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2 years ago

13

Classify each of the following aqueous solutions as a nonelectrolyte, weak electrolyte, or strong electrolyte.

1 answer:

OLEGan [10]2 years ago

6 0

<h3>Answer:</h3>

a) LiClO4 - Strong electrolyte

b) HClO -Weak electrolyte

c) CH3CH2CH2OH - Non-electrolyte

d) HClO3 - Strong electrolyte

e) CuSO4 -strong electrolyte

f) C12H22O11-Non-electrolyte

<h3>Explanation:</h3>

An electrolyte is a substance in an aqueous or molten form which is decomposed by passing an electric current through it. Electrolytes ionize to ions which are responsible for the conduction of electric charge.
Non-electrolytes are substances that do not ionize into cations and anions and thus do not conduct. They include molecular compounds such as gases.
Electrolytes may be weak or strong depending on the level of ionization.
Weak electrolytes are those that undergo partial ionization while strong electrolytes completely ionize.

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Cyclohexane has a freezing point of 6.50 ∘C and a Kf of 20.0 ∘C/m. What is the freezing point of a solution made by dissolving 0

ExtremeBDS [4]

Answer:

The freezing point will be $2.9^{0}C$

Explanation:

The depression in freezing point is a colligative property.

It is related to molality as:

$Depressioninfreezing point=K_{f}Xmolality$

Where

Kf= $20\frac{^{0}C}{m}$

the molality is calculated as:

$molality=\frac{moles_{solute}}{mass_{solvent}}$

$moles=\frac{mass}{molarmass}=\frac{0.694}{154}=0.0045mol$

$massofcyclohexane=25g=0.025Kg$

$molarity=\frac{0.0045}{0.025}=0.18m$

Depression in freezing point = $20X0.18=3.6^{0}C$

The new freezing point = $6.5^{0}C-3.6^{0}C=2.9^{0}C$

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2 years ago

The standard heats of combustion (δh∘) per mole of 1,3-butadiene, c4h6(g); butane, c4h10(g); and h2(g) are −2540.2,−2877.6, and

ryzh [129]

solution:

Hydration is the addition of water; hydrogenation is the addition of hydrogen.

desire rxn: _C4H6(g) + 2 H2(g)-----> C4H10(g)___dHhy = ??

knowns:

__________C4H6 + 11/2 O2 --------> 4CO2 + 3H2O______dHox = -2540.2 kJ/mole

__________4CO2 + 5H2O -----------> C4H10 + 13/2 O2___-dHox = 2877.6 kJ/mole

___________2(1/2 O2 + H2 -------------> H2O)___________2*dHox = 2(-285.8 kJ/mole)

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mote1985 [20]

Arkeisha is correct because the fluid in an alkaline battery has a ph between 7.1 and 14.0

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2 years ago

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Vinvika [58]

The answer to your question is d

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2 years ago

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For each reaction, find the value of ΔSo. Report the value with the appropriate sign. (a) 3 NO2(g) + H2O(l) → 2 HNO3(l) + NO(g)

aev [14]

Answer:

ΔS° = -268.13 J/K

Explanation:

Let's consider the following balanced equation.

3 NO₂(g) + H₂O(l) → 2 HNO₃(l) + NO(g)

We can calculate the standard entropy change of a reaction (ΔS°) using the following expression:

ΔS° = ∑np.Sp° - ∑nr.Sr°

where,

ni are the moles of reactants and products

Si are the standard molar entropies of reactants and products

ΔS° = [2 mol × S°(HNO₃(l)) + 1 mol × S°(NO(g))] - [3 mol × S°(NO₂(g)) + 1 mol × S°(H₂O(l))]

ΔS° = [2 mol × 155.6 J/K.mol + 1 mol × 210.76 J/K.mol] - [3 mol × 240.06 J/K.mol + 1 mol × 69.91 J/k.mol]

ΔS° = -268.13 J/K

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2 years ago