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lora16 [44]
3 years ago
7

How many molecules are in 4.2 miles of water?​

Chemistry
1 answer:
viktelen [127]3 years ago
4 0

Answer:

2.5284 x 10^24

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The vapor pressure of liquid chloroform, CHCl3, is 100. mm Hg at 283 K. A 0.380 g sample of liquid CHCl3 is placed in a closed,
Katyanochek1 [597]

Answer:

a

No

b

100 mm Hg

Explanation:

From the question we are told that

The vapor pressure of CHCl3, is P = 100 \  mmHg =  \frac{100}{760}=  0.13156 \ atm

The temperature of CHCl3 is T  =  283 \  K

The volume of the container is V_c =  380mL =  380 *10^{-3}\  L

The temperature of the container is T_c  =  283 \  K

The mass of CHCl3 is m = 0.380 g

Generally the number of moles of CHCl3 present before evaporation started is mathematically represented as

n  =  \frac{m }{M }

Here M is the molar mass of CHCl3 with the value M  =  119.38 \ g/mol

=> n  =  \frac{ 0.380 }{119.38 }

=> n  =  0.00318 \  mols

Generally the number of moles of CHCl3 gas that evaporated is mathematically represented as

n_g  =  \frac{PV}{RT}

Here R is the gas constant with value R =  0.08206 L \  atm /mol\cdot  K

So

          n_g  =  \frac{0.13156* 380 *10^{-3} }{0.08206 * 283}

          n_g  =  0.00215 \  mols

Given that the number of moles of  CHCl3 evaporated is less than the number of moles of CHCl3  initially present , then it mean s that not all the liquid evaporated

At equilibrium the temperature of CHCl3 will be equal to the pressure of  air so the pressure at equilibrium is  100 mmHg

4 0
3 years ago
A 30ml sample of a liquid has a mass of 50 grams. What is the density of the liquid?
mash [69]

Answer:

<h2>Density = 1.67 g/mL</h2>

Explanation:

The density of a substance can be found by using the formula

Density =  \frac{mass}{volume}

From the question

mass = 50 g

volume = 30 mL

Substitute the values into the above formula and solve for the density

That's

Density =  \frac{50}{30}  \\  =  \frac{5}{3}  \\  = 1.66666...

Wr have the final answer as

<h3>Density = 1.67 g/mL</h3>

Hope this helps you

5 0
3 years ago
One of the compounds used to increase the octane rating of gasoline is toluene (pictured). Suppose 43.3 mL of toluene (d = 0.867
Thepotemich [5.8K]

<u>Answer:</u>

(A)

Density = Mass / Volume

So  

Mass = Density × Volume

= 0.867 g/mL \times 43.3mL = 37.5411 g Toluene

1C_6 H_5 CH_3  + 9 O_2  > 7 CO_2  + 4 H_2 O

Mole ratio of toluene : Oxygen is 1 : 9

$37.5411 g \text { Toluene } \times \frac{1 \text {mol} \text {toluene}}{92 g \text { toluene}} \times \frac{9 {mol} O_{2}}{1 \text {mol} \text { toluene }} \times \frac{32 g O_{2}}{1 {mol} O_{2}}=117 g O_{2}(\text {Answer})$

(B)

1 mole of Toluene produces 7 moles of CO_2 gas and 4 moles of H_2 O Vapour

So the mole ratio is 1 : 11

37.5411 g Toluene $\times \frac{1 \text { mol toluene }}{92 g \text { toluene }} \times \frac{11 \mathrm{mol} \text { gas }}{1 \text { mol toluene }} $$\\\\=4.49 \text { mol gaseous products (Answer) } $

(C)

1mole contains 6.022\times10^{23} molecules

37.5411 g Toluene $\times \frac{1 \text { mol toluene }}{92 g \text { toluene}} \times \frac{4 \mathrm{mol} \mathrm{H}_{2} \mathrm{O}}{1 \mathrm{mol} \text { toluene }} \times \frac{6.022 \times 10^{23} \text { molecules } \mathrm{H}_{2} \mathrm{O}}{1 \mathrm{mol} \mathrm{H}_{2} \mathrm{O}} $\\\\$=9.82 \times 10^{23} \text { molecules } \mathrm{H}_{2} \mathrm{O} \text { (Answer) } $

6 0
3 years ago
The Society of Automotive Engineers has established an accepted numerical scale to measure the viscosity of motor oil. For examp
natta225 [31]

Answer:

Most viscous to least viscous: (c)> (b)> (a)

Explanation:

For hydrocarbons, viscosity increases with increasing molar mass. Because increasing molar mass signifies increase in number of electrons in molecules.

We know that in non-polar hydrocarbons, only van der waal intermolecular force exists. Van der waal force is proportional to number of electrons in a molecule.

Therefore with increasing molar mass, van der waal force increases. hence molecules gets more tightly bind with each other resulting increase in viscosity.

Here molar mass order : (c)> (b)> (a)

Therefore viscosity order : (c)> (b)> (a)

5 0
3 years ago
How many particles are in one mole substance
Karolina [17]

Avogadro's number is defined as the number of elementary particles (molecules, atoms, compounds, etc.) per mole of a substance. It is equal to 6.022×1023 mol-1 and is expressed as the symbol NA.

6 0
3 years ago
Read 2 more answers
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