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Tom [10]
3 years ago
8

Calculate the vapor pressure at 35ºC of a solution made by dissolving 20.2 g of sucrose (C12H22O11)in 60.5 g of water. The vapor

pressure of pure water at 35ºC is 42.2 mmHg. What is the vapor-pressure depressionof the solution(in units of mmHg)? (Sucrose is nonvolatile)
Chemistry
1 answer:
sukhopar [10]3 years ago
5 0

Answer:

P' = 41.4 mmHg → Vapor pressure of solution

Explanation:

ΔP = P° . Xm

ΔP = Vapor pressure of pure solvent (P°) - Vapor pressure of solution (P')

Xm = Mole fraction for solute (Moles of solvent /Total moles)

Firstly we determine the mole fraction of solute.

Moles of solute → Mass . 1 mol / molar mass

20.2 g . 1 mol / 342 g = 0.0590 mol

Moles of solvent → Mass . 1mol / molar mass

60.5 g . 1 mol/ 18 g = 3.36 mol

Total moles = 3.36 mol + 0.0590 mol = 3.419 moles

Xm = 0.0590 mol / 3.419 moles → 0.0172

Let's replace the data in the formula

42.2 mmHg - P' = 42.2 mmHg . 0.0172

P' = - (42.2 mmHg . 0.0172 - 42.2 mmHg)

P' = 41.4 mmHg

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i would say the the first 1

Explanation:

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2 years ago
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. Explain why some desert animals excrete uric acid rather than ammonia.<br>(2 marks)​
avanturin [10]

Answer:

AFAIK

Explanation:

uric acid is much less toxic than ammonia, hence bigger concentrations of it are tolerated in the body. This means you can excrete it while excreting very little water - beneficial wherever water's not abundant.

There's a tradeoff though, uric acid requires more energy to synthesize than ammonia, so pretty much all fish, say, excrete ammonia rather than uric acid - it's no problem to dilute ammonia since there's no water shortage.

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3 years ago
A quantity of gas occupies a volume of 804 mL at a temperature of 27 °C. At what temperature will the
irga5000 [103]

Answer:

T₂ = 150 K

Explanation:

Given data:

Initial volume of gas = 804 mL

Initial temperature = 27°C (27+273=300 K)

Final temperature = ?

Final volume = 402 mL

Solution:

The given problem will be solve through the Charles Law.

According to this law, The volume of given amount of a gas is directly proportional to its temperature at constant number of moles and pressure.

Mathematical expression:

V₁/T₁ = V₂/T₂

V₁ = Initial volume

T₁ = Initial temperature

V₂ = Final volume  

T₂ = Final temperature

Now we will put the values in formula.

V₁/T₁ = V₂/T₂

T₂ = V₂T₁/V₁  

T₂ = 402 mL × 300 K / 804 mL

T₂ = 120,600 mL.K / 804 mL

T₂ = 150 K

3 0
2 years ago
Which statement is true about the relationship between entropy and spontaneity? O Spontaneous reactions tend to lead to higher e
dimulka [17.4K]

Answer:

The true statement is: Spontaneous reactions tend to lead to higher entropy.

Explanation:

The spontaneity of a reaction is linked to the value of Gibbs free energy (ΔG°). The more negative is this value, the more spontaneous is a reaction. At the same time, Gibbs free energy depends on enthalpy (ΔH°) and entropy (ΔS°), according to the following expression:

ΔG° = ΔH° - T.ΔS°

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3 0
2 years ago
A 14.4-gg sample of granite initially at 86.0 ∘C∘C is immersed into 24.0 gg of water initially at 25.0 ∘C∘C. What is the final t
kari74 [83]

Answer:

The final temperature of both substances when they reach thermal equilibrium is 31.2 °C

Explanation:

Step 1: Data given

Mass of sample granite = 14.4 grams

Initial temperature = 86.0 °C

Mass of water = 24.0 grams

The initial temperature of water = 25.0 °C

The specific heat of water = 4.18 J/g°C

The specific heat of granite = 0.790 J/g°C

Step 2: Calculate the final temperature

Heat lost = heat gained

Qgranite = - Qwater

Q = m*c*ΔT

m(granite)*c(granite)*ΔT(granite) = -m(water)*c(water)*ΔT(water)

⇒with m(granite) = the mass of granite = 14.4 grams

⇒with c(granite) = The specific heat of granite = 0.790 J/g°C

⇒with ΔT⇒(granite) = the change of temperature of granite = T2 - T1 = T2 - 86.0 °C

⇒with m(water) = the mass of water = 24.0 grams

⇒with c(water) = The specific heat of water = 4.18 J/g°C

⇒with ΔT(water) = the change of temperature of granite = T2 - T1 = T2 -25.0°C

14.4 grams * 0.790 * (T2 - 86.0°C) = -24.0 *4.18 * (T2 - 25.0°C)

11.376T2 - 978.336 = -100.32T2 + 2508

111.696 T2 = 3486.336

T2 = 31.2 °C

The final temperature of both substances when they reach thermal equilibrium is 31.2 °C

4 0
3 years ago
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