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aalyn [17]
2 years ago
7

An aqueous antifreeze solution is 70.0% ethylene glycol (c2h6o2) by mass. the density of the solution is 1.75 g/cm3. calculate t

he molality, molarity, and mole fraction of the ethylene glycol.
Chemistry
1 answer:
AVprozaik [17]2 years ago
6 0

Molarity = moles of solute / kg solvent

 

Assume you have exactly 1 kg of solution. 70% of that, or 700 grams of the solution is C2H6O2, and 300 g of the solution is water.

 

Moles C2H6O2 = 700 g / 62.07 g/mol = 11.28 mol C2H6O2

 

molarity = 11.28 mol / 0.3 kg H2O = 37.59 molal

__________________________

 

Molarity = moles of solute/L of solution

there is 1 kg of solution has a volume of:

 

1000 g / 1.05 g/mL = 952 mL = 0.952 L

 

Molarity = 11.28 mol / 0.952 L = 11.85 M

___________________________

 

Mole fraction ethylene glycol = moles ethylene glycol / (moles ethylene glycol + moles H2O)

 

moles ethylene glycol = 11.28

moles H2O = 300 g / 18.0 g/mol = 16.67 mol H2O

 

mole fraction = 11.28 /( 11.28 + 16.67) = 0.404

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pickupchik [31]

Answer:

[Ca²⁺] = 1M

[NO₃⁻] = 2M

Explanation:

Calcium nitrate dissociates in water as follows:

Ca(NO₃)₂ ⇒ Ca²⁺ + 2NO₃⁻

The moles of Ca²⁺ can be found using the molar relationship between Ca(NO₃)₂ and Ca²⁺

(0.100mol Ca(NO₃)₂) (Ca²⁺ /Ca(NO₃)₂) = 0.100 mol Ca²⁺

The concentration of Ca²⁺  is then:

[Ca²⁺] = n/V = (0.100mol)/(100.0mL) x (1000ml)/(1L) = 1M

Similarly, moles of NO₃⁻ can be found using the molar relationship between Ca(NO₃)₂ and NO₃⁻:

(0.100mol Ca(NO₃)₂) (2NO₃⁻/Ca(NO₃)₂) = 0.200 mol NO₃⁻

The concentration of NO₃⁻ is then:

[NO₃⁻] = (0.200mol)/(100.0mL) x (1000ml)/(1L) = 2M

6 0
3 years ago
Using this formula, to find K, what does E0 represent?
Marysya12 [62]
The correct answer would be A. The symbol Eo would represent the cell potential of an electrolytic cell. This potential is being created by two metals that possess different properties. The energy per charge that is available from the reaction of the metals is the measure of this potential and is related to the equilibrium constant, K.
6 0
2 years ago
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How many mole of ZnCl2 will be produced from 55.0 g of Zn, assuming HCL is available in excess
Ilia_Sergeevich [38]

Answer:

0.84 mol

Explanation:

Given data:

Moles  of ZnCl₂ produced = ?

Mass of Zn = 55.0 g

Solution:

Chemical equation:

2HCl + Zn  →  ZnCl₂ + H₂

Number of moles of Zn:

Number of moles = mass / molar mass

Number of moles = 55.0 g/ 65.38 g/mol

Number of moles = 0.84 mol

Now we will compare the moles of Zn with ZnCl₂ from balance chemical equation.

                                      Zn          :             ZnCl₂

                                         1          :               1

                                      0.84       :           0.84

So from 55 g of Zn 0.84 moles of zinc chloride will be produced.

8 0
3 years ago
How is an electron orbital similar to a parabola?
AVprozaik [17]

Answer:

Explained below.

Explanation:

First of all, the orbital path of electron is mostly parabolic in electric field.

In an electric field, electrons behave very similar to a projectile. Thus, Electrons have a parabolic path in an electric field simply because the speed of the electrons in a direction which is perpendicular to the electric field is constant since there is no force. Therefore, there will be no acceleration along that perpendicular direction. However there will be an acceleration that is constant in the direction of the electric field which makes it act in a similar manner to a projectile under gravity.

5 0
3 years ago
A 1.04L sample of gas of 759 mm Hg pressure is expanded
Marta_Voda [28]

Answer:

P2 = 352 mm Hg (rounded to three significant figures)

Explanation:

PV = nRT

where P is the pressure,

V is the volume,

n is the moles of gas,

R is the gas constant,

and T is the temperature.

We must relate this equation to a sample of gas at two different volumes however. Looking at the equation, we can relate the change in volume by:

P1V1 = P2V2

where P1 is the initial pressure,

V1 is the initial volume,

P2 is the final pressure,

and V2 is the final volume.

Looking at this relationship, pressure and volume have an indirect relationship; when one goes up, the other goes down. In that case, we can use this equation to solve for the new pressure.

P1V1 = P2V2

(759 mm Hg)(1.04 L) = P2(2.24 L)

P2 = 352 mm Hg (rounded to three significant figures)

8 0
2 years ago
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