Suppose you have been given the task of distilling a mixture of hexane + toluene. Pure hexane has a refractive index of 1.375 and pure toluene has a refractive index of 1.497. You collect a distillate sample which has a refractive index of 1.441. Assuming that the refractive index of the hexane + toluene mixture varies linearly with mole fraction, what is the mole fraction of hexane in your sample?

Answers

Answer 1

Answer:

Answer:

0.4590

Explanation:

How the refractive index of the hexane + toluene mixture varies linearly with mole fraction, it means that the mole fraction is the fraction that each pure index contribute for the mixture index, so, calling xh the mole fraction of hexane and xt the mole fraction of toluene:

1.375xh + 1.497xt = 1.441

And, xh + xt = 1 (because there are only hexane and toluene in the mixture), so xt = 1- xh

1.375xh + 1.497(1-xh) = 1.441

1.375xh + 1.497 - 1.497xh = 1.441

-0.122xh = -0.056

xh = -0.056/(-0.122)

xh = 0.4590

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Suppose 0.10 mol of Cu(NO_3)_2 and 1.50 mol of NH_3 are dissolved in water and diluted to a total volume of 1.00 L. Calculate the concentrations of Cu(NH_3)_4^2+ and of Cu^2+ at equilibrium.

Answers

Explanation:

It is known that the coefficients change in concentration and in the exponents. Hence, the reaction equation will be as follows.

$Cu^(2+)(aq) + 4NH_(3)(aq) \rightarrow Cu(NH_(3))^(2)_(4)(aq)$

According to the ICE table,

$Cu^(2+)(aq) + 4NH_(3)(aq) \rightarrow Cu(NH_(3))^(2)_(4)(aq)$

Initial : 0.10 1.50 0

Change : -x -4x +x

Equilibrium: 0.10 - x 1.50 - 4x x

Hence, the mass action expression is as follows.

$K_(f) = ([Cu(NH3)^(2+)_(4)])/([Cu^(2+)][NH_(3)]_(4))$

= $(x)/((0.10 - x)(1.50 - 4x)^(4))$

As, the value of is huge, it means that the reaction is very product favored. Hence, we need to find the limiting reactant first and then we get to know what x should be.

In the given reaction ammonia is the limiting reactant, because there is less than 4 times the ammonia as the copper cation. Thus, we expect it to run out first, and so, x is approximately equal to 0.25 M.

So, putting the given values into the above equation as follows.

$1.03 * 10^(13) = (0.25)/((0.10 - 0.25)(1.50 - 4x))^(4)$

From here

$[NH_(3)] = 1.50 - 4x = ((2.33)/(1.03 * 10^(13)))^{(1)/(4)$

= M

Therefore, we can "re-solve" for x to get and verify that it is still ≈0.250 M.

x = $[Cu(NH_(3))^(2+)_(4)]$

= $\frac{1.50 - 2.31284 * 10{-4}}{4}]$

= 0.37491425 M

Thus, we can conclude that concentration of ( $Cu^(2+)$ ) is 0.37491425 M.

Be sure to answer all parts. A baseball pitcher's fastballs have been clocked at about 97 mph (1 mile = 1609 m). (a) Calculate the wavelength of a 0.148−kg baseball (in nm) at this speed. × 10 nm (Enter your answer in scientific notation) (b) What is the wavelength of a hydrogen atom at the same speed? nm

Answers

Answer:

a) The wavelength of the baseball is $1.033* 10^(-25) nm$ .

b) 9.131 nm is the wavelength of a hydrogen atom at the 43.35 m/s.

Explanation:

Velocity of the baseball = v = 97 mile/hour

1 mile = 1609 meter

1 hour = 3600 seconds

$v =(97* 1609 m)/(3600 s)=43.35 m/s$

Mass of baseball = m = 0.148 kg

Wavelength of the baseball: $\lambda$

$\lambda =(h)/(mv)$ De Broglie wavelength

h =Planck's constant

$=(6.626* 10^(-34) Js)/(0.148 kg* 43.35 m/s)$

$\lambda =1.033* 10^(-34) m$

$1 m=10^9 nm$

$\lambda = 1.033* 10^(-25) nm$

The wavelength of the baseball is $1.033* 10^(-25) nm$ .

Mass of the hydrogen atom = $m=1.674* 10^(-27) kg$

Velocity of hydrogen atom = u = 43.35 m/s

$\lambda =(h)/(mv)$ De Broglie wavelength

$=(6.626* 10^(-34) Js)/(1.674* 10^(-27) kg* 43.35 m/s)$

$\lambda =9.131* 10^(-9) m$

$1 m=10^9 nm$

$\lambda =9.131 nm$

9.131 nm is the wavelength of a hydrogen atom at the 43.35 m/s.

Final answer:

To calculate the wavelength of the baseball and hydrogen atom, we can use the wavelength formula. However, the wavelengths calculated are extremely small and cannot be practically detected.

Explanation:

To calculate the wavelength of the baseball, we can use the wavelength formula: λ = v/f. In this case, the velocity (v) of the baseball is given as 97 mph, which is equal to 97 * 1609 m/h. The frequency (f) can be calculated by dividing the speed of light (3 * 10^8 m/s) by the wavelength of the baseball.

For the hydrogen atom, we can use the same formula. However, we need to convert the hydrogen atom velocity to m/s. Once we have the velocity in m/s, we can calculate the wavelength by dividing the velocity by the frequency.

It is important to note that the wavelength calculated for the baseball and hydrogen atom are extremely small and cannot be practically detected by our senses or instruments.

