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2 years ago

6

Diffusion is a general term referring to the net movement of atoms and molecules along a concentration gradient, from an area of

________ concentration to an area of _______ concentration.

1 answer:

aniked [119]2 years ago

7 0

Answer:

Higher concentration to an area of lower concentration

Explanation:

When you open a perfume bottle at a corner of a room, after a while, its fragrance can be perceived across a distance at the other end of the room. This is because, molecules of the compound in the fragrance have moved from the area of higher concentration in the perfume bottle, across a concentration gradient to a region of lower concentration at the other end of the room. This is diffusion.

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The smell of hot food moves faster than the smell of cold food. Which property of gas is responsible for this?

miskamm [114]

When pressure is added to a gas the molecules bounce around really fast and push against the walls of its container. That's why when you squeeze an empty water bottle you can crush it all the way. The smaller the container the more the molecules hit the walls faster and that's what creates pressure. While moving around the molecules get hot as they speed up. Heat = Pressure.

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2 years ago

Read 2 more answers

What happens to the electrons in an ionic bond?

Anton [14]

Answer:

In an ionic bonds, the metal loses electrons to become a positively charged cation, In which the nonmetal accepts those electrons to become a negatively charged anion.

Explanation:

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3 years ago

Read 2 more answers

. Write the balanced equation, then outline the steps necessary to determine the information requested in each of the following:

mafiozo [28]

Answer:

See explanation

Explanation:

a) The equation of the reaction is;

2Na + Cl2 ------>2NaCl

Number of moles of sodium = 10g/23 g/mol = 0.43 moles

If 2 moles of sodium reacts with 1 mole of Cl2

0.43 moles reacts with 0.43 * 1/2 = 0.215 moles of Cl2

Mass = 0.215 moles of Cl2 *71 g/mol = 15.265 g

b) Equation of the reaction;

HgO -> Hg + O2

1.252 moles of HgO 1.252/32 gmol = 0.039 moles

1 mole of HgO yields 1 mole of oxygen hence

0.039 moles of HgO yields 0.039 moles of oxygen

Mass of oxygen = 0.039 moles * 32 g/mol = 1.248 g

c) Equation of the reaction;

2NaNO3 -----> 2NaNO2 + O2

Number of moles of 128 g of oxygen = 128g/32 g/mol = 4 moles

2 moles of NaNO3 yields 1 mole of oxygen

x moles of NaNO3 yields 4 moles of oxygen

x = 8 moles of NaNo3

Mass of NaNO3 = 8 * 85 g/mol = 680 g of NaNo3

6 0

2 years ago

Express the van der Waals equation of state as a virial expansion in powers of 1/Vm and obtain expressions for B and C in terms

nirvana33 [79]

Answer:

$PV_{m} = RT[1 + (b-\frac{a}{RT})\frac{1}{V_{m} } + \frac{b^{2} }{V^{2} _{m} } + ...]$

B = b -a/RT

C = b^2

a = 1.263 atm*L^2/mol^2

b = 0.03464 L/mol

Explanation:

In the given question, we need to express the van der Waals equation of state as a virial expansion in powers of 1/Vm and obtain expressions for B and C in terms of the parameters a and b. Therefore:

Using the van deer Waals equation of state:

$P = \frac{RT}{V_{m}-b } - \frac{a}{V_{m} ^{2} }$

With further simplification, we have:

$P = RT[\frac{1}{V_{m}-b } - \frac{a}{RTV_{m} ^{2} }]$

Then, we have:

$P = \frac{RT}{V_{m} } [\frac{1}{1-\frac{b}{V_{m} } } - \frac{a}{RTV_{m} }]$

Therefore,

$PV_{m} = RT[(1-\frac{b}{V_{m} }) ^{-1} - \frac{a}{RTV_{m} }]$

Using the expansion:

$(1-x)^{-1} = 1 + x + x^{2} + ....$

Therefore,

$PV_{m} = RT[1+\frac{b}{V_{m} }+\frac{b^{2} }{V_{m} ^{2} } + ... -\frac{a}{RTV_{m} }]$

Thus:

$PV_{m} = RT[1 + (b-\frac{a}{RT})\frac{1}{V_{m} } + \frac{b^{2} }{V^{2} _{m} } + ...]$ equation (1)

Using the virial equation of state:

$P = RT[\frac{1}{V_{m} }+ \frac{B}{V_{m} ^{2}}+\frac{C}{V_{m} ^{3} }+ ...]$

Thus:

$PV_{m} = RT[1+ \frac{B}{V_{m} }+ \frac{C}{V_{m} ^{2} } + ...]$ equation (2)

Comparing equations (1) and (2), we have:

B = b -a/RT

C = b^2

Using the measurements on argon gave B = −21.7 cm3 mol−1 and C = 1200 cm6 mol−2 for the virial coefficients at 273 K.

$b = \sqrt{C} = \sqrt{1200} = 34.64[tex]cm^{3}/mol$ [/tex] = 0.03464 L/mol

a = (b-B)*RT = (34.64+21.7)*(1L/1000cm^3)*(0.0821)*(273) = 1.263 atm*L^2/mol^2

3 0

2 years ago

Calculate the solubility of benzene in water at 25 c in ppm. the required henry's law constant is 5.6 bar/mol/kg and benzene's s

kap26 [50]

The relationship between pressure and solubility of the gas is given by Henry's law as:

$S_g = kP_g$

where,

$S_g$ is the solubility of the gas.

$k$ is proportionality constant i.e. Henry's constant.

$P_g$ is pressure of the gas.

$k = 5.6 bar/mol/kg$ (given)

$P_g = 0.13 bar$ (given)

Substituting the values,

$S_g = 5.6 bar/mol/kg\times 0.13 bar = 0.728 mole/kg$

To convert $mole/kg$ to $g/kg$ :

Molar mass of benzene, $C_6H_6$ = $6\times 12+6\times 1 = 78 g/mol$

$0.728\times 78 = 56.784 g/kg$

Now for converting into $ppm$ :

Since, $1 ppm = 0.001 g/kg$

So, $56.784\times 1000 = 56784 ppm$ .

Hence, the solubility of benzene in water at $25^{o} C$ in $ppm$ is $56784 ppm$ .

7 0

3 years ago