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2 years ago

What’s the total ionic equation for NaOH reacting with HCl

Chemistry

1 answer:

exis [7]2 years ago

6 0

Answer:

Na⁺ (aq) + OH⁺ (aq) + H⁺ (aq) + Cl⁻ (aq) → Na⁺ (aq) + Cl⁻ (aq) + H₂O (l)

General Formulas and Concepts:

Atomic Structure

Compounds

Aqueous Solutions

Solubility Rules
States of matter

Stoichiometry

Reaction RxN Prediction
Balancing Reactions RxN

Explanation:

Step 1: Define

NaOH reacting w/ HCl

NaOH is soluble

HCl is soluble

[RxN] NaOH (aq) + HCl (aq) → NaCl (aq) + H₂O (l)

Step 2: Total Ionic Equation

Break up soluble compounds into ionic form.

[T.I.E] Na⁺ (aq) + OH⁺ (aq) + H⁺ (aq) + Cl⁻ (aq) → Na⁺ (aq) + Cl⁻ (aq) + H₂O (l)

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What is the Ka value if pka=10.36

siniylev [52]

Pka = - log ka

10.36 = - log ka

Ka = .......

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3 years ago

Read 2 more answers

Which element in Group 17 is the most active nonmetal? (1) Br (2) I (3) Cl (4) F

Maksim231197 [3]

The element of the group 17 that is most active non metal is fluorine.

The group 17 of the periodic table contains bromine(Br), iodine(I), Chlorine(Cl) and fluorine(F).

Among all the elements of the group 17. Fluorine is the smallest in size.

Because of the small size of fluorine it has the highest electronegativity in group 17.

This high electronegativity makes it a very active non metal. It provides a very high oxidizing power and low dissociation energy to the fluorine atom.

Also because of the very small size the source of attraction between the nucleus and the electrons is very high in floor in atom.

It reacts readily to form oxides and hydroxides.

So, we can conclude here that fluorine is the most active non metal of group 17.

To know more about group 17, visit,

brainly.com/question/26440054

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7 0

1 year ago

Consider the Fischer ester synthesis of methyl benzoate from benzoic acid and methanol in the presence of sulfuric acid as a cat

zlopas [31]

Answer:

48.8%

Explanation:

The reaction has a 1:1 mole ratio so;

Number of moles of benzoic acid reacted = mass/molar mass = 3.8 g/122.12 g/mol = 0.03 moles

So;

0.03 moles of methyl benzoate is formed in the reaction

Mass of methyl benzoate formed = 0.03 moles * 136.15 g/mol = 4.1 g

percent yield = actual yield/theoretical yield * 100/1

percent yield = 2.0 g/4.1 g * 100 = 48.8%

4 0

3 years ago

A student isolated 7.2 g of 1-bromobutane reacting equimolar amounts of 1-butanol (10 ml) and NaBr (11.1 g) in the presence of s

Alla [95]

Answer: The percent yield of the 1-bromobutane is 48.65 %

Explanation:

To calculate the number of moles, we use the equation:

$\text{Number of moles}=\frac{\text{Given mass}}{\text{Molar mass}}$ .....(1)

For NaBr:

Given mass of NaBr = 11.1 g

Molar mass of NaBr = 103 g/mol

Putting values in equation 1, we get:

$\text{Moles of NaBr}=\frac{11.1g}{103g/mol}=0.108mol$

The chemical equation for the reaction of 1-butanol and NaBr is:

$\text{1-butanol + NaBr}\rightarrow \text{1-bromobutane}$

By Stoichiometry of the reaction

1 mole of NaBr produces 1 mole of 1-bromobutane

So, 0.108 moles of NaBr will produce = $\frac{1}{1}\times 0.108=0.108$ moles of 1-bromobutane

Now, calculating the mass of 1-bromobutane from equation 1, we get:

Molar mass of 1-bromobutane = 137 g/mol

Moles of 1-bromobutane = 0.108 moles

Putting values in equation 1, we get:

$0.108mol=\frac{\text{Mass of 1-bromobutane}}{137g/mol}\\\\\text{Mass of 1-bromobutane}=(0.108mol\times 137g/mol)=14.80g$

To calculate the percentage yield of 1-bromobutane, we use the equation:

$\%\text{ yield}=\frac{\text{Experimental yield}}{\text{Theoretical yield}}\times 100$

Experimental yield of 1-bromobutane = 7.2 g

Theoretical yield of 1-bromobutane = 14.80 g

Putting values in above equation, we get:

$\%\text{ yield of 1-bromobutane}=\frac{7.2g}{14.80g}\times 100\\\\\% \text{yield of 1-bromobutane}=48.65\%$

Hence, the percent yield of the 1-bromobutane is 48.65 %

5 0

3 years ago

Which describes why Mendeleev succeeded where others failed?

wolverine [178]

Answer:

He realized that some elements had not been discovered.

Explanation:

Some scientists that tried to arrange the list of elements known before Mendeleev include Antoine Lavoisier, Johann Döbereiner, Alexandre Béguyer de Chancourtois, John Newlands, and Julius Lothar Meyer.

Dimitri Mendeleev was so succesful that he is recognized as the most important in such work.

Mendeleev by writing the properties of the elements on cards elaborated by him, and "playing" trying to order them, realized that, some properties regularly (periodically) repeated.

The elements were sorted in increasing atomic weight (which is not the actual order in the periodic table), but when an element did not meet the pattern discovered, he moved it to a position were its properties fitted.

The amazing creativity of Mendeleev led him to leave blanks for what he thought were places that should be occupied by elements yet undiscovered. More amazing is that he was able to predict the properties of some of those elements.

When years after some of the elements were discovered, the genius of Mendeleev was proven because the "new" elements had the properties predicted by him.

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3 years ago