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What mass of Nz will be needed to produce 31.5 grams of N2O5?4N2 + 502 --> 2N2O5
a) 158.3 grams
b) 38.64 grams
c) 4.96 grams
d) 16.34 grams

Answers

Answer: d) 16.34 grams

Explanation:

To calculate the moles :

$\text{Moles of solute}=\frac{\text{given mass}}{\text{Molar Mass}}$

$\text{Moles of} N_2O_5=(31.5g)/(108.01g/mol)=0.292moles$

The balanced chemical reaction given is:

$4N_2+5O_2\rightarrow 2N_2O_5$

According to stoichiometry :

As 2 moles of $N_2O_5$ are produced by= 4 moles of $N_2$

Thus 0.292 moles of $N_2O_5$ are produced by= = $(4)/(2)* 0.292=0.584moles$ of $N_2$

Mass of $N_2=moles* {\text {Molar mass}}=0.584moles* 28g/mol=16.34g$

Thus 16.34 g of $N_2$ will be needed

How many moles of carbondioxide are produced when 0.2mol of sodium carbonate react with excess hydrovhloric acid

Answers

Answer:

0.2 moles of CO₂ are produced

Explanation:

Given data:

Moles of CO₂ produced = ?

Moles of Na₂CO₃ react = 0.2 mol

Solution:

Chemical equation:

Na₂CO₃ + 2HCl → 2NaCl + CO₂ + H₂O

Now we will compare the moles of CO₂ with Na₂CO₃ .

Na₂CO₃ : CO₂

1 : 1

0.2 : 0.2

Thus, 0.2 moles of CO₂ are produced.

Grunge is a rock style from Detroit. True False

Answers

Answer:

FALSE

Explanation:

Grunge refers to the genre of rock music and the fashion inspired by it. It originated in the mid-1980s in Seattle, Washington State.

Grunge was described as the fusion of punk rock and heavy metal.

This genre of music became popular in the early mid-1990s and included lyrics based on the theme of emotional and social alienation, betrayal, abuse, trauma etc.

22. What is the mass in grams of each of the following?a. 3.011 x 1023 atoms F
b. 1.50 x 1023 atoms Mg
c. 4.50 x 1012 atoms Cl
d. 8.42 x 1018 atoms Br
e. 25 atoms W
f. 1 atom Au

Answers

The mass in grams of 3.011 x 10²³ atoms of F is 9.5 g.

The mass in grams of 1.50 x 10²³ atoms of Mg is 5.98 g.

The mass in grams of 4.50 x 10¹² atoms of Cl is 2.65 x 10⁻¹⁰ g.

The mass in grams of 8.42 x 10¹⁸ atoms of Br is 1.12 x 10⁻³ g.

The mass in grams of 25 atoms of W is 3.1 x 10⁻²¹ g.

The mass in grams of 1 atom of Au is 3.27 x 10⁻²² g.

What is the mass in grams of 3.011 x 10²³ atoms F?

The mass in grams of 3.011 x 10²³ atoms of F is calculated as follows;

6.023 x 10²³ atoms = 19 g of F

3.011 x 10²³ atoms F = ?

= (3.011 x 10²³ x 19 g)/(6.023 x 10²³)

= 9.5 g

The mass in grams of 1.50 x 10²³ atoms of Mg is calculated as follows;

6.023 x 10²³ atoms = 24g of Mg

1.5 x 10²³ atoms F = ?

= (1.5 x 10²³ x 24 g)/(6.023 x 10²³)

= 5.98 g

The mass in grams of 4.50 x 10¹² atoms of Cl is calculated as follows;

6.023 x 10²³ atoms = 35.5 g of Cl

4.5 x 10²³ atoms Cl = ?

= (4.5 x 10¹² x 35.5 g)/(6.023 x 10²³)

= 2.65 x 10⁻¹⁰ g

The mass in grams of 8.42 x 10¹⁸ atoms of Br is calculated as follows;

6.023 x 10²³ atoms = 80 g of Br

8.42 x 10¹⁸ atoms Br = ?

= (8.42 x 10¹⁸ x 80 g)/(6.023 x 10²³)

= 1.12 x 10⁻³ g

The mass in grams of 25 atoms of W is calculated as follows;

6.023 x 10²³ atoms = 74 g of W

25 atoms W = ?

= (25 x 74 g)/(6.023 x 10²³)

= 3.1 x 10⁻²¹ g

The mass in grams of 1 atom of Au is calculated as follows;

6.023 x 10²³ atoms = 197 g of Au

1 atom of Au = ?

= (1 x 197 g)/(6.023 x 10²³)

= 3.27 x 10⁻²² g

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Final answer:

This solution provides the calculations necessary to convert the number of atoms of various elements (smallest particle of an element) to grams. It does so by using the molar mass of each element and Avogadro's number.

Explanation:

The mass of atoms can be determined by using Avogadro's number (6.022 x 1023 atoms/mol) and the molar mass of the specific element (g/mol). We use these to create a conversion factor and multiply by the number of atoms given.

For F (fluorine), which has a molar mass of about 18.9984 g/mol, 3.011 x 1023 atoms F is 9.00 g F.
For Mg (magnesium), with molar mass of about 24.3050 g/mol, 1.5 x 1023 atoms Mg is 6.07 g Mg.
For Cl (chlorine), with molar mass of about 35.453 g/mol, 4.50 x 1012 atoms Cl is 2.67 x 10-10 g Cl.
For Br (bromine), with molar mass about 79.904 g/mol, 8.42 x 1018 atoms Br is 0.12 g Br.
For W (tungsten), with molar mass about 183.84 g/mol, 25 atoms W is 7.65 x 10-22 g W.
For Au (gold), with molar mass about 197.0 g/mol, 1 atom Au is 3.28 x 10-22 g Au.

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