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Otrada [13]
3 years ago
9

A chemist dissolves 1.3 mol NaCl in a 2.0-kg sample of water. What is the boiling point of this solution? (Assume that pure wate

r boils at 100°C and that Kb for water = 0.512°C/m.)
Chemistry
2 answers:
mafiozo [28]3 years ago
8 0
For this problem, the solution is exhibiting some colligative properties since the solute in the solution interferes with some of the properties of the solvent. We use equation for the boiling point elevation for this problem. We do as follows:
<span>
ΔT(boiling point)  = (Kb)mi
</span>ΔT(boiling point)  = (0.512)(1.3/2.0)(2)
ΔT(boiling point)  = 0.67 degrees Celsius
<span>
T(boiling point) = 100 + 0.67 = 100.67 degrees Celsius</span>
lara31 [8.8K]3 years ago
7 0

Answer:

100.67°C

Explanation:

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What element is being oxidized in the following redox reaction? Pb 2+(aq) + NH4 +(aq) - Pb(s) + NO3 (aq)
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2 years ago
A mixture of CS2(g) and excess O2(g) is placed in a 10 L reaction vessel at 100.0 ∘C and a pressure of 3.10 atm . A spark causes
ziro4ka [17]

Answer:

PCO2  = 0.6 25 atm

PSO2  = 1.2 75 atm

PO2 = 0.6  atm

Explanation:

Step 1: Data given

Volume = 10.0 L

Temperature = 100.0 °C

Pressure = 3.10 °C

After reaction, the temperature returns to 100.0 ∘C, and the mixture of product gases (CO2, SO2, and unreacted O2) is found to have a pressure of 2.50 atm

Step 2: The balanced equation

CS2(g)+3O2(g)→CO2(g)+2SO2(g)

Step 3: Name the reactants and products

a = CS2

b = O2 before reaction

c = CO2

d = SO2

e = nS O2 after reaction with n = the number of moles

Step 4: Calculate moles before reaction

PV = nRT

n = PV/(RT)

(na + nb) = (3.10atm) * (10.0L) / ((0.08206 Latm/moleK) * (373.15K))

(na + nb) = 1.0124

Step 5: Calculate moles after reaction

PV = nRT

n = PV/(RT)

nc + nd + ne) = PV/(RT) = (2.50 atm)*(10.0L) / ((0.08206 Latm/moleK)*(373.15K))

(nc + nd + ne) = 0.816 moles

Step 6: Calculate mol fraction

For  1 mole CS2 we need 3 moles O2  to produce 1 mole of CO2 and 2 moles of SO2

moles O2 remaining = ne = nb - 3na

moles CO2 produced = nc = na

moles SO2 producted = nd = 2na

(nc + nd + ne) = 0.816 moles = nb - 3na + na + 2na = 0.816

nb = 0.816

. (na + nb) = 1.0124

na = 1.0124 moles - 0.816 moles = 0.208

which leads to  

nc = na = 0.208

nd = 2na = 2*0.208 = 0.416

ne = 0.816 - 3*0.208 = 0.192

mole fraction CO2 = 0.208 / (0.208 + 0.416 + 0.192) = 0.25

mole fraction SO2 = 0.416 / (0.208 + 0.416 + 0.192) = 0.5 1

mole fraction O2 = 0.192 /(0.208 + 0.416 + 0.192) = 0.24

Step 6: Calculate partial pressure

PCO2 = 0.25 * 2.50 atm = 0.6 25 atm

PSO2 = 0.51 * 2.50 atm = 1.2 75 atm

PO2 = 0.24 * 2.50 atm = 0.6  atm

Step 7: Control results

now let's verify a couple of things

PV = nRT

P = nRT/V

before rxn

P = (0.208 + 0.816) * (0.08206 L*atm/mole*K) * (373.15K) / (10.0L) ≈ 3.10 atm

after rxn

P = ((0.208 +0.416+0.192) * (0.08206 L*atm/mole*K) * (373.15K) / (10.0L) ≈ 2.50 atm

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Diano4ka-milaya [45]

Answer:

A C D I believe

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If a gas has a proportionality constant of 4.32 x 10-4 mol at room temperature for a particular solvent, what will the
inysia [295]

0.0467 X 10^{-4} M/kPa is the solubility of the gas when it exerts a partial pressure of 92.4kPa.

<h3>What is Henry's law?</h3>

Mathematically, we can get this from Henry's law

From Henry law;

Concentration = Henry constant × partial pressure

Thus Henry constant = \frac{Concentration}{partial \;pressure}

Henry constant = \frac{4.32 \;X \;10^{-4} mol}{92.4kPa}

= 0.0467 X 10^{-4} M/kPa

Hence, 0.0467 X 10^{-4} M/kPa is the solubility of the gas when it exerts a partial pressure of 92.4kPa.

Learn more about the Henry's law here:

brainly.com/question/16222358

#SPJ1

6 0
1 year ago
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