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vagabundo [1.1K]
3 years ago
11

The vapor pressure of pure water at 110 °C is 1070 torr. A solution of ethylene glycol and water has a vapor pressure of 1.00 at

m at 110 °C. Assuming that Raoult’s law is obeyed, what is the mole fraction of ethylene glycol in the solution? g
Chemistry
2 answers:
malfutka [58]3 years ago
8 0

Answer:

0.291

Explanation:

Given data

  • Vapor pressure of the pure solvent (Psolvent): 1.41 atm

1070 torr × (1 atm/ 760 torr) = 1.41 atm

  • Vapor pressure of the solvent above the solution (Psolution): 1.00 atm

According to Raoult's law, the vapor pressure of a solvent above a solution is equal to the vapor pressure of the pure solvent at the same temperature scaled by the mole fraction of the solvent present.

Psolution = Psolvent × Χsolvent

Χsolvent = Psolution/Psolvent

Χsolvent = 1.00 atm/1.41 atm

Χsolvent = 0.709

The sum of the mole fraction of the solvent (water) and the solute (ethylene glycol) is 1.

Χsolvent + Χsolute = 1

Χsolute = 1 - Χsolvent = 1 - 0.709

Χsolute = 0.291

Kitty [74]3 years ago
7 0

Answer:

idkidkidkidkidkidkidkidkidkidklookitup

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10.0 grams of water are heated during the preparation of a cup of coffee 1.0x 103 j of the heat are added to the water. which is
katovenus [111]

<u>Answer:</u> The final temperature of the coffee is 43.9°C

<u>Explanation:</u>

To calculate the final temperature, we use the equation:

q=mC(T_2-T_1)

where,

q = heat released = 1.0\times 10^3J=1000J

m = mass of water = 10.0 grams

C = specific heat capacity of water = 4.184 J/g°C

T_2 = final temperature = ?

T_1 = initial temperature = 20°C

Putting values in above equation, we get:

1000J=10.0g\times 4.184J/g^oC\times (T_2-20)\\\\T_2=43.9^oC

Hence, the final temperature of the coffee is 43.9°C

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3 years ago
Determine the molar mass of CuSO4 (the solute) in a 1.0M aqueous solution of CuSO4
inna [77]

Answer:

See explanation.

Explanation:

Hello,

In this case, we could have two possible solutions:

A) If you are asking for the molar mass, you should use the atomic mass of each element forming the compound, that is copper, sulfur and four times oxygen, so you can compute it as shown below:

M_{CuSO_4}=m_{Cu}+m_{S}+4*m_{O}=63.546 g/mol+32.00g/mol+4*16.00g/mol\\\\M_{CuSO_4}=159.546g/mol

That is the mass of copper (II) sulfate contained in 1 mol of substance.

B) On the other hand, if you need to compute the moles, forming a 1.0-M solution of copper (II) sulfate, you need the volume of the solution in litres as an additional data considering the formula of molarity:

M=\frac{n_{solute}}{V_{solution}}

So you can solve for the moles of the solute:

n_{solute}=M*V_{solution}

Nonetheless, we do not know the volume of the solution, so the moles of copper (II) sulfate could not be determined. Anyway, for an assumed volume of 1.5 L of solution, we could obtain:

n_{solute}=1mol/L*1.5L=1.5mol

But this is just a supposition.

Regards.

4 0
2 years ago
How many moles of carbonate are there in sodium carbonate​
vaieri [72.5K]

There are 0.566 moles of carbonate in sodium carbonate.

<h3>CALCULATE MOLES:</h3>
  • The number of moles of carbonate (CO3) in sodium carbonate (Na2CO3) can be calculated by dividing the mass of carbonate in the compound by the molar mass of the compound.

  • no. of moles of CO3 = mass of CO3 ÷ molar mass of Na2CO3

  • Molar mass of Na2CO3 = 23(2) + 12 + 16(3)

  • = 46 + 12 + 48 = 106g/mol

  • mass of CO3 = 12 + 48 = 60g

  • no. of moles of CO3 = 60/106

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Learn more about number of moles at: brainly.com/question/1542846

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 Thermodynamic quantity equivalent to the total heat content of a system It is equal to the internal energy of the system plus the product of pressure and volume
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expeople1 [14]

Answer:

54g of water

Explanation:

Based on the reaction, 1 mole of methane produce 2 moles of water.

To solve this question we must find the molar mass of methane in order to find the moles of methane added. With the moles of methane and the chemical equation we can find the moles of water produced and its mass:

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1C = 12g/mol*1

4H = 1g/mol*4

12g/mol + 4g/mol = 16g/mol

<em>Moles methane: </em>

24g CH₄ * (1mol / 16g) = 1.5 moles methane

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1.5moles CH₄ * (2mol H₂O / 1mol CH₄) = 3.0moles H₂O

<em>Molar mass water:</em>

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2g/mol + 16g/mol = 18g/mol

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3.0moles H₂O * (18g / mol) =

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