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What volume (mL) of 0.135 M NaOH is required to neutralize 13.7 mL of 0.129 M HCl? a: 0.24 b: 13.1 c: 0.076 d: 6.55 e: 14.3

Len [333]

Answer:

The volume (mL) of 0.135 M NaOH that is required to neutralize 13.7 mL of 0.129 M HCl is 13.1 mL (option b).

Explanation:

The reaction between an acid and a base is called neutralization, forming a salt and water.

Salt is an ionic compound made up of an anion (positively charged ion) from the base and a cation (negatively charged ion) from the acid.

When an acid is neutralized, the amount of base added must equal the amount of acid initially present. This base quantity is said to be the equivalent quantity. In other words, at the equivalence point the stoichiometry of the reaction is exactly fulfilled (there are no limiting or excess reagents), therefore the numbers of moles of both will be in stoichiometric relationship. So:

V acid *M acid = V base *M base

where V represents the volume of solution and M the molar concentration of said solution.

In this case:

V acid= 13.7 mL= 0.0137 L (being 1,000 mL= 1 L)
M acid= 0.129 M
V base= ?
M base= 0.135 M

Replacing:

0.0137 L* 0.129 M= V base* 0.135 M

Solving:

$V base=\frac{0.0137 L*0.129 M}{0.135 M}$

V base=0.0131 L = 13.1 mL

The volume (mL) of 0.135 M NaOH that is required to neutralize 13.7 mL of 0.129 M HCl is 13.1 mL (option b).

4 0

3 years ago

12. Which compound can act as both a BrønstedLowry

oksian1 [2.3K]

1) The compound which can act as a Bronsted-Lowry acid and a Bronsted-Lowry base is definitely water - H2O. Remember that water is amphoteric which means it can either accept protons or donate them, so it is the most proper option among other represented. Here are examples of both base and acid with water : HCl+H20=H30+Cl ; NH3+H2O=NH4+OH

2) The acids in this equilibrium reaction CN– + H2O HCN + OH. Acid species always donate H+ to the species with which they react. In the second option you can see how H2O donates an H+ to CN-. If the reaction gets reversed we will obtain HCN that donates an H+ to OH that shows that it is an acid.

3) The products of self-ionization of water are OH⁻ and H₃O⁺. Self-ionization is an ionization reaction during which H2O deprotonates its hydrogen atoms to become a hydroxide ion -- OH−. After this process OH- protonates another water molecule forming H3O+.

4) The type of solution which is one with a pH of 8 is acidic. Here is a little table that can be a prompt for you if you ever come across such tasks - ph : 7 is neutral. pH lower than 7 are acidic, and pH higher than 7 basic ones.

5) The acid dissociation constant for an acid dissolved in water is equal to the equilibrium constant. I consider this option correct because we can obtan Kw only when dealing with Kb, and we can conclude that the hydrolysis constant of the conujugate base.

6) A 0.12 M solution of an acid that ionizes only slightly in solution would be termed dilute and weak. You can determine it depending on its concentration. Such value as 0.12M usually defined as a dilute solution of a weak acid due to the fact that acid represents its partial ionization which is a direct characteristic of a weak acid.

7) To solve this task we should appeal to Henry's law that says the solubility of a gaz is proportional to its partial pressure. And according to this we can understand that 202kPa is the half of 404kPa which means that the needed solubility must be divided by 2 7.5/2=3.75 g/L and that's all.

8) I think that the most important points which best show how the addition of a solute affects the boiling point, the freezing
point, and the vapor pressure of the solvent are : BOILING: additional attractive forces can only exist between solute and solvent and in order to boil they must be overcome for the solution;we should add KE to overcome the forces. FREEZING : to freeze we have to withdrawn KE as the solute particles are surrounded by solvent molecules. VAPOR : WHen solvent shells are being formed the solute particles reduces the number of solvent particles that have sufficient KE to vaporize.

9) $[H+][OH-]= Kw = 1.0 * 10^-14$
$[H+]= Kw/ [OH-]= 1.0x 10^-14 / 1 x 10^-11 =1 x 10^-3 mol/L 

pH = - log [H+]= - log 1 x 10^-3 = 3$
Since we got Ph of 3 in a result we can define solution as an acidic one, as I mentioned before.

10) Since the formula of the given acid is HA it undergoes like that : HA<=> H+ + A- .
ka = [H][A] / [HA].
Now we have only [H+] and to go further you need to write electroneutrality equation for the reaction :
[H+] = [OH-] + [A-] (since [H]>>>[OH]), then
[H+] = [A-]
Then mass balance equation :
Ct = 0.5M = [A-] + [HA]
[HA] = 0.5 - [A-] = 0.5 - [H+]
Finally here is what we have done and get :
$ka = [H]^2 / (0.5 - [H+]) 
$
$ka = 0.0001*0.0001/(0.5-0.0001) = 2.00x10^-8$

11) The main points that are common for acids : they form Hydrogen ions when dissloved in water, - Ex. Vinegar and Lemon, Ph >7, they have Increased hydrogen ions (H+). The facts about bases : they reduce the concentration of hydgoren ions in a solution which is opposite to asids,- Ex. Antiacid,and Ammonia ,Ph valuse above 7, they form hydrogen (OH-).
- The common points of both acids and bases : Hydrogen ions ,
both not neutral and water based.

5 0

3 years ago

Scientists in 1.legal services 2.research and development 3.marketing 4.corporate management

Maslowich

2. research and development

4 0

3 years ago

Aerobic cellular respiration requires an adequate supply of

Varvara68 [4.7K]

Answer:

oxygen ?

Explanation:

not sure but I think so

3 0

3 years ago

A nuclear waste site. cesium-137 is a particularly dangerous by-product of nuclear reactors. it has a half-life of 30 years. it

hram777 [196]

The mass decay rate is of the form
$m(t) = m_{0} e^{-kt}$
where
m₀ = 3000 g,the initial mass
k = the decay constant
t = time, years.

Because the half-life is 30 years, therefore
$e^{-30k} = \frac{1}{2} \\\ -30k = ln(0.5) \\ k = \frac{ln(0.5)}{-30} =0.0231$

After 60 years, the mass remaining is
$m = 3000 e^{-0.0231*60} = 750 \, g$

Answer: 750 g

4 0

3 years ago