Classify each substance based on the intermolecular forces present in that substance.A. H2OB. CH4C. CO D. CH3Cl1. Hydrogen bonding, dipole dipole, and dispersion2. Dipole dipole and dispersion only 3. Dispersion only

Answers

Answer 1

Answer:

H2O - Hydrogen bonding, dipole dipole, and dispersion

CH4 - Dispersion only

CO - Dipole dipole and dispersion only

CH3Cl - Dipole dipole and dispersion only

Explanation:

Hydrogen bonding can only exist when hydrogen is bonded to a small highly electronegative atom such as oxygen hence hydrogen bonds, dipole interactions and dispersion by are present in water.

CH4 is a nonpolar molecule hence only dispersion forces are present.

CH3Cl and CO both possess dipoles in the molecule hence both dipole interactions and dispersion forces exist in the molecule.


Related Questions

Which tool can be used to measure the volume of liquid?
an electronic balance
a meniscus
a caliper
a beaker

Answers

Answer:

The answer is D a beaker

Explanation:

Hope this helps :D

Beaker is the answer

Ocean water becomes colder; the density of ocean water increases or decreases?

Answers

Answer:

.............increases

Use the ruler to determine the length of this object. Record your answer to the nearest tenth. The object is ____ long.

Answers

The object is 2.7 cm long

Answer: 2.7 cm

Explanation: ~~~~~ there ya go

What is the maximum length of a polyethene chain with molar mass 10^6 and monomer length 2.5?

Answers

Answer:

L=7.0125*10^-5

Explanation:

LET  

 Length of polyethene chain= L

Molecular weight OF polyethene= MW

Molecular weight of monomer =mw

Monomer length= l

L=(MW/mw) *l

              L= (28.05g/mol10^6gmol-1) * 2.5

             L=70.125*10^-6

            L=7.0125*10^-5

Hence, the length will be L=7.0125*10^-5 .

Is burning sugar a chemical or physical change? Please provide explanation

Answers

Answer:

It is a chemical change

Explanation:

the reason is because the will not be able to materialize back into it's original form.

It be a chemical change my dude

What is the A# of fluoride-19?​

Answers

Answer:

atomic number of fluorine

is 9

A buffer contains 0.010 moles of lactic acid (pKa = 3.86) and 0.050 mol of sodium lactate per liter.
(a) Calculate the pH of the buffer.
(b) Calculate the change in pH when 5 mL of 0.5 M HCl is added to 1 L of the buffer.
(c) What pH change would you expect if you added the same quantity (5 mL) of HCl to 1 L of pure water?

Answers

Answer:

a. pH = 4.56

b. Change in pH = 0.85

c. Change in pH = 5.4

Explanation:

a. The pKa of lactic buffer is: 3.86.

Using Henderson-Hasselbalch formula for the lactic buffer:

pH = 3.86 + log [Lactate] / [Lactic acid]

Where [] is molarity of each compound but could be taken as moles

Replacing:

pH = 3.86 + log [0.050 moles] / [0.010 moles]

pH = 4.56

b. The HCl added reacts with Lactate producing lactic acid. Moles of HCl are:

5x10⁻³L * (0.5mol /L) = 0.025 moles HCl

Moles of lactate: 0.050moles - 0.025 moles = 0.025 moles

Moles lactic acid: 0.010 moles + 0.025 moles = 0.035 moles

pH = 3.86 + log [0.025 moles] / [0.035 moles]

pH = 3.71

Change in pH = 4.56 - 3.71 = 0.85

c. 1L of pure water has a pH of 7. 0.025 moles of HCl = 0.025 moles H⁺ in 1.005L:

0.025 mol / 1.005L = 0.0249M = [H⁺]

As pH = -log [H⁺]

pH = 1.6

Change in pH = 7.0 - 1.6 = 5.4

A liquid has a volume of 4mL and a mass of 24 grams. What is the density of the liquid? ​

Answers

Answer:

the density is 6

Explanation:

mass divided by volume equals density

What is the density of a silver coin from the following data:
Mass of silver coin 6.581 g
Volume of coin and water 23.7 mL
Volume of water alone 23.1 mL

The answer is assumed to be in g/mL. Please type a numeric value only.​

Answers

Answer:

you can use math wa, it helps

Which of the following DOES NOT have 2 significant
figures?
11,000,000,000
0.11
1.001
0.0000011

