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kherson [118]
4 years ago
8

Acids & bases/concentration A mixture is made by mixing 425 mL of 0.94 M H2SO4 with 750. mL of 0.83 M NaOH. (a) What is the

limiting reactant? (b) How much excess reactant will be left over? (c) Is this mixture acidic or basic? Why? (d) What is the concentration of the [H+] or [OH-] ions that remain in this solution? (e) Calculate the pH of this solution.
Chemistry
1 answer:
Ksenya-84 [330]4 years ago
4 0

Answer:

I'm really sorry but I didn't understand your question

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Treatment of ethyl acetoacetate with NaOEt (2 equiv) and BrCH2CH2Br forms compound X. This reaction is the first step in the syn
valkas [14]

Answer:

See explanation below

Explanation:

In this reaction we have the ethyl acetoacetate which is reacting with 2 eq of sodium etoxide. The sodium etoxide is a base and it usually behaves as a nucleophyle of many reactions. Therefore, it will atract all the acidics protons in a molecule.

In the case of the ethyl acetoacetate, the protons that are in the methylene group (CH3 - CO - CH2 - COOCH2CH3) are the more acidic protons, therefore the etoxide will substract these protons instead of the protons of the methyl groups. This is because those hydrogens (in the methylene group) are between two carbonile groups, which make them more available and acidic for any reaction. As we have 2 equivalents of etoxide, means that it will substract both of the hydrogen atoms there, and then, reacts with the Br - CH2CH2 - Br and form a product of an aldolic condensation.

The mechanism of this reaction to reach X is shown in the attached picture.

3 0
4 years ago
In the reaction below, 22 g of H2S with excess
BartSMP [9]

Answer:

24%

Explanation:

From the question, the limiting reactant is H2S.

The reaction equation is;

2H2S + O2 → 2S + 2H2O

Number of moles of H2S reacted = 22g/34 g/mol = 0.647 moles

According to the reaction equation;

2 moles of H2S yields 2 moles of sulphur

0.647 moles of H2S yields 0.647 moles of sulphur

So;

Theoretical yield of sulphur = 0.647 moles * 32 g/mol = 20.7 g

Actual yield = 5 g

% yield = actual yield/theoretical yield * 100

% yield =5 g/20.7 g * 100

% yield = 24%

7 0
3 years ago
Peepeepoopoo? peepeepoopoo?
Keith_Richards [23]

Answer:

I just need points

Explanation:

4 0
3 years ago
Read 2 more answers
How many electrons will each ele-
Andre45 [30]

<h2>Explanation:</h2><h3 /><h3>Oxygen- gains 2 electrons to form ions</h3><h3>Fluorine- gains 1 electron to form negative ions</h3><h3>Aluminum - loses three electrons to form ions</h3><h3>Calcium- loses 2 electrons in order to form ions</h3>

<h3>*Non metals gain electrons to form ions</h3><h3>*Metals loses electrons to form ions</h3>

6 0
3 years ago
What would the products be when aluminum chloride (which contains Al3+ and Cl- ions) is melted and electrolyzed? Write half equa
Hoochie [10]

The product of electrolysis : Al at cathode and Cl₂(chlorine) at anode

<h3>Further explanation</h3>

Given

Aluminum chloride compound

Required

The product of electrolysis

Solution

The rule :

The reaction at the cathode(the negative pole) :

1. the reduced active metal is water, other than that the metal will be reduced

2. H⁺ of the acid will be reduced

The reaction at the anode((the positive pole) :

1. if the electrodes are not inert then the metal is oxidized

2. If inert then:

a. OH⁻ from the base will be oxidized

b. The halogen metal will oxidize

In the electrolysis of molten AlCl₃ with an inert electrode, the cation will be reduced at the cathode and the anion will be oxidized at the anode

Cathode : Al³⁺ + 3e⁻ ⇒ Al(s)

Anode : 2Cl⁻⇒Cl₂(g) + 2e⁻

6 0
3 years ago
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