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3 years ago

What is the driving force for the reaction of HCl(aq) and NaOH(aq)?

1 answer:

8 0

HCl and NaOH react based on the following equation:
HCl + NaOH ..............> NaCl + H2O

This reaction is known as a double replacement reaction. In this types of reactions, the electrolyte (and the solvent) is water and the driving force is mainly the formation of a non-electrolyte produce.

Based on this, the driving force in the above neutralization reaction is the formation if the non-electrolyte salt (NaCl)

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How many moles of H2O2 are needed to react with 1.07 moles of N2H4?

I am Lyosha [343]

Answer:

2.14 moles of H₂O₂ are required

Explanation:

Given data:

Number of moles of H₂O₂ required = ?

Number of moles of N₂H₄ available = 1.07 mol

Solution:

Chemical equation:

N₂H₄ + 2H₂O₂ → N₂ + 4H₂O

now we will compare the moles of H₂O₂ and N₂H₄

N₂H₄ : H₂O₂

1 : 2

1.07 : 2×1.07 = 2.14 mol

6 0

3 years ago

Pressure gauge at the top of a vertical oil well registers 140 bars. The oil well is 6000 m deep and filled with natural gas dow

andreyandreev [35.5K]

Explanation:

(a) The given data is as follows.

Pressure on top ( $P_{o}$ ) = 140 bar = $1.4 \times 10^{7} Pa$ (as 1 bar = $10^{5}$ )

Temperature = $15^{o}C$ = (15 + 273) K = 288 K

Density of gas = $\frac{PM}{ZRT}$

$\frac{dP}{dZ} = \rho \times g$

$\frac{dP}{dZ} = \int \frac{PM}{ZRT}$

$\int_{P_{o}}^{P_{1}} \frac{dP}{dZ} = \frac{Mg}{ZRT} \int_{0}^{4700} dZ$

$ln (\frac{P_{1}}{P_{o}}) = \frac{18.9 \times 10^{-3} \times 9.81 \times 4700 m}{0.80 \times 8.314 J/mol K \times 288 K}$

= 0.4548

$P_{1} = P_{o} \times e^{0.4548}$

= $1.4 \times 10^{7} Pa \times 1.5797$

= $2.206 \times 10^{7} Pa$

Hence, pressure at the natural gas-oil interface is $2.206 \times 10^{7} Pa$ .

(b) At the bottom of the tank,

$P_{2} = P_{1} + \rho \times g \times h$

= 2.206 \times 10^{7} Pa + 700 \times 9.81 \times (6000 - 4700)[/tex]

= $309.8 \times 10^{5} Pa$

= 309.8 bar

Hence, at the bottom of the well at $15^{o}C$ pressure is 309.8 bar.

6 0

3 years ago

PLEASE HELP! ILL GIVE BRAINLIEST, EXPLAIN YOUR ANSWER PLS

Pavel [41]

Answer:

arteries carry blood from the heart to the lungs and veins carry blood from the lungs to the heart

Explanation:

Arteries carry oxygenated blood from the heart, while veins carry oxygen-depleted blood back to the heart. ... The pulmonary veins transport oxygenated blood back to the heart from the lungs, while the pulmonary arteries move deoxygenated blood from the heart to the lungs.)

4 0

2 years ago

Find the pH of a 0.100 molar H2C6O6 solution with ka, where KA is equal 8.0×10–5

lubasha [3.4K]

The pH of the solution is 2.54.

Explanation:

pH is the measure of acidity of the solution and Ka is the dissociation constant. Dissociation constant is the measure of concentration of hydrogen ion donated to the solution.

The solution of C₆H₂O₆ will get dissociated as C₆HO₆ and H+ ions. So the molar concentration of 0.1 M is present at the initial stage. Lets consider that the concentration of hydrogen ion released as x and the same amount of the base ion will also be released.

So the dissociation constant Kₐ can be written as the ratio of concentration of products to the concentration of reactants. As the concentration of reactants is given as 0.1 M and the concentration of products is considered as x for both hydrogen and base ion. Then the

$K_{a}=\frac{[H^{+}][HB] }{[reactant]}$

[HB] is the concentration of base.

$8 * 10^{-5} =\frac{x^{2} }{0.1}\\\\\\x^{2} = 8 * 10^{-5}*0.1$

$x^{2} = 0.08 * 10^{-4}\\ \\x = 0.283*10^{-2}$

Then

$pH = - log [x] = - log [ 0.283 * 10^{-2}]\\ \\pH = 2 + 0.548 = 2.54$

So the pH of the solution is 2.54.

4 0

2 years ago

Small molecules that bind with self-proteins to produce antigenic substances are called ________. Small molecules that bind with

Leona [35]

Answer:

Haptens.

Explanation:

Haptens are known as small molecules that help to stimulate the production of antibody molecules when attached with a large molecule known as a carrier molecule such as proteins.

Haptens are used to study the mechanism of inflammatory bowel disease to help induce the autoimmune type of responses and allergic contact dermatitis.

4 0

3 years ago