Answers

Answer:

1.001

Explanation:

The Significant Figures are 1 0 0 1, This answer has 4 Significant figures, while the other three have only 2 significant figures

If 6.09 g of sodium chloride are mixed with 35.50 g of water, what is the mass % (w/w%) salt in the solution

Answers

Answer:

14.6 %

Explanation:

Step 1: Given data

Mass of sodium chloride (solute): 6.09 g

Mass of water (solvent): 35.50 g

Step 2: Calculate the mass of solution

We will use the following expression.

m(solution) = m(solute) + m(solvent)

m(solution) = 6.09 g + 35.50 g

m(solution) = 41.59 g

Step 3: Calculate the percent by mass of the salt

We will use the following expression.

%w/w = mass of NaCl / mass of solution × 100%

%w/w = 6.09 g / 41.59 g × 100%

%w/w = 14.6 %

An atom's Lewis dot structure has four dots. Which of the following elements could it be, and why?

Answers

Answer: Carbon, because it is in group 14 and has four valence electrons

Explanation: Just did this quiz

Answer:carbon

Explanation:

For the following amino acid, the name, three-letter abbreviation, or one-letter abbreviation is given. Complete the missing information.Name: proline Three letter abbreviation:____________One letter abbreviation:________

Answers

Answer:

The correct answer is "Pro; P".

Explanation:

Amino acids are not only identified by its full name, there are three-letter and one-letter abbreviations for each amino acid that helps to annotate sequences and biological processes. In the case of proline, the three letter abbreviation is "Pro" and the one letter abbreviation is "P". Proline got its name for its cyclic structure, that resembles the structure of pyrrole.

What is a factual statement about how nature behaves or functions?
a hypothesis
O a law
O a theory
an experiment

Answers

I think it would be a law

Answer:its a law

Explanation:

I passed that quiz 100%, my answer for that question was law

so its law

In reading a line drawing, how do you know where atoms of these elements are in the structure if they are missing from the drawing?

Answers

Answer:

The atoms of an element are represented in a chemical line drawing with its chemical formula.

Explanation:

chemical structural drawing helps to represent the pattern for which an element is formed. Chemical elements are made up of atoms that represent their single state.

The line drawing is made up of lines (representing the chemical bond between atoms) and the atoms or various atoms that make up the element.

Photosynthesis by land plants leads to the fixation each year of about 1 kg of carbon on the average for each square meter of an actively growing forest. The atmosphere is approximately 20% O2 and 80% N2, but contains 0.039% CO2 by weight.
A) How much carbon is present in the entire atmosphere lying above each square meter of the earth's surface?
B) At the current rate of utilization, how long would it take to use all the CO2
in the entire atmosphere directly above a forest?

Answers

Answer:

a) mass of carbon directly above 1 ( each)  square meter of the earth is 1.65kg

b) all CO₂ will definitely be used up from the atmosphere directly above a forest in 1.65 years

Explanation:

first we calculate the moles of carbon

moles = mass/molar mass

= 1kg/12gmol⁻¹

= 1000g/12gmol⁻¹

= 83.33 mol

now using the ideal gas equation

we find the volume of co₂required based on 83.33 moles

PVco₂ = nRT

Vco₂ = nRT/P

Vco₂ = (83.22mol × 0.0821L atm k⁻¹ mol⁻¹ 298 K) / 1 atm

Vco₂ = 2083.73 L

so since CO₂ in air is 0.0390% by volume in the atmosphere, we find the the total amount of air required to obtain 1kg carbon

therefore

Vair × 0.0390/100 = 2038.73L

Vair = (2038.73L × 100) / 0.0390

Vair = 5.23 × 10⁶L

therefore 5.23 × 10⁶ L of air will be required to obtain 1kg carbon

a)

Here we calculate the mass of air over 1 square meter of surface.

