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Wittaler [7]
2 years ago
8

The mole fraction of hcl in a solution prepared by dissolving 5.5 g of hcl in 200 g of c2h6o is ________. the density of the sol

ution is 0.79 g/ml.
Chemistry
1 answer:
Thepotemich [5.8K]2 years ago
5 0
The answer:
all that we search for is the number of mole of HCl and the number of mole of C2H6O

M(HCl) = 5.5g/ mole of HCl , so mole of HCl = 5.5/M(HCl), where M(HCl) is the molar mass.
M(HCl) = 1+ 36.5= 37.5 

moles of HCl = 5.5/37.5=0.14 

M(C2H6O) = 200g / moles of C2H6O, so moles of C2H6O=200g / M(C2H6O)

M(C2H6O)= 2x12+ 6 + 16=46,

moles of C2H6O=200g / 46 =<span>4.35 </span><span> moles 
</span>
the sum of the moles is    0.14 + <span>4.35 </span> = 4.501 moles

finally,  <span>The mole fraction of hcl in a solution prepared by dissolving 5.5 g of hcl in 200 g of c2h6o is 0.031
</span>
because it can be found by  0.14 / 4.501= 0.031

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Which of the following compounds has the lowest boiling point?
sergejj [24]

Answer : The correct answer is option A : Diethyl ether

Explanation :

The boiling point of a compound depends on the intermolecular forces (IMF) of attractions present among its molecules.

Stronger the IMF, more difficult it is to separate the molecules. And hence the compound shows higher boiling point.

The most common types of IMF are

a) Hydrogen bonding : Hydrogen bonding occurs when a compound has hydrogen atom directly attached to strongly electro-negative atoms like O, N and F

Hydrogen bondings are the strongest IMF. Therefore compounds that have hydrogen bondings show higher boiling point.

In the given examples, 2-butanol and 4-octanol both have -OH group where H is directly attached to highly electro-negative O atom. As a result hydrogen bonding is present in these compounds.

Therefore they show higher boiling point.

b) Dipole interactions : These are seen in case of polar compounds.

c) London dispersion forces : These are present in all the compounds but they are predominant in case of non polar compounds.

Both diethyl ether and diphenyl ether predominantly show London dispersion forces. Since these forces are weaker as compared to other IMF, the molecules having london dispersion tend to have lower boiling points.

But the magnitude of dispersion forces increases as the molecular weight of the compound increases.

Therefore diphenyl ether which has a molecular weight of 170 g/mol has much stronger london dispersion forces as compared to diethyl ether which has a molecular weight of 74 g/mol

From above discussion, we can conclude that diethyl ether has the weakest intermolecular forces of attraction. Hence it has the lowest boiling point.


6 0
2 years ago
Water is dissolved into n-butanol (a polar liquid). Which is the second step at the molecular level as water dissolves into n-bu
hodyreva [135]
Keeping it nice and simple, it is D- Water molecules surrounded by water molecules who are then carried into n-butanol

Hope this helps

the real answer is D n butanol molecules are attracted to the surface of the water molecules

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5 0
2 years ago
Read 2 more answers
The rate law for the reaction
juin [17]

Explanation:

Since, the given reaction is as follows.

              2NO(g) + Cl_{2}(g) \rightarrow 2NOCl(g)

Hence, rate law of the reaction is as follows.

               R = k[NO][Cl_{2}]

As it is known that rate of a reaction depends on the initial concentration of products. So here, the rate of reaction will depend on the concentration of NO and Cl_{2}. Since, power of the concentrations of each of these is equal to 1. Therefore, order of the reaction is equal to 1 + 1 = 2.

According to the rate law, reactants involved in the rate determining step are NO and Cl_{2}. Hence, first step of the mechanism is the rate determining step.

Also, according to the rate of reaction doubling the concentration of NO will double the rate of reaction.

The number of reactants taking part in a single step of the reaction is known as molecularity of the reaction. Therefore, molecularity of the first step of the reaction is 2.

Both the given steps are not termolecular.

5 0
2 years ago
Calculate the molarity of each solution.
weeeeeb [17]
Q1)
molarity is defined as the number of moles of solute in 1 L solution 
the number of moles of LiNO₃ - 0.38 mol
volume of solution - 6.14 L
since molarity is number of moles in 1 L 
number of moles in 6.14 L - 0.38 mol
therefore number of moles in 1 L - 0.38 mol / 6.14 L = 0.0619 mol/L
molarity of solution is 0.0619 M

Q2)
the mass of C₂H₆O in the solution is 72.8 g
molar mass of C₂H₆O is 46 g/mol 
number of moles = mass present / molar mass of compound
the number of moles of C₂H₆O - 72.8 g / 46 g/mol 
number of C₂H₆O moles - 1.58 mol
volume of solution - 2.34 L
number of moles in 2.34 L - 1.58 mol
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molarity of C₂H₆O is 0.675 M

Q3)

Mass of KI in solution - 12.87 x 10⁻³ g
molar mass - 166 g/mol
number of mole of KI = mass present / molar mass of KI
number of KI moles = 12.87 x 10⁻³ g / 166 g/mol = 0.0775 x 10⁻³ mol
volume of solution - 112.4 mL 
number of moles of KI in 112.4 mL - 0.0775 x 10⁻³ mol
therefore number of moles in 1000 mL- 0.0775 x 10⁻³ mol / 112.4 mL x 1000 mL
molarity of KI - 6.90 x 10⁻⁴ M
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If an electronic balance reports an object weighing 35.9920g what will be displayed when
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Answer:

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Explanation:

It will show the number out to the 4th decimal place.  Exactly 2 g will be displayed as 2.0000 g.

8 0
2 years ago
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