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lana66690 [7]
2 years ago
12

A solution contains 0.182 molmol NaClNaCl and 0.897 molH2OmolH2O. Calculate the vapor pressure of the solution at 55 ∘C∘C. The v

apor pressure of pure water at 55 ∘C∘C is 118.1 torrtorr. (Assume that the solute completely dissociates.)
Chemistry
1 answer:
solong [7]2 years ago
5 0

Answer:

Vapor pressure of solution is 78.2 Torr

Explanation:

This is solved by vapor pressure lowering:

ΔP =  P° . Xm . i

Vapor pressure of pure solvent  (P°) - vapor pressure of solution  = P° . Xm . i

NaCl  →  Na⁺  +  Cl⁻     i = 2

Let's determine the Xm (mole fraction) These are the moles of solute / total moles.

Total moles = moles of solvent + moles of solute

Total moles = 0.897 mol + 0.182 mol → 1.079 mol

0.182 / 1.079 = 0.168

Now we replace on the main formula:

118.1° Torr - P' = 118.1° Torr . 0.168 . 2

P' = - (118.1° Torr . 0.168 . 2 - 118.1 Torr)

P' =  78.2 Torr

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What is the empirical formula for the following molecular formula: C10H5O2
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The empirical formula is the same as the molecular formula : C₁₀H₅O₂

<h3>Further explanation</h3>

Given

Molecular formula : C₁₀H₅O₂

Required

The empirical formula

Solution

The empirical formula (EF) is the smallest comparison of atoms of compound forming elements.  

The molecular formula (MF) is a formula that shows the number of atomic elements that make up a compound.  

(empirical formula) n = molecular formula  

<em>(EF)n=MF </em>

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If we divide by the number of moles of Oxygen (the smallest) which is 2 then the moles of Hydrogen will be a decimal number (not whole), which is 2.5, then the empirical formula is the same as the molecular formula

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Hydrogen peroxide can decompose to water and oxygen by the following reaction
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2 years ago
A 85.2 g copper bar was heated to 221.32 degrees Celsius and placed in a coffee cup calorimeter containing 4250 mL of water at 2
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Answer:

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Explanation:

A 85.2 g copper bar was heated to 221.32 degrees Celsius and placed in a coffee cup calorimeter containing 425 mL of water at 22.55 degrees Celsius. The final temperature of the water was recorded to be 26.15 degrees Celsius. What is the specific heat of the copper?

Step 1: Data given

Mass of copper = 85.2 grams

Temperature of copper = 221.32 °C

Volume of water = 425 mL

Temperature of water = 22.55 °C

Final temperature = 26.15 °C

Specific heat of water = 4.184 J/g°C

Step 2: Calculat the specific heat of copper

Heat lost = heat gained

Q = m*c*ΔT

Qcopper = -Qwater

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

⇒ m(copper) = 85.2 grams

⇒ c(copper) = TO BE DETERMINED

⇒ ΔT(copper) = the change in temeprature = T2 -T1 = 26.15 -221.32 = -195.17 °C

⇒ m(water) = The mass of water = 425 mL * 1g/mL = 425 grams

⇒ c(water) = The specific heat of water = 4.184 J/g°C

⇒ ΔT(water) = The change of temperature of water = 26.15 - 22.55 = 3.6

85.2 * c(copper) * (-195.17) = -425 * 4.184 * 3.6

c(copper) = 0.385 J/g°C

The specific heat of copper is 0.385 J/g°C

(Note, The original question says the volume of the water is 4250 mL. IF this is not an error, the specific heat of copper is 3.85 J/g°C (10x higher than the normal value).

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