answer.
Ask question
Login Signup
Ask question
All categories
  • English
  • Mathematics
  • Social Studies
  • Business
  • History
  • Health
  • Geography
  • Biology
  • Physics
  • Chemistry
  • Computers and Technology
  • Arts
  • World Languages
  • Spanish
  • French
  • German
  • Advanced Placement (AP)
  • SAT
  • Medicine
  • Law
  • Engineering
Sergeeva-Olga [200]
2 years ago
9

Classify each property as associated with a liquid that has strong or weak intermolecular forces Drag the appropriate items to t

heir respective bins. View Available Hint(s) low viscosity high surface tension low vapor pressure high vapor pressure low surface tension high high boiling point point Strong inter r forces Weak intermolecular forces rch
Chemistry
1 answer:
Ivan2 years ago
3 0

Answer: intermolecular forces are described as forces that either cause attraction or repulsion between neighbouring particles or molecules

Explanation: liquids with strong intermolecular forces has the following properties:

1.) High boiling point

2.) high surface tension

3.) Low viscosity

4.) Low vapour pressure

While liquids with weak intermolecular forces has the following properties:

1.) Low boiling point

2.)low surface tension

3.) High viscosity

4.) High vapour pressure.

You might be interested in
A 126-gram sample of titanium metal is heated from 20.0°C to 45.4°C while absorbing 1.68 kJ of heat. What is the specific heat o
Radda [10]

Answer:

The specific heat for the titanium metal is 0.524 J/g°C.

Explanation:

Given,

Q = 1.68 kJ   = 1680 Joules

mass = 126 grams

T₁ = 20°C

T₂ = 45.4°C

The specific heat for the metal can be calculated by using the formula

Q = (mass) (ΔT) (Cp)

Here, ΔT =  T₂ - T₁ = 45.4 - 20 = 25.4°C.

Substituting values,

1680 = (126)(25.4)(Cp)

By solving,

Cp = 0.524 J/g°C.

The specific heat for the titanium metal is 0.524 J/g°C.

3 0
2 years ago
A 0.2-mm-thick wafer of silicon is treated so that a uniform concentration gradient of antimony is produced. One surface contain
krek1111 [17]

Answer:

- 0.0249% Sb/cm

-1.2465 * 10^9 \frac{atoms}{cm^3.cm}

Explanation:

Given that:

One surface contains 1 Sb atom per  10⁸  Si atoms and the other surface contains 500 Sb atoms per  10⁸ Si atoms.

The concentration gradient in atomic percent (%) Sb  per cm can be calculated as follows:

The difference in concentration = \delta_c

The distance \delta_x = 0.2-mm = 0.02 cm

Now, the concentration of silicon at one surface containing  1 Sb atom per 10⁸ silicon atoms and at the outer surface that has 500 Sb atom per   10⁸ silicon atoms can be calculated as follows:

\frac{\delta_c}{\delta_c} = \frac{(1/10^8 -500/10^8)}{0.02cm} *100%

= - 0.0249% Sb/cm

b) The concentration (c_1) of Sb in atom/cm³ for the surface of 1 Sb atoms can be calculated by using the formula:

c_1 = \frac{(8 si atoms/unit cells)(1/10^3)}{(lattice parameter)^3/unit cell}

Lattice parameter = 5.4307 Å;  To cm ; we have

= 5.4307A^0* \frac{10^{-8}cm}{ A^0}

c_1 = \frac{(8 si atoms/unit cells)(1/10^8)}{(5.4307*10^{-8}cm)^3/unit cell}

= 0.00499*10^{17}atoms/cm^3

The concentration (c_2) of Sb in atom/cm³ for the surface of 500 Sb can be calculated as follows:

c_1 = \frac{(8 si atoms/unit cells)(500/10^8)}{(5.4307*10^{-8}cm)^3/unit cell}

   =  \frac{4*10^{-3}}{1.601*10^{-22}}

   = 2.4938*10^{17}atoms/cm^3

Finally, to calculate the concentration gradient

(\frac{\delta _c}{\delta_ x}) = \frac{c_1-c_2}{\delta_x}

(\frac{\delta _c}{\delta_ x}) = \frac{0.00499*10^{17}-2.493*10^{17}}{0.02}

= -1.2465 * 10^9 \frac{atoms}{cm^3.cm}

8 0
2 years ago
Calculate the daily aluminum production of a 150,000 [A] aluminum cell that operates at a faradaic efficiency of 89%. The cell r
Gala2k [10]

Explanation:

It is known that in one day there are 24 hours. Hence, number of seconds in 24 hours are as follows.

