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lana66690 [7]
3 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]3 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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Explanation:

NH₃+NO=N₂+H₂O;

NH₃+NO=N₂+6H₂O;

NH₃+6NO=N₂+6H₂O;

4NH₃+6NO=N₂+6H₂O;

4NH₃+6NO=5N₂+6H₂O.

3 0
1 year ago
A gas exerts a pressure of 1.35 atm. What is the equivalent measures in mmHg?
Pie
1.35atm(760mmHg/1atm)=1026mmHg
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The following reaction, catalyzed by iridium, is endothermic at 700 K: CaO(s) + CH4(g) + 2H2O (g) → CaCO3 (s) + 4H2 (g) For the
arsen [322]

Answer:

A. Increasing the temperature will favor forward reaction and more CaCo3 formed.

B. More CaCo3 will be formed.

C. CaCo3 will decrease and more react ants formed.

D. Less CaCo3 will be formed.

E. Iridium is a catalyst so there is no effect

Explanation:

A. Temperature will increase because it's an endothermic reaction.

B. Adding Cao will favor forward reaction and more CaCo3 formed.

C. Removing methane, more react ants are formed and CaCo3 decreases.

D. Irridi is a catalyst so it has no effect on the CaCo3 but only speeds its rate of reaction.

3 0
2 years ago
One gram of liquid benzene is burned in a bomb calorimeter. The temperature before ignition was 20.826 C, and the temperature af
Licemer1 [7]

Answer:

fH = - 3,255.7 kJ/mol

Explanation:

Because the bomb calorimeter is adiabatic (q =0), there'is no heat inside or outside it, so the heat flow from the combustion plus the heat flow of the system (bomb, water, and the contents) must be 0.

Qsystem + Qcombustion = 0

Qsystem = heat capacity*ΔT

10000*(25.000 - 20.826) + Qc = 0

Qcombustion = - 41,740 J = - 41.74 kJ

So, the enthaply of formation of benzene (fH) at 298.15 K (25.000 ºC) is the heat of the combustion, divided by the number of moles of it. The molar mass od benzene is: 6x12 g/mol of C + 6x1 g/mol of H = 78 g/mol, and:

n = mass/molar mass = 1/ 78

n = 0.01282 mol

fH = -41.74/0.01282

fH = - 3,255.7 kJ/mol

4 0
3 years ago
The solubility of a salt in a given volume of water depends primarily on the A) surface area of the salt crystals B) temperature
Darina [25.2K]

Answer:

The correct answer to the following question will be Option B (temperature of the water).

Explanation:

  • The solubility of that same salt which appears whenever the substance is added in some kind of a fluid which already incorporates another origin of many of its atoms.
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7 0
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