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inn [45]
2 years ago
9

Choose the aqueous solution below with the highest freezing point. These are all solutions of nonvolatile solutes and you should

assume ideal van't Hoff factors where applicable. Choose the aqueous solution below with the highest freezing point. These are all solutions of nonvolatile solutes and you should assume ideal van't Hoff factors where applicable.
a. 0.200 m Ba(NO3)2
b. 0.200 m Mg(ClO4)2
c. 0.200 m HOCH2CH2OH
d. 0.200 m Na3PO3
Chemistry
1 answer:
Inessa [10]2 years ago
3 0

Answer: c. 0.200 m HOCH2CH2OH

Explanation:

The collective properties of solvents expressed that the freezing point of water is lowered when solute particles are dissolved in it.

When more particles are dissolved, the more the freezing point is deepened and lowered.

Here, all the molarities are equal, the deciding factor in measuring the lowering will be which substance produces the fewest particles in solution; then we relate it to lowering of the freezing point.

Ba(NO3)2 ---> Ba2+ + 2NO3(1-)

1 mol releases 3 mol of ions.

Mg(ClO4)2 ---> Mg2+ + 2ClO4(1-)

1 mol releases 3 mol of ions.

HOCH2CH2OH is covalent and doesn't ionize.

1 mol in water is just 1 mol of molecules.

Na3PO3 ---> 3Na+ + PO3(1-)

1 mol releases 4 mol ions.

The substance that produces the fewest particles in solution is the 0.200 m HOCH2CH2OH

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In each reaction box, place the best reagent from the list below. draw the intermediate compound.
Fynjy0 [20]

Answer:

Reagent A: PBr₃

Reagent B: Mg in Et₂O.

Explanation:

Hello,

In this case, your facing a problem in which a carboxylic acid is produced starting by an alcohol. More specifically, cyclopentanol must react with phosphorous tribromide in order to yield bromocyclopentane which is more likely to produce a carboxylic acid, therefore, reagent A is PBr₃.

On the other hand, by means of the production of the specified product, bromocyclopentane must react with carbon dioxide and magnesium in diethyl ether in acidic media to promote the production of the cyclopentanoic acid via the grignard reaction (substitution of the bromine by the carboxyle group), therefore, reagent B is Mg in Et₂O.

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8 0
2 years ago
A solution is made by mixing 55.g of thiophene C4H4S and 65.g of acetyl bromide CH3COBr.
EleoNora [17]

Answer:

Mole fraction of C₄H₄S = 0.55

Explanation:

Mole fraction is moles of solute / Total moles

Total moles are the sum of moles of solute + moles of solvent.

Let's find out the moles of our solute and our solvent.

Mass of solute: 55g

Mass of solvent: 65g

Mol = Mass / molar mass

55 g / 84.06 g/mol = 0.654 moles of C₄H₄S

65 g /123 g/mol = 0.529 moles of C₂H₃BrO

Total moles = 0.654 + 0.529 = 1.183 moles

Mole fraction of thiophene = Moles of tiophene / Total moles

0.654 / 1.183 = 0.55

4 0
3 years ago
Is mountain lions a herbivore
egoroff_w [7]
Mountain lion have to hunt other animals up in the Mountains. Other animals are meat. Therefore mountain lions are carnivores, not herbivores. 

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7 0
3 years ago
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dolphi86 [110]

Answer:

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3 years ago
Magnesium metal (0.100 mol) and a volume of aqueous hydrochloric acid that contains 0.500 mol of HCl are combined and react to c
jok3333 [9.3K]

Answer:

2.24 L of hydrogen gas, measured at STP, are produced.

Explanation:

Given, Moles of magnesium metal, Mg = 0.100 mol

Moles of hydrochloric acid, HCl = 0.500 mol

According to the reaction shown below:-

Mg_{(s)} + 2HCl_{(aq)}\rightarrow MgCl_2_{(aq)} + H_2_{(g)}

1 mole of Mg reacts with 2 moles of HCl

0.100 mol of Mg reacts with 2*0.100 mol of HCl

Moles of HCl must react = 0.200 mol

Available moles of HCl = 0.500 moles

Limiting reagent is the one which is present in small amount. Thus, Mg is limiting reagent.

The formation of the product is governed by the limiting reagent. So,

1 mole of Mg on reaction forms 1 mole of H_2

0.100 mole of Mg on reaction forms 0.100 mole of H_2

Mole of H_2 = 0.100 mol

At STP,  

Pressure = 1 atm  

Temperature = 273.15 K

Volume = ?

Using ideal gas equation as:

PV=nRT

where,  

P is the pressure

V is the volume

n is the number of moles

T is the temperature  

R is Gas constant having value = 0.0821 L.atm/K.mol

Applying the equation as:

1 atm × V L = 0.100 × 0.0821 L.atm/K.mol × 273.15 K  

<u>⇒V = 2.24 L</u>

2.24 L of hydrogen gas, measured at STP, are produced.

4 0
3 years ago
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