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Vlada [557]
3 years ago
13

Arrange the following aqueous solutions in order of decreasing freezing point: 0.10 m Na3PO4, 0.35 m NaCl, 0.20 m MgCl2, 0.15 m

C6H12O6, 0.15 m CH3COOH. (Introduce only the formulas, not the concentrations, of each substance in the appropriate box, beginning with the highest freezing point. Click in the answer boxes to activate the palette.) > > > > highest freezing point lowest freezing point
Chemistry
1 answer:
GarryVolchara [31]3 years ago
8 0

Answer: 0.35 m NaCl > 0.20 m MgCl_2 >0.10 m Na3PO_4 > 0.15 m CH_3COOH > 0.15 m C_6H_{12}O_6

Explanation:

Depression in freezing point:

T_f^0-T_f=i\times k_b\times m

where,

T_f= freezing point of solution

T^o_f = freezing point of solvent

k_f = freezing point constant

m = molality

1. For 0.10 m Na3PO_4

Na_3PO_4\rightarrow 3Na^++PO_4^{3-}

, i= 4 as it is a electrolyte and dissociate to give 4 ions. and concentration of ions will be 4\times 0.10=0.40

2. For 0.35 m NaCl

NaCl\rightarrow Na^++Cl^-

, i= 2 as it is a electrolyte and dissociate to give 2 ions. and concentration of ions will be 2\times 0.35=0.70 

3. For 0.20 m MgCl_2

MgCl_2\rightarrow Mg^{2+}+2Cl^-

, i= 3 as it is a electrolyte and dissociate to give 3 ions. and concentration of ions will be 3\times 0.20=0.60 

4. For 0.15 m C_6H_{12}O_6

, i= 1 as it is a non electrolyte and does not dissociate to give ions.

5. For 0.15 m CH_3COOH

CH_3COOH\rightarrow CH_3COO^-+H^+

, i= 2 as it is a electrolyte and dissociate to give 2 ions and thus have 2\times 0.15=0.30 

As concentration is highest for 0.35 m NaCl , freezing point depression will be highest and thus has lowest freezing point. As concentration is lowest for 0.15 m C_6H_{12}O_6 , freezing point depression will be lowest and thus has highest freezing point

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The first, third and fourth statements are correct.

Explanation:

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⇒ Le Chatellier says As the CaCO3 concentration is increased, the system will attempt to undo that concentration change by shifting the balance to the right. <u>This statement is true.</u>

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2) For the following reaction at equilibrium: CaCO3(s)⇌ CaO(s) + CO2(g) increasing the total pressure by adding Ar(g) will shift the equilibrium to the right.

⇒ Le chatellier says that if we increase the pressure, the equilibrium will shift to the side with the least number of particles.

Since the molar densities of CaO and CaCO3 are constant, they don't appear in the equilibrium expression. This is why only changes to the pressure (concentration) of CO2 affect the position of the equilibrium.

If the pressure in the container is increased by adding an inert or non-reacting gas, nothing happens to the amounts of CO2, CaO or CaCO3. The added gas won't affect the partial pressure of CO2. <u>This statement is false. </u>

3)For the following reaction at equilibrium: 2 H2(g) + O2(g) ⇌ 2 H2O(g) the equilibrium will shift to the left if the volume is doubled.

⇒ Le Chatellier says if we increase the pressure, the equilibrium will shift to the side with the most particles.

In this case we have 2 moles of H2 and 1 mole of O2 on the left side and 2 mole of H2O on the right side. This means on the left side are more particles. So the equilibrium will shift to the left, so <u>this statement is true.</u>

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The features that are shared by the rocks include the melting and cooling of rocks.

<h3>What is a rock? </h3>

It should be noted that a rock simply means a relatively hard naturally occurring mineral material.

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4 0
2 years ago
a sample of carbon dioxide contains 3.8 moles of oxygen atoms, how many moles of carbon dioxide are in the sample?
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<h3>Answer:</h3>

1.9 moles

<h3>Explanation:</h3>

Carbon dioxide (CO₂) is a compound that is made up of carbon and oxygen elements.

It contains 2 moles of oxygen atoms and 1 mole of carbon atoms

Therefore;

We would say, 1 mole of CO₂ → 2 moles of Oxygen atoms + 1 mole of carbon atoms

Thus;

If a sample of CO₂ contains 3.8 moles of oxygen atoms we could use mole ratio to determine the moles of CO₂

Mole ratio of CO₂ to Oxygen is 1 : 2

Therefore;

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Hence, the moles of CO₂ present in a sample that would produce 3.8 moles of Oxygen atoms is 1.9 moles

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