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Juliette [100K]
3 years ago
7

What does viscosity describe?

Chemistry
1 answer:
aev [14]3 years ago
5 0
The density of fluid
You might be interested in
C2F4 effuses through a barrier at a rate of 4.6x10-6 mol/hour, while an unknown gas effuses at a rate of 5.8x10-6 mol/hour. What
umka21 [38]
The  molar mass  of  the unknown  compound  is   calculated   as   follows

let the unknown  gas be represented by   letter  Y

Rate of C2F4/  rate of  Y  = sqrt of   molar  mass of gas Y/ molar mass of  C2F4

 =  (4.6  x10^-6/ 5.8  x10^-6)  = sqrt  of  Y/ 100

remove  the  square  root  sign  by  squaring  in both  side

(4.6  x  10^-6 / 5.8  x10^-6)^2 =  Y/100

= 0.629 =Y/100

multiply  both side  by  100

Y=  62.9 is  the molar  mass of unknown  gas



5 0
3 years ago
Which best explains the statement "It has no definite volume”?
NeX [460]
The correctanswer is D hope this helps
5 0
3 years ago
Read 2 more answers
Determine Z and V for steam at 250°C and 1800 kPa by the following: (a) The truncated virial equation [Eq. (3.38)] with the foll
makvit [3.9K]

Answer:

Explanation:

Given that:

the temperature T_1 = 250 °C= ( 250+ 273.15 ) K = 523.15 K

Pressure = 1800 kPa

a)

The truncated viral equation is expressed as:

\frac{PV}{RT} = 1 + \frac{B}{V} + \frac{C}{V^2}

where; B = - 152.5 \ cm^3 /mol   C = -5800 cm^6/mol^2

R = 8.314 × 10³ cm³ kPa. K⁻¹.mol⁻¹

Plugging all our values; we have

\frac{1800*V}{8.314*10^3*523.15} = 1+ \frac{-152.5}{V} + \frac{-5800}{V^2}

4.138*10^{-4}  \ V= 1+ \frac{-152.5}{V} + \frac{-5800}{V^2}

Multiplying through with V² ; we have

4.138*10^4  \ V ^3 = V^2 - 152.5 V - 5800 = 0

4.138*10^4  \ V ^3 - V^2 + 152.5 V + 5800 = 0

V = 2250.06  cm³ mol⁻¹

Z = \frac{PV}{RT}

Z = \frac{1800*2250.06}{8.314*10^3*523.15}

Z = 0.931

b) The truncated virial equation [Eq. (3.36)], with a value of B from the generalized Pitzer correlation [Eqs. (3.58)–(3.62)].

The generalized Pitzer correlation is :

T_c = 647.1 \ K \\ \\ P_c = 22055 \  kPa  \\ \\ \omega = 0.345

T__{\gamma}} = \frac{T}{T_c}

T__{\gamma}} = \frac{523.15}{647.1}

T__{\gamma}} = 0.808

P__{\gamma}} = \frac{P}{P_c}

P__{\gamma}} = \frac{1800}{22055}

P__{\gamma}} = 0.0816

B_o = 0.083 - \frac{0.422}{T__{\gamma}}^{1.6}}

B_o = 0.083 - \frac{0.422}{0.808^{1.6}}

B_o = 0.51

B_1 = 0.139 - \frac{0.172}{T__{\gamma}}^{ \ 4.2}}

B_1 = -0.282

The compressibility is calculated as:

Z = 1+ (B_o + \omega B_1 ) \frac{P__{\gamma}}{T__{\gamma}}

Z = 1+ (-0.51 +(0.345* - 0.282) ) \frac{0.0816}{0.808}

Z = 0.9386

V= \frac{ZRT}{P}

V= \frac{0.9386*8.314*10^3*523.15}{1800}

V = 2268.01 cm³ mol⁻¹

c) From the steam tables (App. E).

At T_1 = 523.15 \  K \ and  \ P = 1800 \ k Pa

V = 0.1249 m³/ kg

M (molecular weight) = 18.015 gm/mol

V  =  0.1249 × 10³ × 18.015

V = 2250.07 cm³/mol⁻¹

R = 729.77 J/kg.K

Z = \frac{PV}{RT}

Z = \frac{1800*10^3 *0.1249}{729.77*523.15}

Z = 0.588

3 0
3 years ago
A solution of phosphoric acid was made by dissolving 10.8 g of H3PO4 in 133.00 mL of water. The resulting volume was 137 mL. Cal
Nesterboy [21]

Answer:

Density is: 1.05 g/ml

Mole fraction solute: 0.015

Mole fraction solvent:  0.095

Molarity: 0.80 M

Molality: 0.82 m

Explanation:

A typical excersise of solution.

It is more confortable to make a table for this.

                |   masss  |  volume  |  mol

solute       |                |                |          

solvent     |                |                |  

solution    |                |                |

Let's complete, what we have.

                 |   masss  |  volume  |  mol

solute       |  10.8g     |                |          

solvent     |                |  133 mL   |  

solution    |                |  137 mL    |

We can first, know how many moles are 10.8 g

Molar Mass H3PO4 = 97.99 g/mol

Mass / Molar mass = mol

10.8 g / 97.99 g/m = 0.110 mol

Density of water is 1 g/ml (it is a very knowly value)

From this data, we can know water mass, solvent.

Density = mass / volume

1 g/ml = mass / 133 mL

Mass = 133 g

We can also have the moles, by the molar mass of water 18 g/m

133 g / 18 g/m = 7.39 mol

                 |   masss  |  volume  |  mol

solute       |   10.8g     |                |   0.110 mol      

solvent     |   133g      |  133 mL   |  7.39 mol

solution    |   143.8g   |  137 mL   | 7.50 mol

Mass of solution will be solute mass + solvent mass

Moles of solution will be solute moles + solvent moles

Now we can calculate everything.

Molarity means mol of solute in 1 L of solution. (mol/L)

We have to convert 137 mL in L (/1000)

0.137L so → 0.110 m / 0.137L = 0.80 M

Molality means mol of solute in 1kg of solvent.

We have to convert 133g in kg (/1000)

0.133 kg so → 0.110 m/0.133 kg = 0.82 m

Density is mass / volume

Solution density will be solution mass / solution volume

143.8 g/137 mL = 1.05 g/m

Molar fraction is : solute moles / total moles  or  solvent moles/total moles.

You can also (x 100%) to have a percent of them.

Remember sum of molar fraction = 1

Molar fraction of solute = 0.110 mol / 7.50mol = 0.015

Molar fraction of solvent = 7.39 mol / 7.50 mol = 0.985

5 0
3 years ago
Morgan wants to compare the weather conditions at her school each day during the week. She decides to measure and record weather
Firlakuza [10]

Answer: Air Temperature

8 0
3 years ago
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