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4 years ago

A chemist test new mixtures for paint samples and decides one mixtuees stickiness makes is worthless

Chemistry

1 answer:

monitta4 years ago

4 0

Answer:

known are those proposed by Patrick,1 and Howard,2 and the group scheme devised by ... are unsatisfactory for complicated food mixtures and entirely useless so far as ... Congdon's procedure is evidently based on qualitative tests made on the ... it was decided at the outset that on account of the complex food mixtures.

Explanation:

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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.

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.

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

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

What compound requires the most energy to melt

FinnZ [79.3K]

Answer: Tungsten. This requires a lot of energy. Molecules are held together by covalent bonds, which are strong.

7 0

3 years ago

What is the relationship between the classification of a reaction as endothermic or exothermic and the enthalpy of the reaction?

Aleks04 [339]

If Enthalpy is positive, the reaction is endothermic.
If Enthalpy is negative, the reaction is exothermic.

3 0

3 years ago

The number of molecules of ammonia, NH₃, in water solution required for complete ionization of one molecule of sulfuric acid, H₂

Natalija [7]

Answer:

6 that will be the answer

6 0

3 years ago

Describe how you would obtain a beaker of pure water from salty water

Maslowich

Answer:

You can boil or evaporate the water and the salt will be left behind as a solid. If you want to collect the water, you can use distillation. This works because salt has a much higher boiling point than water. One way to separate salt and water at home is to boil the salt water in a pot with a lid.

Explanation:

4 0

3 years ago