It takes 60 mL of 0.20 M of sodium hydroxide (NaOH) to neutralize 25 mL of carbonic acid (H2CO3) for the following chemical reaction:

2 NaOH + H2CO3 → N2CO3 + 2 NaOH

The concentration of the carbonic acid is _____.
A) 0.48m
B) 0.10m
C) 0.96m
d)0.24m

Answers

Answer 1

If It takes 60mL of 0.20M of sodium hydroxide (NaOH) to neutralize 25 mL of carbonic acid ([tex]H_2CO_3[/tex]), then the concentration of the carbonic acid is 0.24M

The reaction between NaOH solution and [tex]H_2CO_3[/tex] is written below

[tex]2NaOH + H_2CO_3 \rightarrow Na_2CO_3 + 2H_2O[/tex]

Volume of NaOH, [tex]V_B[/tex] = 60 ml

Volume of [tex]H_2CO_3[/tex], [tex]V_A=25 ml[/tex]

Molarity of [tex]H_2CO_3[/tex], [tex]C_A=?[/tex]

Molarity of NaOH, [tex]C_B=0.20M[/tex]

Number of moles of [tex]H_2CO_3[/tex], [tex]n_A=1[/tex]

Number of moles of NaOH, [tex]n_B=2[/tex]

The mathematical equation for neutralization reaction is:

[tex]\frac{C_AV_A}{C_BV_B} =\frac{n_A}{n_B}[/tex]

Substitute  [tex]C_B=0.2 M[/tex],  [tex]n_A=1[/tex],  [tex]n_B=2[/tex],  [tex]V_B[/tex] = 60ml, and  [tex]V_A=25 ml[/tex] into the equation above in order to solve for [tex]C_A[/tex]

[tex]\frac{C_A \times 25}{0.2 \times 60}=\frac{1}{2} \\\\50C_A=12\\\\C_B=\frac{12}{50} \\\\C_B=0.24M[/tex]

Therefore, the concentration of the carbonic acid is 0.24M

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Related Questions

A 6.00 L container of N_2 has a temperature of 273 K. Calculate the volume if the temperature is doubled.


a
24.0 L

b
12.0 L

c
6.0 L

d
3.0 L

Answers

V1/T1 =V2/T2 (using charles law)

V1=6.00
V2=?
T1=273
T2=273

Making V2 the subject of formula the equation then becomes

V2= V1xT2/T1

6.00x263/273=6.0L

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Answers

Answer:

methyl 2-methyl-ethane

Explanation:

Check all of the following below which describe experimental errors that would affect the percent chloride you report in your experimental results: ____________
a) The solid product was wet when you weighed the final mass
b) You added excess silver nitrate into the reaction mixture
c) The liquid collected from the filtration was cloudy
d) The transfer from the reaction beaker to the filter funnel was imperfect (you lost some of the reaction solution or some was left in the beaker)
e) Extra water was used to dissolve the original unknown solid

Answers

The options that describe the experimental errors that would affect the percent of chloride reported are:

B (You added excess silver nitrate into the reaction mixture) and D (The transfer from the reaction beaker to the filter funnel was imperfect).

While performing experiments, there are some errors that occur which are generally known as experimental errors.

These experimental errors are the difference between the true measurement and what we measured.

There are three types of experimental errors they include:

systematic,

random, and

human error.

Human errors occur due to carelessness of the observer.

Examples are:

when you added excess silver nitrate into the reaction mixture

when the transfer from the reaction beaker to the filter funnel was imperfect due to lost of some of the reaction solution.

Therefore, the options describe the experimental errors that would affect the percent of chloride reported are B and D.

Learn more about experimental errors here:

https://brainly.com/question/2764830

Pls answer this question

Answers

Answer:

A carbon dioxide and oxygen

Convert 12.044×10²² molecules of sulphur dioxide into moles​

Answers

Answer:

➜ Properties of liquids and gases. Liquids have a fixed volume but no fixed shape. Gases have no fixed volume and no fixed shape. Gases expand to fill the space available. They can also be compressed into a very small space.

Explanation:

12.044 × 10 ²³molecules of sulphur dioxide are 0.2 mole.

How many moles of chlorine could be produced by decomposing 157g NaCl? 2NaCl --> 2Na+Cl2

Answers

Here is the answer for your question which the mole of chlorine is 1.34

By decomposing 157g of NaCl, approximately 1.35 moles of chlorine can be produced.

To determine the number of moles of chlorine that could be produced by decomposing 157g of NaCl, we need to use the molar mass of NaCl and apply stoichiometry.

The molar mass of NaCl is the sum of the atomic masses of sodium (Na) and chlorine (Cl), which is approximately 22.99 g/mol + 35.45 g/mol = 58.44 g/mol.

Now, we can calculate the number of moles of NaCl:

Moles of NaCl = Mass of NaCl / Molar mass of NaCl

Moles of NaCl = 157 g / 58.44 g/mol

Moles of NaCl ≈ 2.69 mol

According to the balanced equation 2NaCl → 2Na + Cl₂, we can see that for every 2 moles of NaCl, we produce 1 mole of Cl₂.

Therefore, using the stoichiometric ratio, we can calculate the number of moles of chlorine produced:

Moles of Cl₂ = (Moles of NaCl / 2) × 1

Moles of Cl₂ = 2.69 mol / 2

Moles of Cl₂ ≈ 1.35 mol

Thus, by decomposing 157g of NaCl, approximately 1.35 moles of chlorine can be produced.

Learn more about chlorine from the link given below.

https://brainly.com/question/19460448

#SPJ2

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Answers

Answer:

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Answer:

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Answers

I think correct answer is C.Jet stream would be displaced southwards causing heavy rain and flooding

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