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Snezhnost [94]
3 years ago
10

What is the mole fraction of ethanol when 10.00 ml of pure ethanol is combined with 2.00 ml of water? the volume of 10 ml of the

solution must be measured?
Chemistry
1 answer:
GenaCL600 [577]3 years ago
3 0
The mole fraction is calculated using the formula:

mole fraction of component A = # of moles of component A / # of total moles of the solution.

A) number of moles of ethanol

To calculate the number of moles of ethanol, you need its density, which will permit you to determine the mass of the 10.00 ml, and then convert into moles using the molar mass of ethanol.

The normal density of ethanol is 0.789 g/ml

density = mass / volume => mass = density  * volume = 0.789 g/ml * 10.00 ml = 7.890 g

Molar mass of ethanol = 46.07 g/mol

number of moles = mass / molar mass = 7.890g / 46.07 g/mol = 0.1713 mol

B) number of moles of water

density of water = 1.00 g/mol

mass of water = density * volume = 1.00 g/mol * 2.00 ml = 2.00 g

number of moles of water = mass / molar mass = 2.00 g / 18.0 g/mol = 0.111  mol

C) mole fraction

mole fraction of ethanol = number of moles of ethanol / number of moles of solution

number of moles of ethanol = 0.1713 / (0.1713 + 0.111) = 0.1713 / 0.2824 = 0.607

Answer: 0.607

The volume of the final solution may  be calculated by adding the volume of the two components. This is 10.00 ml of ethanol + 2.00 ml of water makes 12.00 ml of solution.

It is not clear what the second question is meant for. Some context is missing. If you know density and you know maqss (or can calculate the mass from other data) you do not need to measure the volume.
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padilas [110]
First, draw the 2-hexene. Th is is a molecule of six carbons with a double bond in the second carbon:

      CH3 - CH = CH2 - CH2 - CH2 - CH3

Secong, put one Br on the second carbon and one Br on the third carbon:

    CH3 - CBr = CBr - CH2 - CH2 - CH3

Third, cis means that the two Br are placed in opposed positions, this is drawn with one Br up and the other down. So, you need to represent the position of the Br in the space:


      H   Br          H     H      H
       |    |             |       |       |
H - C - C = C -  C  -  C  -  C - H
       |           |      |       |       |
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The important fact to realize is that the two Br are in opposed sides of the molecule.
4 0
3 years ago
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The metabolic oxidation of glucose, C6H12O6, in our bodies produces CO2, which is expelled from our lungs as a gas.
enot [183]

Answer:

\large \boxed{\text{21.6 L}}

Explanation:

We must do the conversions

mass of C₆H₁₂O₆ ⟶ moles of C₆H₁₂O₆ ⟶ moles of CO₂ ⟶ volume of CO₂

We will need a chemical equation with masses and molar masses, so, let's gather all the information in one place.

Mᵣ:        180.16

         C₆H₁₂O₆ + 6O₂ ⟶ 6CO₂ + 6H₂O

m/g:      24.5

(a) Moles of C₆H₁₂O₆

\text{Moles of C$_{6}$H$_{12}$O}_{6} = \text{24.5 g C$_{6}$H$_{12}$O}_{6}\times \dfrac{\text{1 mol C$_{6}$H$_{12}$O}_{6}}{\text{180.16 g C$_{6}$H$_{12}$O}_{6}}\\\\= \text{0.1360 mol C$_{6}$H$_{12}$O}_{6}

(b) Moles of CO₂

\text{Moles of CO}_{2} =\text{0.1360 mol C$_{6}$H$_{12}$O}_{6} \times \dfrac{\text{6 mol CO}_{2}}{\text{1 mol C$_{6}$H$_{12}$O}_{6}} = \text{0.8159 mol CO}_{2}

(c) Volume of CO₂

We can use the Ideal Gas Law.

pV = nRT

Data:

p = 0.960 atm

n = 0.8159 mol

T = 37  °C

(i) Convert the temperature to kelvins

T = (37 + 273.15) K= 310.15 K

(ii) Calculate the volume

\begin{array}{rcl}pV &=& nRT\\\text{0.960 atm} \times V & = & \text{0.8159 mol} \times \text{0.082 06 L}\cdot\text{atm}\cdot\text{K}^{-1}\text{mol}^{-1} \times \text{310.15 K}\\0.960V & = & \text{20.77 L}\\V & = & \textbf{21.6 L} \\\end{array}\\\text{The volume of carbon dioxide is $\large \boxed{\textbf{21.6 L}}$}

7 0
2 years ago
A substance has a volume of 10.0 cm3 and a mass of 89 grams. What is its density?
Irina18 [472]
Remembering that
d = m ÷ v

d = ?
m = 89 g
v = 10 cm³

Therefore:

d = 89 ÷ 10

d = 8,9 g÷cm³
5 0
3 years ago
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Mg3N2(s)+6H2O(l)→3Mg(OH)2(s)+2NH3(g) When 36.0 g of H2O react, how many grams of NH3 are produced? When 36.0 g of H2O react, how
Afina-wow [57]

Answer:

11.3 g of NH_{3} are produced from 36.0 g of H_{2}O

Explanation:

1. The balanced chemical equation is the following:

Mg_{3}N_{2}(s)+6H_{2}O(l)=3Mg(OH)_{2}(s)+2NH_{3}(g)

2. Use the molar mass of the H_{2}O, the molar mass of the NH_{3} and the stoichiometry of the balanced chemical reaction to find how many grams of NH_{3} are produced:

Molar mass H_{2}O = 18\frac{g}{mol}

Molar mass NH_{3} = 17\frac{g}{mol}

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Therefore 11.3 g of NH_{3} are produced from 36.0 g of H_{2}O

3 0
3 years ago
How many moles of atoms are contained in 15.7 g of Bromine
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1 mol of Br = 79.9 g

15.7 g / 79.9 g = 0.196 moles of atoms
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3 years ago
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