What volume of 0.215 M HCl is required to neutralize 50.0 mL of
0.800 M NaOH?

Answers

Answer 1

Answer:

186 mL HCl

Explanation:

M1V1 = M2V2

M1 = 0.215 M HCl

V1 = ?

M2 = 0.800 M NaOH

V2 = 50.0 mL

Solve for V1 --> V1 = M2V2/M1

V1 = (0.800 M)(50.0 mL) / (0.215 M) = 186 mL HCl


Related Questions

How many grams are in 32.2 L of CO2?

Answers

Answer:

63.25 grams of CO₂

Explanation:

To convert from liters to grams, we first need to convert from liters to moles. To do this, we divide the liters by 22.4, the amount of liters of a gas per mole.

32.2 / 22.4

= 1.4375 moles of CO₂

Now we want to convert from moles to grams. To do this, we multiply the moles by the molar mass of CO₂. The total molar mass can be found on the periodic table by adding up the molar mass of carbon (12) and two oxygen (32).

12 + 32 = 44

Now we want to multiply the moles by the molar mass.

1.4375 • 44

= 63.25 grams of CO₂

This is your answer.

Hope this helps!

Is liquid ammonia a household acid?​

Answers

Yes it is an cleaning product

Answer:

yes

Explanation:

it cleans in the house

what are possible source of error for rusting of a nail​

Answers

Answer:

A nail can rust when exposed to oxygen. the molecules of iron on the surface of the nail exchange atoms with the oxygen in the air and produce a new substance, the reddish brown ferrous oxide i.e rust.

Which example is a mixture? alcohol water mercury orange soda

Answers

Answer:

water is an example of mixture

I don’t know how to do that cause that’s what I’m having trouble with

On the line provided, write the formula for the compound made by each of the following. (3 pts each) 20. Mgand NO3-1, 21. CIO3-1 and K 22. Ca and So4-2, 23. NH4+1 and SO4-2, 24. C2H30-1, ' and H 25. Hand CO, t

Answers

Answer:

20. Mg(NO3)2

21. KClO3

22. CaSO4

23. (NH4)2SO4

24. HC2H3O

25.  CO is carbon dioxide, which is neutral and would not combine with a hydrogen ion. Did you mean H and CO3(2-)? That would be H2CO3 (carbonic acid)

Explanation:

You have to match the charges of the Cations and Anions.

What scientifically goes on with a plasma ball?

Answers

Answer:

the electrode at the center of a plasma ball emits a high-frequency,high-voltage alternating electric current. This current flows through the plasma filaments to create colorful tendrils of light.

Explanation:

what colors depend upon the gases used inside the plasma ball? common gases include neon, argon, xenon, and krypton.

2 NaClO3→ 2 NaCl + 3 O2

How many moles of O2 are produced when 40g of NaCl are formed?

Answers

Answer:

75.6

Explanation:

Please give me the answer please

Answers

Answer:

A. 30cm³

Explanation:

Based on the chemical reaction:

CaCO₃ + 2HCl → CaCl₂ + H₂O + CO₂

1 mol of calcium carbonate reacts with 2 moles of HCl to produce 1 mol of CO₂

To solve this question we must convert the mass of each reactant to moles. With the moles we can find limiting reactant and the moles of CO₂ produced. Using PV = nRT we can find the volume of the gas:

Moles CaCO₃ -Molar mass: 100.09g/mol-

1.00g * (1mol / 100.09g) = 9.991x10⁻³ moles

Moles HCl:

50cm³ = 0.0500dm³ * (0.05 mol / dm³) = 2.5x10⁻³ moles

For a complete reaction of 2.5x10⁻³ moles HCl there are necessaries:

2.5x10⁻³ moles HCl * (1mol CaCO₃ / 2mol HCl) = 1.25x10⁻³ moles CaCO₃. As there are 9.991x10⁻³ moles, HCl is limiting reactant.

The moles produced of CO₂ are:

2.5x10⁻³ moles HCl * (1mol CO₂ / 2mol HCl) = 1.25x10⁻³ moles CO₂

Using PV = nRT

Where P is pressure = 1atm assuming STP

V volume in L

n moles = 1.25x10⁻³ moles CO₂

R gas constant = 0.082atmL/molK

T = 273.15K at STP

V = nRT / P

1.25x10⁻³ moles * 0.082atmL/molK*273.15K / 1atm = V

0.028L = V

28cm³ = V

As 28cm³ ≈ 30cm³

Right option is:

A. 30cm³

A 50.0 mL sample of buffer solution contains 0.120 M acetic acid and 0.150 M sodium acetate. If 5.55 mL of 0.092 M NaOH is added to this solution, identify the resulting number of moles of acetic acid, sodium acetate, and NaOH.

