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azamat
3 years ago
13

An aqueous solution containing 100.0 g of nacl and 100.0 g of cacl2 has a volume of 1.00 l and a density of 1.15 g/ml. the vapor

pressure of pure water at 25 ∘c is 23.8 mmhg, and you can assume complete dissociation for both solutes
Chemistry
1 answer:
Lemur [1.5K]3 years ago
6 0
Missing question: What is the vapor pressure of the solution at 25°<span>C?
n(NaCl) = 100 g </span>÷ 58,4 g/mol.
n(NaCl) = 1,71 mol. 
NaCl → Na⁺ + Cl⁻, amount of ions are 2 · 1,71 mol = 3,42 mol.
n(CaCl₂) = 100 g ÷ 111 g/mol = 0,9 mol.
CaCl₂ → Ca²⁺ + 2Cl⁻, amount of ions 3 · 0,9 mol = 2,7 mol.
m(solution) = 1000 ml (1,00 L) · 1,15 g/ml = 1150 g.
m(H₂O) = 1150 g - 100 g - 100 g = 950 g.
n(H₂O) = 950 g ÷ 18 g/mol = 118,75 mol.
<span>water's mole fraction = 118,75 mol </span>÷ (118,75 mol + 2,7 mol + 3,42 mol).
water's mole fraction = 0,95.
p(solution) = 0,95 · 23 mmHg = 21,85 mmHg.

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Hello!

The chemical reaction for the dissolving of calcium fluoride is the following:

CaF₂(s) ⇄ Ca⁺²(aq) + 2F⁻(aq)

In this reaction, and according to Le Chatelier's principle, the action that would shift this reaction away from solid calcium fluoride and towards the dissolved ions is the removing of fluoride ions.

Le Chatelier's principle
states that in an equilibrium reaction, the system would shift in the opposite direction of the changes. If we remove fluoride ions from the system, it will shift towards the formation of more fluoride ions by dissolving more Calcium Fluoride to achieve equilibrium again.

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7 0
3 years ago
If 88.0 L of natural gas, which is essentially methane (CH4), undergoes complete combustion at 720. mm Hg and 22ºC, how many gra
mr_godi [17]

126 grams of H2O is formed.

Explanation:

Data given:

volume of the gas = 88 Liters

pressure = 720 mm Hg or 0.947 atm

temperature T = 22 Degrees or 295.15 K

R = 0.08021 atm L/mole K

n =?

The formula is used is of ideal gas law to know the number of moles of CH4 undergoing combustion.

PV = nRT

n = \frac{PV}{RT}

putting the values in the equation

   = 0.947 X 88/ 0.08021 X 295.15

n = 3.5 moles

balanced reaction for combustion of methane

CH4 + O2  ⇒ CO2 + 2H20

1 mole of CH4 undergoes combustion to form 2 moles of water

3.5 moles will give x moles of water

2/1 = x/3.5

x =  7 moles of water  (atomic mass of water = 18 gram/mole)

mass = atomic mass x number of moles

mass = 18 x 7

          =126 grams of water is formed.

7 0
3 years ago
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