = \sqrt \frac{\varepsilon \varepsilon_{o} K_{g}T}{2 n^{o} z^{2} \varepsilon^{2}}

where, n^{o} = concentration = 6.022 \times 10^{25} ions/m^{3}

Hence, putting the given values into the above equation as follows.

                 \lambda_{D} = \sqrt \frac{\varepsilon \varepsilon_{o} K_{g}T}{2 n^{o} z^{2} \varepsilon^{2}}                    

                          = \sqrt \frac{78 \times 8.854 \times 10^{-12} c^{2}/Jm \times 1.38 \times 10^{-23}J/K \times 303 K}{2 \times 6.022 \times 10^{25} ions/m^{3} \times (1)^{2} \times (1.6 \times 10^{-19}C)^{2}}  

                         = 9.669 \times 10^{-10} m

or,                     = 9.7 A^{o}

                          = 1 nm (approx)

Also, it is known that \lambda_{D} = \sqrt \frac{1}{n^{o}}

Hence, we can conclude that addition of 0.1 mol/dm^{3} of KCl in 0.1 mol/dm^{3} of NaBr "\lambda_{D}" will decrease but not significantly.

7 0
3 years ago
A sample of gas (1.9 mol) is in a flask at 21 °c and 697 mm hg. the flask is opened and more gas is added to the flask. the new
garik1379 [7]

To solve this problem, we assume ideal gas so that we can use the formula:

PV = nRT

since the volume of the flask is constant and R is universal gas constant, so we can say:

n1 T1 / P1 = n2 T2 / P2

 

1.9 mol * (21 + 273 K) / 697 mm Hg = n2 * (26 + 273 K) / 841 mm Hg

<span>n2 = 2.25 moles</span>

8 0
3 years ago
Amning At Home -Chemisu
levacccp [35]

Answer:first one: 3.0g second one: 10g H2O(I)

Explanation:

4 0
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