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nikklg [1K]
3 years ago
6

A 3.50 g sugar cube (sucrose: C12H22O11) is dissolved in a 350 mL teapot containing 80∘ C water (density of water at 80∘ C= 0.97

5 g/mL). What is the molality of the sugar solution?
Chemistry
2 answers:
KIM [24]3 years ago
6 0

Answer: 0.04893mole/kg

Explanation:

Molality is defined as the mole of solute per kilogram of the solvent.

Sugar is the solute while water is the solvent.

Molar mass of sugar =

1*12+1*22+11*16

12+22+176=210g/mol.

Mole of solute= reacting mass /molar mass

Reacting mass 3.50g and molar mass 210g/mol

Mole= 3.50/210=0.0167moles.

For solvent

Density=mass/volume

Mass= density *volume

0.975*350

341.25g

Converting gram back to kilogram

341.25/1000= 0.34125kg

Molality =0.0167/0.34125

=0.04893mole/kg

mestny [16]3 years ago
6 0

Answer:

Molality is 0.03 m

Explanation:

Molality indicates the moles of solute that are contained in 1000g of solvent.

We have solvent volume and density, so let's find out the mass:

Solvent density = Solvent mass / Solvent volume

Solvent density . Solvent volume = Solvent mass → 0.975 g/mL . 350 mL

Solvent mass = 341.25g

Now we need the moles of solute → mass of solute / molar mass

3.50 g / 342 g/mol = 0.0102 moles

So finally we can make a rule of three:

341.25 g of solute contain 0.0102 moles of solute

1000 g of solute, may contain (1000 . 0.0102) / 341.25 = 0.03 m

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Sindrei [870]

Answer:

Distillation.

Explanation:

If we are heating a mixture of two miscible liquids and collecting the vapors it means we are separating the two mixtures from each other based on their boiling point differences.

This technique of separation of two liquids based on the difference in boiling point is known as Distillation.

Alcohol will evaporate easily as compared to water as water has stronger influence of hydrogen bond making the inter-molecular forces stronger.

4 0
3 years ago
Which of the following are true statements about equilibrium systems? For the following reaction at equilibrium: CaCO3(s) ⇌ CaO(
Grace [21]

Answer:

The first, third and fourth statements are correct.

Explanation:

1) For the following reaction at equilibrium: CaCO3(s) ⇌ CaO(s) + CO2(g) adding more CaCO3 will shift the equilibrium to the right.

⇒ 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>

<u />

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>

4) For the following reaction at equilibrium: H2(g) + F2(g) ⇌ 2HF(g) removing H2 will increase the amount of F2 present once equilibrium is reestablished. Increasing the temperature of an endothermic reaction shifts the equilibrium position to the right.

⇒ Le chatellier says if H2 will be removed (this means the left side will get less particles) so the equilibrium will shift to the left, to increase the amount of F2.

⇒Le chatelier says if we increase the temperature of an exotherm reaction , there will be less energy released. The equilibrium will shift to the side of the reactants (the left side).

If we increase the temperature of an endotherm reaction, the equilibrium will shift to the side of the products (the right side). <u>This statement is true.</u>

4 0
3 years ago
Nitrous oxide gas is a form of
zepelin [54]
Its a form of anaesthesia used by dentists 

hope that helsp 
6 0
2 years ago
Mrs. Nogaki tested four black markers in rubbing alcohol and in water. Each filter paper had four evenly spaced dots. She found
ExtremeBDS [4]

Answer:

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Explanation:

dependent viable = output

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8 0
2 years ago
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saw5 [17]
By  use  of  combined  gas  law
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v1=5L
P1=540  torr
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T2=32+273=305  k

P2  is   therefore=(  305k  x  540 torr  x5 L) /( 15L  x  298)=  184.23  torr
6 0
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