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lianna [129]
3 years ago
11

Cu + 2 AgNO3 + Cu(NO3)2 + 2Ag

Chemistry
1 answer:
Allushta [10]3 years ago
5 0

Answer:

7.28 moles Ag°

Explanation:

              Cu°          +    2 AgNO₃  => Cu(NO₃)₂ + 2Ag°

Given 7.28 moles     7.28 moles

To determine limiting reactant, divide the mole values by the respective coefficient of balanced equation. The resulting smallest value is the limiting reactant. Note: this is a short cut method for determining limiting reactant only. Once the limiting reactant is determined one must use the given mole values of the limiting reactant to solve problem. That is ...

Limiting reactant determination:

              Cu°          +    2 AgNO₃  => Cu(NO₃)₂ + 2Ag°

Cu: 7.28 / 1  = 7.28

AgNO₃ : 7.28 / 2 = 3.64 => Limiting Reactant is AgNO₃

Solving Problem depends on AgNO₃; Cu will be in excess.

Since coefficients of AgNO₃ & Ag° are equal, then the moles AgNO₃ used equals moles Ag° produced and is therefore 7.28 moles Ag°.

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Incomplete contribution happens when there is a limited supply of air so only half is much oxygen adds to the carbon forming carbon monoxide
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To separate a mixture of benzoic acid and fluorene, we are going to use an acid-base extraction technique. Most carboxylic acids
pentagon [3]

Answer: please find attached to see the structure.

1. THE STRUCTURE OF BENZOIC ACID AND FLUORENE, soluble in ether and insoluble in water.

2. THE STRUCTURE OF CARBOXYLIC ACID BEEN EXTRACTED.

Explanation: the mixture of benzoic acid and fluorene are the first diagrams which shows the carboxylic acid attached to the benzene ring, which are soluble in ether and insoluble in water. When dissolved in NaOH(aq) is the carboxy ion becomes soluble in water but insoluble in ether, this is seen in the second diagram.

The third diagram shows the carboxylic acid been precipitated and soluble in ether but insoluble in water.

NOTE THE TWO MAIN DIAGRAM IS THE FIRST AND THE LAST DIAGRAM, WHERE CARBOXYLIC ACID DISSOLVES IN AQUEOUS SODIUM HYDROXIDE, AND WHEN THE ACID IS BEEN PRECIPITATED IN AQUEOUS HCl.

ALSO NOTE THE CHANGE IN BENZOIC RING MIXED WITH FLUORENE TO THAT OF THE ACID BEEN EXTRACTED.

Hope together with the picture, this has helped you.

8 0
4 years ago
In performing this week's bromination reaction, if you were to start with 126 mg of acetanilide (135.17 g/mol), calculate the th
Artemon [7]

Answer:

Theoretical yield of C8H8BrNO:

In moles

0.000945

In grams

0.204

Explanation:

Theoretical yield of a reaction is defined as the quantity of the product obtained from the complete conversion of a limiting reactant in a chemical reaction. Theoretical yield can be expressed as grams or moles.

Equation of reaction:

C8H9NO + Br2 --> C8H8BrNO + HBr

Since C8H9NO is the limiting reagent, 1 mole of C8H9NO reacted to form 1 mole of C8H8BrNO

Mass of C8H9NO = 129 mg

= 0.129 g.

Molar mass of C8H9NO = 135.17 g/mol.

Number of moles of C8H9NO = mass/molar mass.

= 0.129/135.17

= 0.00095 moles of C8H9NO

Since 1 mole of C8H9NO yielded 1 mole of C8H8BrNO

Therefore, 0.000954 moles of C8H8BrNO

Theoretical yield (in grams) = molar mass * number of moles

= 214.06 * 0.00095

= 0.204 of C8H8BrNO

3 0
3 years ago
How many moles of HCl are present in 40.0 mL of a 0.035 M solution? 1. 0.0014 mol 2. 0.0060 mol 3. 0.25 mol?
Vinil7 [7]
40.0mL(1 L/1000 mL) = 0.040 L 

<span>then plug into the formula M = moles/liters </span>

<span>0.035 M = moles/ 0.040L </span>

<span>multipy both sides by 0.040L, and you get 0.0014 moles </span>

<span>so the answer is 1</span>
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Wet land isnt it obvious 
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