A state government is trying to decide between using biomass or solar power

Grey Goose ® vodka has an alcohol content of 40.0 % (v/v). Assuming that vodka is composed of only ethanol and water answer the

Vitek1552 [10]

Explanation:

Grey Goose vodka has an alcohol content of 40.0 % (v/v).

Volume of vodka = V = 100 mL

This means that 40.0 mL of alcohol is present 100 mL of vodka.

Volume of ethanol=V' = 40.0 mL

Mass of ethanol = m

Density of the ethanol = d = 0.789 g/mL

$m=d\times V' = 0.789 g/ml\times 40.0 mL=31.56 g$

Volume of water = V''= 100 ml - 40.0 mL = 60.0 mL

Mass of water = m'

Density of the water = d' = 1.00 g/mL

$m'=d'\times V'' = 1.00 g/ml\times 60.0 mL=60.0 g$

a.)

Moles of ethanol = n= $\frac{31.56 g}{46g/mol}=0.6861 mol$

Volume of vodka = V = 100 mL = 0.100 L ( 1mL=0.001 L)

Molarity of the ethanol:

$=\frac{0.6861 mol}{0.100 L}=6.861 M$

6.861 M the molarity of ethanol in this vodka.

b) Mass of ethanol = 31.56 g

Moles of ethanol = n= $\frac{31.56 g}{46g/mol}=0.6861 mol$

Volume of vodka = V = 100 mL

Mass of vodka = m

Density of the water = D = 0.935 g/mL

$M=D\times V=0.935 g/ml\times 100 ml=93.5 g$

The percent by mass of ethanol % (m/m):

$\frac{31.56 g}{93.5 g}\times 100=33.75\%$

33.75% is the percent by mass of ethanol % (m/m) in this vodka.

c)

Moles of ethanol = n= $\frac{31.56 g}{46g/mol}=0.6861 mol$

Mass of solvent that is water = 60.0 g = 0.060 kg ( 1g = 0.001 kg)

Molality of ethanol in vodka :

$m=\frac{0.6861 mol}{0.060 kg}=11.435 m$

11.435 m is the molality of ethanol in this vodka.

d)

Moles of ethanol = $n_1=\frac{31.56 g}{46g/mol}=0.6861 mol$

Moles of water = $n_2=\frac{60.0 g}{18 g/mol}=3.333 mol$

Mole fraction of ethanol = $\chi_1$

$\chi_1=\frac{n_1}{n_1+n_2}=\frac{0.6861 mol}{0.6861 mol+3.333 mol}$

= 0.1707

Mole fraction of water = $\chi_2$

$\chi_2=\frac{n_2}{n_1+n_2}=\frac{3.3333 mol}{0.6861 mol+3.333 mol}$

= 0.8290

e)

The vapor pressure of vodka = P

Mole fraction of ethanol = $\chi_1=0.1707$

Mole fraction of water = $\chi_2=0.8290$

The vapor pressures of ethanol = $p_1=45.0 Torr$

The vapor pressures of pure water = $p_2=23.8Torr$

$P=\chi_1\times p_1+\chi_2\times p_2$

$P=0.1707\times 45.0torr+0.8290\times 23.8 Torr=27.41 torr$

The vapor pressure of vodka is 27.41 Torr.

5 0

3 years ago

How much heat, in joules and in calories, must be added to a 75.0Â–g iron block with a specific heat of 0.449 j/g Â°c to increas

lianna [129]

Answer:

$Q=50,849.25J\\\\Q=12,153.3cal$

Explanation:

Hello!

In this case, given the mass, temperature change and specific heat, it is possible to compute the required heat in joules as shown below:

$Q=mC(T_2-T_1)\\\\Q=75.0g*0.449\frac{J}{g\°C}(1535\°C-25\°C)\\\\Q=50,849.25J$

Now, since 1 cal =4.184 J, this result in calories is:

$Q=50,849.25J*\frac{1cal}{4.184J}\\\\Q=12,153.3cal$

Best regards!

7 0

3 years ago

Calculate the following:

d1i1m1o1n [39]

Answer:

Explanation:

1) Given data:

Number of moles of lead = 4.3×10⁻³ mol

Mass of lead = ?

Solution:

Mass = number of moles × molar mass

Molar mass of lead = 207.2 g/mol

Mass = 4.3×10⁻³ mol × 207.2 g/mol

Mass = 890.96 g

2) Given data:

Number of atoms of antimony = 3.8×10²² atoms

Mass of antimony = ?

Solution:

1 mole contain 6.022 ×10²³ atoms

3.8×10²² atoms × 1 mol / 6.022 ×10²³ atoms

0.63×10⁻¹ mol

0.063 mol

Mass = number of moles × molar mass

Molar mass of lead = 121.76 g/mol

Mass = 0.063 mol × 121.76 g/mol

Mass = 7.67 g

3) Given data:

Mass of tungsten = 15.5 Kg (15.5 kg × 1000 g/ 1kg = 15500 g)

Number of atoms = ?

Solution:

Number of moles of tungsten:

Number of moles = mass/molar mass

Number of moles = 15500 g / 183.84 g/mol

Number of moles = 84.3 mol

1 mole contain 6.022 ×10²³ atoms

84.3 mol × 6.022 ×10²³ atoms / 1mol

507.65 ×10²³ atoms

8 0

3 years ago

I need help please!!!!!!!!!!!!!!!!!

AleksAgata [21]

Answer:I think its both

Explanation:

7 0

3 years ago

What is the speed of a wave on a guitar string with a mass of

jok3333 [9.3K]

Answer: 4.5

Explanation:

6 0

3 years ago