Remember that atmospheric pressure is the consequence of the force exerted by all the air above the surface; 1 bar is equivalent to 1.020×10⁴kgm⁻²

NOW

mass of air = 1.020×10⁴kgm⁻² × 1m²

= 1.020×10⁴kg

= 1.020×10⁷g    [1kg = 10³g]

we now find the moles of air associated with it

moles = mass/molar mass

= 1.020 × 10⁷g / ( 20%×Mo₂ + 80%×Mn₂)

= 1.020 × 10⁷g / ( 20%×32gmol⁻¹ + 80%×28gmol⁻¹)

= 1.020 × 10⁷g / 28.8 gmol⁻¹

= 354166.67mol

so based on the question, for each mole (air), there is 0.0390% of CO₂

now to calculate the moles of CO₂ we say;

MolesCo₂ = 0.0390/100 × 354166.67mol

= 138.125 moles

Now we calculate mass of CO₂ from the above findings

Moles = mass/molar mass

mass = moles × molar mass

= 138.125 moles × 12gmol⁻¹

= 1657.5g

we covert to KG

= 1657.5g / 1000

mass = 1.65kg

therfore mass of carbon directly above 1 ( each)  square meter of the earth is 1.65kg

b)

to find the number years required to use up all the CO₂, WE SAY

Number of years = total carbon per m² of the forest / carbon used up per m² from the forest per year

Number of years = 1.65kgm⁻² / 1kg²year⁻¹

Number of years = 1.65 years

Therefore all CO₂ will definitely be used up from the atmosphere directly above a forest in 1.65 years

When 1 carbon atom combines with 2 oxygen atoms, the resulting substance is called a​

Answers

Answer:

Carbon Dioxide

Explanation:

Which is a compound:)

How does the government control scientific research?
O A. Scientists are dependent upon the government to set ethical
standards.
B. The government can take away money if a scientist is not
following ethical standards.
O C. The government determines which experiments are important.
D. Scientists can do only research that the government supports.

Answers

Answer: B) The government can take away money if a scientist is not following ethical standards

Explanation: hope this helped :)

The government control scientific research by taking  away money if a scientist is not following ethical standards. Therefore, the correct option is option B.

While it is crucial to recognise that the government's role in scientific research is varied and extends beyond the control of ethical standards, funding agencies and institutions frequently require researchers to follow particular ethical norms and standards when conducting scientific investigations. Failure to adhere to these ethical norms may result in the loss or denial of research funding.

However, it is critical to recognise that the government's role in scientific research extends beyond simply enforcing ethical norms. Governments can also fund and support specific research areas or priorities, create regulations and guidelines for research practises, encourage researcher collaboration, provide infrastructure and resources, and establish policies that shape the overall direction and focus of scientific research.

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It takes 330 joules of energy to raise the temperature of 24.6 gbenzene from 21 degrees Celsius to 28.7 degrees Celsius at constantpressure. What is the molar hear capacity of benzene at constantpressure?

Answers

Given :

Energy , E = 330 J .

Initial temperature , [tex]T_i=21^oC[/tex] .

Final temperature , [tex]T_f=24.6^oC[/tex] .

Mass of benzene , m = 24.6 g .

To Find :

The molar hear capacity of benzene at constant pressure .

Solution :

Molecular mass of benzene , M = 78 g/mol .

Number of moles of benzene :

[tex]n=\dfrac{24.6}{78} \ mol\\\\n=0.32 \ mol[/tex]

Energy required is given by :

[tex]q=nC_p\Delta T\\\\330=0.32\times C_p\times (28.7-21)\\\\C_p=\dfrac{330}{0.32\times 7.7}\ J\ mol^{-1}^oC^{-1} \\\\C_p=133.9\ J\ mol^{-1}^oC^{-1}[/tex]

Hence , this is the required solution .

hey can anyone help me with organic chemestry?​

Answers

Organic chemistry is

What sequence should be followed when conducting a laboratory investigation? Make observations, gather experimental data, form a conclusion, state a problem Define a problem, form a hypothesis, gather experimental data, form a conclusion. Form a hypothesis, form a conclusion, gather experimental data, define a problem Gather experimental data, make observations, form a conclusion, for a hypothesis.​

Answers

Answer:

Define a problem, form a hypothesis, gather experimental data, form a conclusion

Answer:

Make observations, gather experimental data, form a conclusion, state a problem Define a problem, form a hypothesis, gather experimental data, form a conclusion.

Explanation:

Describe how matter is classified into mixtures, pure substances, elements, and compounds.

Answers

Answer:

Matter can be broken down into two categories: pure substances and mixtures.

Pure substances are further broken down into elements and compounds. ... A chemical substance is composed of one type of atom or molecule.

