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3 years ago

Can somebody that is good at science help me with these two questions? Please and thank you!

Chemistry

1 answer:

laiz [17]3 years ago

8 0

1. The forces make the train float because the poles are the same on each side causing them to repell (north and north repell)

2. B is more stable. If you pushed A it would fall and that would cause more disturbance

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Solve as soon as possible with detailed step by step explanation

scoundrel [369]

Mr. Jones's prescription calls for 1.04 tablets per day. Based on this information, how many tablets should Mr. Jones take per day? a) 1.25 O b) 1.5 c) 1 O d) 2

8 0

3 years ago

In chemistry, what would the chemical reaction look like if you added graphite to a solution of lead (II) nitrate?

aleksandrvk [35]

Can i guess, am i gonna give a suspension, so the graphite is gonna settle at the bottom

6 0

3 years ago

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

At what temperature would a 1.15 m NaCl solution freeze, given that the van't Hoff factor for NaCl is 1.9? Kf for water is 1.86

MariettaO [177]

Answer:

-4.1°C

Explanation:

The freezing-point depression (ΔTf) is a colligative property that can be calculated using the following expression:

ΔTf = i × Kf × m

where,

i: van't Hoff factor

Kf: freezing-point depression constant

m: molality

ΔTf = 1.9 × 1.86°C/m × 1.15 m = 4.1 °C

The freezing-point of pure water is 0°C. The freezing-point of this solution is:

0°C - 4.1°C = -4.1°C

8 0

4 years ago

Which of the following would be the best technique to use in order to separate a liquid from a solution? (2 points)

tiny-mole [99]

Simple distillation :) hope I helped!

4 0

4 years ago

Read 2 more answers