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Iteru [2.4K]
3 years ago
8

What is the vapor pressure at 35°C of a solution made by dissolving 44.6 g of sucrose, C12H22O11, in 62.3 g of water? The vapor

pressure of pure water at 35°C is 42.2 torr.
Chemistry
1 answer:
lianna [129]3 years ago
5 0

Answer:

The correct answer is 40.861 torr.

Explanation:

Based on the given information, the weight of the solute (sucrose) given is 44.6 grams, the weight of the solvent, that is, water given is 62.3 grams, and 42.2 torr is the vapor pressure of pure water.

Now the molecular weight of sucrose is 342 grams per mole, and the molecular weight of water is 18 grams per mole.

The mole fraction of solute will be,

Mole fraction of solute = Weight of solute/Molecular weight of the solute/Weight of the solvent/molecular weight of the solvent + Weight of the solute/Molecular weight of the solute

= 44.6 grams/342 grams per mole/62.3 grams/18 grams per mole + 44.6 grams/342 grams per mole

= 0.1304/3.461 + 0.1304

= 0.0363

Now using relative lowering of vapor pressure,

Vapor pressure of pure water-Vapor pressure of solution/Vapor pressure of pure water = Mole fraction of solute

42.4 torr - PS/42.4 torr = 0.0363

Ps = 42.4 -1.5395

Ps = 40.861 torr

Hence, the vapor pressure of the solution is 40.861.

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Consider the reaction. mc014-1.jpg How many grams of methane should be burned in an excess of oxygen at STP to obtain 5.6 L of c
garri49 [273]
The reaction for the combustion of methane can be expressed as follows.
                           CH4 + 2O2 --> CO2 + 2H2O
We solve first for the amount of carbon dioxide in moles by dividing the given volume by 22.4L which is the volume of 1 mole of gas at STP.
                            moles of CO2 = (5.6 L) / (22.4 L/1 mole)
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Then, we can see that every mole of carbon dioxide will need 1 mole of methane
                              moles methane = (0.25 moles CO2) x (1 moles O2/1 mole CO2)
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5 0
4 years ago
Read 2 more answers
Write a ground state electron configuration for each neutral atom
Gre4nikov [31]

Answer:

Pb[lead] [Xe]4f^145d^106s^26p^2

U[uranium] 1s^2 2s^2 2p^6 3s^2 3p^6 4s^2 3d^10 4p^6 5s^2 4d^10 5p^6 6s^2 4e^14 5d^10 6p^6

7s^2 5f^4

This notation can be written in core notation or noble gas notation by replacing the

1s^2 2s^2 2p^6 3s^2 3p^6 4s^2 3d^10 4p^6 5s^2 4d^10 5p^6 6s^2 4e^14 5d^10 6p^6

7s^2 5f^4

with the noble gas [Rn].

[Rn]7s25f4

N[nitrogen] The full electron configuration for nitrogen is 1s^2 2s^2 2p^3.

Ti[titanium] Ti2+:[Ar]3d^2

Ti:1s^2 2s^2 2p^6 3s^2 3p^6 3d^2 4s^2

1s^2 2s^2 2p^6 3s^2 3p^5 = 17 electrons

(1) electron gain will result to a

negative charge (−), and

(2) electron loss will result to a positive charge (+),

1s^2 2s^2 2p^6 3s^2 3p^6 = 18 electrons

Hg[mercury] You should then find its atomic number is 80. It has a Xe core, so in shorthand notation, you can include [Xe]instead of

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for 54 electrons. For the 6th row of the periodic table, we introduce the 4f orbitals, and proceed to atoms having occupied 5d orbitals. We, as usual, have the ns orbitals, and n=? for the 6th period?

Mercury has a regular electron configuration. It becomes:

[Xe]4f145d106s2

Explanation:

socratic.org helped me! I'm really sorry if this is wrong!

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