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ratelena [41]
3 years ago
8

Which aqueous solution has the lowest boiling point?

Chemistry
1 answer:
soldi70 [24.7K]3 years ago
8 0

Answer : The correct option is, (a) 1.25 M C_6H_{12}O_6

Explanation :

Formula used for Elevation in boiling point :

\Delta T_b=i\times k_b\times m

where,

\DeltaT_b = change in boiling point

k_b = boiling point constant

m = molality

i = Van't Hoff factor

As per question, we conclude that the molality of the given solution are same. So, the boiling point depends only on the Van't Hoff factor.

Now we have to calculate the Van't Hoff factor for the given solutions.

(a) C_6H_{12}O_6 is a non-electrolyte solute that means they retain their molecularity, an not undergo association or dissociation.

So, Van't Hoff factor = 1

(b) The dissociation of KNO_3 will be,

KNO_3\rightarrow K^++NO_3^-

So, Van't Hoff factor = Number of solute particles = 1 + 1 = 2

(C) The dissociation of KNO_3 will be,

AlCl_3\rightarrow Al^[3+}+3Cl^-

So, Van't Hoff factor = Number of solute particles = 1 + 2 = 3

The boiling point depends only on the Van't Hoff factor.That means lower the Van't Hoff factor, lower will be the boiling point and vice-versa.

Hence, the correct option is, (a) 1.25 M C_6H_{12}O_6

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Answer:

<h2> 2 is the coefficient of CO₂</h2>

Explanation:

The <u>coefficient</u> is the number in front of a substance that indicates how many molecules or atoms are. For example, when we see 5He, it shows that there are 5 atoms of helium. When we see 5H₂O, it indicates that there are 5 molecules of water, each having 2 atoms of hydrogen and 1 atom of oxygen, so a total of 10 hydrogen atoms and 5 oxygen atoms.

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According to the following electron structure, how many valence electrons do these fluorine atoms have? F= 1s12s22p5
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The highest sequence for this shell is the number 2, preceding both the s and p in the diagram. This means that the outermost shell is the second level shell. In this shell, there are 7 electrons, 2 in the 2s orbital and 5 in the 2p orbital.

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A chemical compound has a molecular weight of 89.05 g/mole. 1.400 grams of this compound underwent complete combustion under con
Nataly_w [17]

Answer:

\Delta _{comb}H=-2,265\frac{kJ}{mol}

Explanation:

Hello!

In this case, for such calorimetry problem, we can notice that the combustion of the compound releases the heat which causes the increase of the temperature by 11.95 °C, it means that we can write:

Q _{comb}=-C_{calorimeter}\Delta T_{calorimeter}

In such a way, we can compute the total released heat due to the combustion considering the calorimeter specific heat and the temperature raise:

Q _{comb}=-2980\frac{J}{\°C} *11.95\°C\\\\Q _{comb}=-35,611J

Next, we compute the molar heat of combustion of the compound by dividing by the moles, considering 1.400 g were combusted:

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Thus, we obtain:

\Delta _{comb}H=\frac{Q_{comb}}{n}=\frac{-35,611J}{0.01572mol}  \\\\\Delta _{comb}H=-2,265,331\frac{J}{mol}*\frac{1kJ}{1000J}  \\\\\Delta _{comb}H=-2,265\frac{kJ}{mol}

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3 years ago
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Answer:

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Explanation:

El pH de un ácido débil, HX, se obtiene haciendo uso de su equilibrio:

HX(aq) ⇄ H⁺(aq) + X⁻(aq)

Donde la constante de equilibrio, Ka, es

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Como los iones H⁺ y X⁻ vienen del mismo equilibrio podemos decir:

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