A mixture is composed of different types of atoms or molecules that are not chemically bonded.

hope this helped

Matter is divided in two categories: pure substances and mixtures.Pure substances are further broken down into elements and compounds.Chemical substance is composed of one type of atom or molecule.A mixture is composed of different types of atoms or molecules .

What is matter?

Matter in chemistry, is defined as any kind of substance that has mass and occupies space that means it has volume .Matter is composed up of atoms which may or not be of same type.

Atoms are further made up of sub atomic particles which are the protons ,neutrons and the electrons .The matter can exist in various states such as solids, liquids and gases depending on the conditions of temperature and pressure.

The states of matter are inter convertible into each other by changing the parameters of temperature and pressure.

Learn more about matter,here:

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At a certain temperature this reaction follows first-order Kinetics with a rate constant of 0.0660
2H1 (g)----------> H2, (g)+I2
Suppose a vessel contains HI at a concentration of 0.310 M. Calculate how long it takes for the concentration of HI to decrease to 0.0558 M. You may assume no other reaction is important. Round your answer to 2 significant digits.

Answers

Answer:

After 26.0s, the concentration of HI decreases from 0.310M to 0.0558M.

Explanation:

Based on the reaction of the problem, you have as general kinetic law for a first-order reaction:

ln[HI] = -kt + ln [HI]₀

Where [HI] is actual concentration after time t,

k is rate constant

and [HI]₀ is initial concentration of the reactant.

Initial concentration of HI is 0.310M,

K is 0.0660s⁻¹,

And the actual concentration is 0.0558M:

ln[HI] = -kt + ln [HI]₀

ln[0.0558M] = -0.0660s⁻¹*t + ln [ 0.310M]

-1.7148 = -0.0660s⁻¹*t

26.0s = t

After 26.0s, the concentration of HI decreases from 0.310M to 0.0558M

Suppose now that you wanted to determine the density of a small crystal to confirm that it is graphite. From the literature, you know that graphite has a density of 2.25 g/cm^3. How would you prepare 20.0 mL of the liquid mixture having that density from pure samples of CHCl3 (d = 1.492 g/mL) and CHBr3 (d = 2.890 g/mL)? (Note: 1 mL = 1 cm^3.)

Answers

Answer:

The volume of first and second compound are 9.15 ml and 10.85 ml.

Explanation:

Given that,

Density of graphite = 2.25 g/cm³

Volume of mixture = 20.0 mL

Density of first compound = 1.492 g/ml

Density of second compound = 2.890 g/ml

Let the volume of first mixture = x

The volume of second mixture = (20-x)....(I)

We need to calculate the volume of first compound

Using formula of  density of mixture

[tex]\rho=\dfrac{V_{1}\rho_{1}+V_{2}\rho_{2}}{V_{1}+V_{2}}[/tex]

Where, [tex]V_{1}[/tex] = volume of first compound

[tex]V_{2}[/tex] = volume of second compound

[tex]\rho_{1}[/tex] =density of first compound

[tex]\rho_{1}[/tex] = density of first compound

Put the volume into the formula

[tex]2.25=\dfrac{x\times1.492+(20-x)\times2.890}{x+20-x}[/tex]

[tex]45=1.492x+57.8-2.890x[/tex]

[tex]45-57.8=1.492x-2.890x[/tex]

[tex]12.8=1.398x[/tex]

[tex]x=\dfrac{12.8}{1.398}[/tex]

[tex]x=9.15\ ml[/tex]

We need to calculate the volume of second compound

Using equation (I)

[tex]V_{2}=20-x[/tex]

Put the value of x

[tex]V_{2}=20-9.15[/tex]

[tex]V_{2}=10.85\ ml[/tex]

Hence, The volume of first and second compound are 9.15 ml and 10.85 ml.

Question 13 of 16
The mass of a piece of metal is 25.4253 grams. When the piece of
metal was dropped into a graduated cylinder that had an initial water
volume of 6.85 cm, the total volume was 9.84 cm? What is the
density of the piece of metal in g/cm??

Answers

Answer:

[tex]\rho =8.50g/cm^3[/tex]

Explanation:

Hello,

In this case, due to the volume difference caused by the addition of the metal, one could notice that the volume of the metal is:

[tex]V_{metal}=9.84cm^3-6.85cm^3=2.99cm^3[/tex]

In such a way, given the mathematical definition of density, it turns out:

[tex]\rho =\frac{m}{V}=\frac{25.4253g}{2.99cm^3}\\\\\rho =8.50g/cm^3[/tex]

Regards.

