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3 years ago

Why do weak acid-weak base reactions not go to completion?

Chemistry

2 answers:

dezoksy [38]3 years ago

7 0

Answer: Option (c) is the correct answer.

Explanation:

Weak acid and weak bases are the species that dissociate weakly in a solvent or water.

For example, acetic acid ( $CH_{3}COOH$ ) is a weak acid and it dissociates poorly in water.

Hence, we can say that due to weak dissociation there will less formation of hydrogen or hydroxide ions upon dissociation of weak acid or weak base. Therefore, due to lack of sufficient number of ions the reaction will not go to completion.

Thus, we can conclude that weak acid-weak base reactions not go to completion because neither a weak acid nor a weak base has a strong tendency to transfer $H^+$ ions.

vladimir2022 [97]3 years ago

4 0

Neither a weak acid nor a weak base has a strong tendency to transfer H+ ions that is why<span> weak acid-weak base reactions not go to completion.</span>

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A student constructs a coffee cup calorimeter and places 50.0 mL of water into it. After a brief period of stabilization, the te

Grace [21]

Answer:

The calorimeter constant is = 447 J/°C

Explanation:

The heat absorbed or released (Q) by water can be calculated with the following expression:

Q = c × m × ΔT

where,

c is the specific heat

m is the mass

ΔT is the change in temperature

The water that is initially in the calorimeter (w₁) absorbs heat while the water that is added (w₂) later releases heat. The calorimeter also absorbs heat.

The heat absorbed by the calorimeter (Q) can be calculated with the following expression:

Q = C × ΔT

where,

C is the calorimeter constant

The density of water is 1.00 g/mL so 50.0 mL = 50.0 g. The sum of the heat absorbed and the heat released is equal to zero (conservation of energy).

Qabs + Qrel = 0

Qabs = - Qrel

Qcal + Qw₁ = - Qw₂

Qcal = - (Qw₂ + Qw₁)

Ccal . ΔTcal = - (cw . mw₁ . ΔTw₁ + cw . mw₂ . ΔTw₂)

Ccal . (30.31°C - 22.6°C) = - [(4.184 J/g.°C) × 50.0 g × (30.31°C - 22.6°C) + (4.184 J/g.°C) × 50.0 g × (30.31°C - 54.5°C)]

Ccal = 447 J/°C

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3 years ago

What is the formula for germicide that is sufficient to kill blood-borne pathogens?

Colt1911 [192]

It is 1 ounce. Bloodborne Pathogens can be transmitted when blood or body liquid from a tainted individual enters someone else's body by means of needle-sticks, human chomps, cuts, scraped areas, or through mucous films. Likewise, semen, vaginal discharges and salivation in dental methods are considered conceivably tainted body liquids.

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3 years ago

A sample of HI (9.30×10^−3mol) was placed in an empty 2.00 L container at 1000 K. After equilibrium was reached, the concentrati

Sedaia [141]

Answer:

The answer is "29.081"

Explanation:

when the empty 2.00 L container of 1000 kg, a sample of HI (9.30 x 10-3 mol) has also been placed.

$\text{calculating the initial HI}= \frac{mol}{V}$

$=\frac{9.3 \times 10 ^ -3}{2}$

$=0.00465 \ Mol$

$\text{Similarly}\ \ I_2 \ \ \text{follows} \ \ H_2 = 0 }$

Its density of I 2 was 6.29x10-4 M if the balance had been obtained, then we have to get the intensity of equilibrium then:

$HI = 0.00465 - 2x\\\\ I_{2} \ eq = H_2 \ eq = 0 + x \\\\$

It is defined that:

$I_2 = 6.29 \times 10^{-4} \ M \\\\x = I_2 \\\\$

$HI \ eq= 0.00465 - 2x \\$

$=0.00465 -2 \times 6.29 \times 10^{-4} \\\\ = 0.00465 -\frac{25.16 }{10^4} \\\\ = 0.003392\ M$

Now, we calculate the position:

For the reaction $H 2(g) + I 2(g)\rightleftharpoons 2HI(g)$ , you can calculate the value of Kc at 1000 K.

data expression for Kc

$2HI \rightleftharpoons H_2 + I_2 \\\\\to Kc = \frac{H_2 \times I_2}{HI^2}$

$= \frac{6.29\times10^{-4} \times 6.29 \times 10^{-4}}{0.003392^2} \\\\= \frac{6.29\times 6.29 \times 10^{-8}}{0.003392^2} \\\\= \frac{39.564 \times 10^{-8}}{1.150 \times 10-5} \\\\= 0.034386$