Answers

Answer:

see calculations below

Explanation:

Given:     HOAc     ⇄    [H⁺]     +    [OAc⁻]            

C(i)          0.12M             0M           0.15M

mix  => 5.55ml(0.092M NaOH) / (50ml + 5.55ml)

            = 0.00555(0.092)mole NaOH / 0.0555 L Soln

            = 0.0092M in NaOH is added into the initial buffer solution

            = 0.0092M in OH⁻ (NaOH is a strong base => 100% ionized)

Rxn => Addition of 0.0092M OH⁻ will react with 0.0092M H⁺ shifting buffer            .           equilibrium to the right decreasing [HOAc] and increasing [OAc⁻] by    .           0.0092M each.

Therfore ...

Given:           HOAc      ⇄    [H⁺]     +       [OAc⁻]            

C(i)                0.12M             0M               0.15M

ΔC           - 0.0092M           +x            +0.0092M

C(f)             0.1108M             x                0.1592M  =>  New Concentrations            .                                                                                     after adding 0.0092M             .                                                                                     NaOH

Substituting new acid and ion concentrations into Ka expression ...

Ka = [H⁺][OAc⁻]/[HOAc] = (x)(0.1592M)/(0.1108M) = 1.75 x 10⁻⁵M

=> x = [H⁺](new) = (1.75 x 10⁻⁵M*)(0.1108M)/(0.1592M) = 1.22 x 10⁻⁵M in H⁺ ions

*units of Ka are Molar

FYI => Adding a strong base to a buffer solution will shift pH to more basic.

          Adding a strong acid   to a buffer solution will shift pH to more acidic.

=> (such is a good way to check that your buffer calculations are correct.)

NOTE => Question asks for moles of HOAc, Na⁺OAc⁻ & NaOH after adding base. Giving answers in terms of Molarity (moles/Liter) is same as moles. Therefore ...

[HOAc] = 0.1108M

[NaOAc] = 0.1592M

[NaOH] = ∅M (from rxn of H⁺ + OH⁻ => H₂O, all NaOH was consumed in acid/base reaction.  Remaining are only Na⁺ as a spectator ion and OH⁻ as a function of the new concentration of H⁺ => [OH⁻] = Kw/[H⁺] = 1 x 10⁻¹⁴/1.22 x 10⁻⁵ = 8.2 x 10⁻¹⁰M.

Hope this helps. :-)

Part A. Classify each of these soluble solutes as a strong electrolyte, a weak electrolyte, or a nonelectrolyte. Solutes Formula Nitric acid HNO3 Potassium hydroxide KOH Formic acid HCOOH Ethyl amine CH3CH2NH2 Sodium bromide NaBr Butanol C4H9OH Sucrose C12H22O11
Part B. Enter a molecular equation for the reaction that occurs between aqueous HBr and aqueous LiOH.

Answers

Answer:

1. Strong electrolytes = Nitric acid HNO3,Potassium hydroxide KOH, Sodium bromide NaBr

Weak electrolytes = Formic acid HCOOH Ethyl amine CH3CH2NH2, Butanol C4H9OH

Non-electrolyte = Sucrose C12H22O11

2. The reaction between aqueous HBr and aqueous LiOH is shown innthe equation below:

HBr + LiOH ----> NaBr + H₂O

Explanation:

Electrolytes are substances that when in molten state or in aqueous solution dissociate into ions both positively-charged and negatively-charged ions known as actions and anions, which are then able conduct electricity.

Strong electrolytes are aqueous or molten solutions of compounds which ionize completely to produce anions and actions. Ionic compounds such as organic salts, acids and bases are strong electrolytes.

Weak electrolytes only ionize partially, that is, they produce few ions. Some organic salts, acids and bases are weak elctrolytes

From the compounds given above;

Strong electrolytes = Nitric acid HNO3,Potassium hydroxide KOH, Sodium bromide NaBr

Weak electrolytes = Formic acid HCOOH Ethyl amine CH3CH2NH2, Butanol C4H9OH

Non-electrolyte = Sucrose C12H22O11

2. The reaction between aqueous HBr and aqueous LiOH is shown innthe equation below:

HBr + LiOH ----> NaBr + H₂O

PLEASEEEE HELLPPP I CANT FAILL PLSS

Answers

It’s the second won the air layer one

How does the entropy change in the reaction 2C3H6(g) + 9O2(g) → 6C02(g) + 6H2O (g)?

I will mark brainliest!! Thank you so much!!

Answers

Answer:

The entropy increases!!!

Explanation:

a pex

The entropy increases in the reaction.

What is entropy?Entropy is defined as the measure of the disorder of a system.Entropy is an extensive property of a thermodynamic system, to put it in simple words, its value changes depending on the amount of matter that is present.Entropy is denoted by the letter S and has units of joules per kelvin    (JK−1)

The entropy increases in the reaction if the total number of product molecules are greater than the total number of reactant molecules.

2C3H6(g) + 9O2(g) → 6C02(g) + 6H2O (g)

In the above reaction, the product molecules are greater than the reactant molecules. Hence, entropy increases.

Hence, we can conclude that option A is the answer.

To learn more about entropy here

https://brainly.com/question/22861773

#SPJ2

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