                             24 \times 3600 sec

Hence, total charge passed daily is calculated as follows.

                      150,000 \times 24 \times 3600 sec

And, number of Faraday of charge is as follows.

                    \frac{150,000 \times 24 \times 3600 sec}{96500}

                     = 134300.52 F

The oxidation state of aluminium in Al_{2}O_{3} is +3.

                       Al^{3+} + 3e^{-} \rightarrow Al(s)

So, if we have to produce 1 mole of Al(s) we need 3 Faraday of charge.

Therefore, from 134300.52 F the moles of Al obtained with 89% efficiency is calculated as follows.

                \frac{134300.52 F}{3} \times \frac{89}{100}

                   = 39842.487 mol

or,               = 3.9842 \times 10^{4} mol

Molar mass of Al = 27 g/mol

Therefore, mass in gram will be calculated as follows.

            Mass in grams = 3.9842 \times 10^{4} mol \times 27

                                     = 107.57 \times 10^{4} g

                                     = 1075.7 kg/day

Thus, we can conclude that the daily aluminum production of given aluminium is 1075.7 kg/day.

8 0
2 years ago
Which is the correct statement regarding the relative Rf values of the starting methyl benzoate vs the product, methyl m-nitrobe
Marianna [84]

Answer:

1. The product has a higher Rf value on a silica gel TLC plate because it is more polar than the starting methyl benzoate.

2. False

3. True

Explanation:

In chromatography, there is a stationary phase and a mobile phase. The ratio of the distance moved by a component and the distance moved by the solvent gives the retention factor (Rf).

Since silica gel is a polar solvent, it will retain the more polar product methyl m-nitrobenzoate compared to the methyl benzoate starting material.

In comparing the electrophillic aromatic substitution of m-nitrobenzoate  and methyl benzoate, we must remember that the presence of electron withdrawing groups (such as -NO2 and -CHO) on the aromatic compound deactivates the compound towards electrophillic aromatic substitution hence, methyl m-nitrobenzoate is less reactive than methyl benzoate in Electrophilic Aromatic Substition and Methyl benzoate is less reactive than benzene in Electrophilic Aromatic Substition

5 0
2 years ago
How many moles of Sodium Hydroxide - NaOH are in a 35.65g sample of NaOH?
Travka [436]

Answer:

40.01

Explanation:

3 0
2 years ago
Other questions:
  • Which statement explains why sulfur is classified as a Group 16 element?
    5·2 answers
  • There are two types of nucleic acids, dna and rna. nearly all organisms use dna, not rna, as the central repository for genetic
    15·1 answer
  • If 35.50 cm3 of a NaOH solution are required for the complete neutralization of a 25.00cm3 sample of 0.200mol dm-3 H2SO4, what i
    5·1 answer
  • A new potential heart medicine, code-named X-281, is being tested by a pharmaceutical company, Pharma-pill. As a research techni
    13·1 answer
  • In the electrochemical cell using the redox reaction below, the anode half reaction is ________. Sn4+ (aq) + Fe (s) → Sn2+ (aq)
    13·1 answer
  • 0.658 g of a compound containing only carbon, hydrogen, and oxygen is burned in excess O2. CO2(1.285 g) and H20 (0.658g) are pro
    5·1 answer
  • A swimmer enters a gloomier world (in one sense) on diving to greater depths. Given that the mean molar absorption coefficient o
    10·1 answer
  • In a voltaic cell made of a Ag strip and an Al strip immersed in 1.0 M aqueous AgNO_3 and Al(NO_3)_3 solutions, respectively, wi
    6·1 answer
  • sekiranya kutu tinggal di badan arnab dan menghisap darah arnab , apakah jenis interaksi antara kutu dengan arnab ?​
    5·1 answer
  • What is the concentration in %m/v of a 0.617 M aqueous solution of methanol (MM = 32.04 g/mol)?
    11·1 answer
Add answer
Login
Not registered? Fast signup
Signup
Login Signup
Ask question!