To prepare 100.0 mL of a 0.0525 M solution of NaCl in water, what mass of sodium chloride is needed? The
molar mass of NaCl is 58.44 g/mol.

Answers

Answer:

0.307 g

Explanation:

Step 1: Given data

Volume of solution: 100.0 mL

Molarity of the NaCl solution: 0.0525 M

Molar mass of NaCl: 58.44 g/mol

Step 2: Calculate the moles of NaCl required

The molarity is equal to the moles of solute (NaCl) divided by the liters of solution.

M = moles of NaCl / liters of solution

moles of NaCl = M × liters of solution

moles of NaCl = 0.0525 mol/L × 0.1000 L = 5.25 × 10⁻³ mol

Step 3: Calculate the mass corresponding to 5.25 × 10⁻³ moles of NaCl

5.25 × 10⁻³ mol × 58.44 g/mol = 0.307 g

Sort the following analytical techniques as either classical methods or instrumental methods.
a) surface analysis
b) precipitation titration
c) gravimetric analysis
d) high performance liquid chromatography
e) potentiometry
f) atmoic spectroscopy

Answers

Answer:

Instrumental methods

surface analysis

high performance liquid chromatography

atomic spectroscopy

potentiometry

Classical methods

precipitation titration

gravimetric analysis

Explanation:

Instrumental methods of analysis are those analytical methods in which the responsibility of detection has been removed from human beings and placed on automated instruments while classical methods are those analytical methods in which the responsibility of detection remains the responsibility of human beings.

Many instrumental methods such as HPLC rely on computer screens as readout devices.

Which of the following is in intensive property a. mass b. magnetism c shape D. volume

Answers

Answer:

b. Magnetism (sorry im very late)

Explanation:

Intensive properties do not depend on size, no matter what it doesn't. For example, magnetism, density, melting and boiling points, and color. All of those support intensive property.

Intensive properties are physical properties that do not depend on the amount or size of the material being measured. In other words, they remain constant regardless of the quantity of the substance.

The correct answer is b. magnetism.

Out of the options provided:

a. mass is an extensive property because it depends on the amount of the substance. If you have more of a substance, you will have a greater mass.

b. magnetism is an intensive property because it remains the same regardless of the size or amount of the material with the magnetic property.

c. shape is not a standard property used to classify intensive or extensive properties. It is more of a description of the object's form.

d. volume is an extensive property because it depends on the size and amount of the substance. If you have more of a substance, you will have a larger volume.

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I need these questions answered

Answers

Because the there’s not enough inertia to keep the bouncy ball going at the same rate

Answer:

Three objects with kinetic energy

A ball rolling down the street

Moving Car

Bullet

Law of Conservation of Energy states that the total energy of an isolated system remains constant; it is said to be conserved over time.

Though technically there are limitless forms of both types of energy, officially there's five types of kinetic energy: radiant, thermal, sound, electrical and mechanical and potential energy adds gravitational, nuclear, and elastic.

The stock solution is 100 μM the protocol calls for 10 mL solution of 80, 60, 40 and 20 μM. Compute the table.Target Concentration(μM) Volume of stock(mL) Volume of water(mL)20406080

Answers

Answer:

See the answer below

Explanation:

Using the dilution equation:

m1v1 = m2v2

m1 =100 μM, v1 = ?, v2 = 10 mL

In order to prepare 80 μM, then m2 = 80 μM

v1 = [tex]\frac{80*10}{100}[/tex] = 8 mL

In order to prepare 60 μM

v1 = [tex]\frac{60*10}{100}[/tex] = 6 mL

In order to prepare 40 μM

v1 = [tex]\frac{40*10}{100}[/tex] = 4 mL

In order to prepare 20 μM

v1 = [tex]\frac{20*10}{100}[/tex] = 2 mL

Since the final solution has to be up to the 10 mL mark, the complete table would be:

Target Concentration(μM)       Volume of stock(mL)        Volume of water(mL)

              20                                           2                                            8

              40                                           4                                             6

              60                                           6                                             4

              80                                           8                                             2

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