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Anna35 [415]
3 years ago
13

For this exercise, we will consider a solution of the nonvolatile solvent ethylene glycol, C2H4(OH)2, in water. If the solution

contains 1 mol of C2H4(OH)2 for each 4 mol of H2O, what is the mole fraction of water?
Chemistry
1 answer:
Lemur [1.5K]3 years ago
3 0

Answer:

Mole fraction of water is 0.8.

Explanation:

Mole fraction is defined as ratio of moles of single component to the sum of moles of all components present in the solution.

Moles of ethylene glycol = n_1=1 mol

Moles of water = n_2=4 mol

Mole fraction of ethylene glycol = \chi_1=\frac{n_1}{n_1+n_2}

\chi_1=\frac{1 mol}{1 mol+4 mol}=0.2

Mole fraction of water  = \chi_2=\frac{n_2}{n_1+n_2}

\chi_2=\frac{4 mol}{1 mol+4 mol}=0.8

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If you were preparing a defined, minimal medium for an aerobic, chemolithotrophic bacterium to grow on, which of the following c
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Answer:

I believe it's D. FE2+ and MgCl2...

Explanation:

Hope this helps!!

7 0
3 years ago
The ph of 0.015 m hno2 (nitrous acid) aqueous solution was measured to be 2.63. what is the value of pka of nitrous acid?
Licemer1 [7]
Nitrous acid<span> dissociates as follows:
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HNO₂(s) ⇄ H⁺(aq) + NO₂⁻(aq) 
           
According to the equation, an acid constant has the following form:

Ka = [H⁺] × [NO₂⁻ ] / [HNO₂] 

From pH, we can calculate the concentration of H⁺ and NO₂⁻:

[H⁺] = 10^-pH = 10^-2.63 = 0.00234 M = [NO₂⁻]

Now, the acid constant can be calculated:

Ka = 0.00234 x 0.00234 / 0.015  = 3.66 x 10⁻⁴

And finally,

pKa = -log Ka = 3.44 


7 0
3 years ago
How many moles of glucose (C6H12O6) are in 1.5 liters of a 4.5 M C6H12O6 solution?
lisov135 [29]
Moles of glucose = Molarity x volume solution 
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3 0
3 years ago
fig. 15 shows a simplified diagram of the electrolysis of a molten electrolyte containing lithium chloride. b. Describe how the
melisa1 [442]

During the electrolysis of the molten lithium chloride, the Lithium ions (Li⁺) at the cathode undergoes reduction, and the electron configuration of lithium becomes 1s²2s¹.

<h3>What is electrolysis?</h3>

Electrolysis can be described as the process in which the electric current is passed through the chemical compound to break them.  In this process, the atoms and ions are interchanged by the addition or removal of electrons.

The ions are allowed to move freely in this process. When an ionic compound is melted or dissolved in water then ions are produced which can move freely.

During the electrolysis of molten lithium chloride, the lithium ions reach the cathode and accept the electrons while chloride ions reach at anode and loss electrons to become chlorine gas.

At anode :  2 Cl⁻ →    Cl₂ + 2e⁻

At cathode:   2 Li⁺   +   2e⁻  →  Li

Learn more about electrolysis, here:

brainly.com/question/12054569

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8 0
1 year ago
The unit cell for cr2o3 has hexagonal symmetry with lattice parameters a = 0.4961 nm and c = 1.360 nm. If the density of this ma
IRINA_888 [86]

To calculate the packing factor, first calculate the area and volume of unit cell.

Area is calculated as:

A=6R^{2}\sqrt{3}

Here, R is radius and is related to a as follows:

R=\frac{a}{2}

Putting the value in expression for area,

A=6(\frac{a}{2})^{2}\sqrt{3}=1.5a^{2}\sqrt{3}

The value of a is 0.4961 nm

Since, 1 nm=10^{-7}cm

Thus, 0.4961 nm=4.961\times 10^{-8} cm

Putting the value,

Area=1.5(4.961\times 10^{-8}cm)^{2}\sqrt{3}=6.39\times 10^{-15}cm^{2}

Now, volume can be calculated as follows:

V=Area\times c

The value of c is 1.360 nm or 1.360\times 10^{-7} cm

Putting the value,

V=(6.39\times 10^{-15}cm^{2})\times (1.360\times 10^{-7} cm)=8.7\times 10^{-22}cm^{3}

now, number of atom in unit cell can be calculated by using the following formula:

n=\frac{\rho N_{A}V_{c}}{A}

Here, A is atomic mass of Cr_{2}O_{3} is 151.99 g/mol.

Putting all the values,

n=\frac{(5.22 g/cm^{3})(6.023\times 10^{23} mol^{-1})(8.7\times 10^{-22}cm^{3})}{(151.99 g/mol)}\approx 18

Thus, there will be 18 Cr_{2}O_{3} units in 1 unit cell.

Since, there are 2 Cr atoms and 3 oxygen atoms thus, units of chromium and oxygen will be 2×18=36 and 3×18=54 respectively.

The atomic radii of Cr^{3+} and O^{2-} is 62 pm and 140 pm respectively.

Converting them into cm:

1 pm=10^{-10}cm

Thus,

r_{Cr^{3+}}=6.2\times 10^{-9}cm

and,

r_{O^{2-}}=1.4\times 10^{-8}cm

Volume of sphere will be sum of volume of total number of cations and anions thus,

V_{S}=V_{Cr^{3+}}+V_{O^{2-}}

Since, volume of sphere is V=\frac{4}{3}\pi r^{3},

V_{S}=36\left ( \frac{4}{3}\pi (r_{Cr^{3+})^{3}} \right )+54\left ( \frac{4}{3}\pi (r_{O^{2-})^{3}} \right )

Putting the values,

V_{S}=36\left ( \frac{4}{3}(3.14) (6.2\times 10^{-9} cm)^{3}} \right )+54\left ( \frac{4}{3}(3.14) (1.4\times 10^{-8} cm)^{3}} \right )=6.6\times 10^{-22}\times 10^{-8}cm^{3}

The atomic packing factor is ratio of volume of sphere and volume of crystal, thus,

packing factor=\frac{V_{S}}{V_{C}}=\frac{6.6\times 10^{-22}cm^{3}}{8.7\times 10^{-22}cm^{3}}=0.758

Thus, atomic packing factor is 0.758.

6 0
3 years ago
Read 2 more answers
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