Ask question

Ask question

All categories

3 years ago

12

Double replacement: NaOH(aq) HCl(aq)→ _ _ Express your answer as a chemical equation. Your answer does not need to inclu

de the phases of the product(s).

1 answer:

mrs_skeptik [129]3 years ago

4 0

Answer:

NaOH + HCl → NaCl + H₂O

Explanation:

In double replacement reactions, we have 2 ionic compounds that exchange their cations and anions. Let's write the ionic equation for the reactants.

Na⁺(aq) + OH⁻(aq) + H⁺(aq) + Cl⁻(aq) →

Cl⁻(aq) will replace OH⁻(aq) in NaOH forming NaCl(aq), and OH⁻(aq) will replace Cl⁻(aq) in HCl forming H₂O(l). The resulting chemical equation is:

NaOH(aq) + HCl(aq) → NaCl(aq) + H₂O(l)

You might be interested in

A tempture of 20°C is equivalent to apporximately

olya-2409 [2.1K]

<h3>Heya</h3><h3>Here is your answer</h3><h3>________________</h3><h3></h3><h3>A temperature of 20°C is equivalent to approximately 68 degree Fahrenheit (standard room temperature).</h3><h3></h3><h3>==================</h3><h3></h3><h3>Hope this helped!!!</h3><h3>Happy to help :)</h3>

4 0

3 years ago

An unknown salt is either NaF, NaCl, or NaOCl. When 0.050 mol of the salt is dissolved in water to form 0.500 L of solution, the

victus00 [196]

<u>Answer:</u> The unknown salt is NaF

<u>Explanation:</u>

To calculate the molarity of solution, we use the equation:

$\text{Molarity of the solution}=\frac{\text{Moles of solute}}{\text{Volume of solution (in L)}}$

Moles of salt = 0.050 moles

Volume of solution = 0.500 L

Putting values in above equation, we get:

$\text{Molarity of salt}=\frac{0.050mol}{0.500L}\\\\\text{Molarity of salt}=0.1M$

To calculate the hydroxide ion concentration, we first calculate pOH of the solution, which is:

pH + pOH = 14

We are given:

pH = 8.08

$pOH=14-8.08=5.92$

To calculate pOH of the solution, we use the equation:

$pOH=-\log[OH^-]$

Putting values in above equation, we get:

$5.92=-\log[OH^-]$

$[OH^-]=10^{-5.92}=1.202\times 10^{-6}M$

The unknown salt given are formed by the combination of weak acid and strong acid which is NaOH

The chemical equation for the hydrolysis of $X^-$ ions follows:

$X^-(aq.)+H_2O(l)\rightleftharpoons HX(aq.)+OH^-(aq.);K_b$

<u>Initial:</u> 0.1

<u>At eqllm:</u> 0.1-x x x

Concentration of $OH^-=x=1.202\times 10^{-6}M$

The expression of $K_b$ for above equation follows:

$K_b=\frac{[OH^-][HX]}{[X^-]}$

Putting values in above expression, we get:

$K_b=\frac{(1.202\times 10^{-6})\times (1.202\times 10^{-6})}{(1-(1.202\times 10^{-6}))}\\\\K_b=1.445\times 10^{-11}M$

To calculate the acid dissociation constant for the given base dissociation constant, we use the equation:

$K_w=K_b\times K_a$

where,

$K_w$ = Ionic product of water = $10^{-14}$

$K_a$ = Acid dissociation constant

$K_b$ = Base dissociation constant = $1.445\times 10^{-11}$

Putting values in above equation, we get:

$10^{-14}=1.445\times 10^{-11}\times K_a\\\\K_a=\frac{10^{-14}}{1.445\times 10^{-11}}=6.92\times 10^{-4}$

We know that:

$K_a\text{ for HF}=6.8\times 10^{-6}$

$K_a\text{ for HCl}=1.3\times 10^{6}$

$K_a\text{ for HClO}=3.0\times 10^{-8}$

So, the calculated $K_a$ is approximately equal to the $K_a$ of HF

Hence, the unknown salt is NaF

6 0

3 years ago

In a oxidation–reduction reaction, oxidation and reduction must occur simultaneously. Is this statement true or false?

bogdanovich [222]

Answer:

True

Explanation:

Oxidation and reduction always occur simultaneously. Gave example when a substance is oxidized or reduced? 1) The substance gaining oxygen or losing electron is oxidized. 2) The substance losing oxygen ( or gaining electron) is reduced. hope this helps you :)

5 0

4 years ago

Dr. Arenas is a scientist who has been trying to determine if increasing pressure will increase the sound that occurs when two c

lozanna [386]

C.Experimental thats the answer

3 0

3 years ago

Read 2 more answers

Cuales son los nombres de la distribución de los electrones en un átomo ?

cluponka [151]

Un nombre es configuración de los electrones. Lo siento, no sé más nombres

7 0

3 years ago