What is the volume of 1.50 lb (pounds) of mercury? The density of mercury is 13.546 g/mL. Use the conversion that 1 kg = 2.20 lb. mL.

Answers

Answer 1

Answer:

50.3mL of mercury are in 1.50lb

Explanation:

Punds are an unit of mass. To convert mass to volume we must use density (13.546g/mL). Now, As you can see, density is in grams but the mass of mercury is in pounds. That means we need first, to convert pounds to grams to use density and obtain volume of mercury.

Mass mercury in grams:

1.50lb * (1kg / 2.20lb) = 0.682kg = 682g of mercury.

Volume of mercury:

682g Mercury * (1mL / 13.546g) =

50.3mL of mercury are in 1.50lb

Related Questions

10 grams of compound J is found to have a mass ratio of 8:2 of lithium (Li): Hydrogen (H). How many grams
of hydrogen would found in 42.0 grams of compound J?

Answers

Answer:

Mass of hydrogen in 42 gram of J = 8.4

Explanation:

Given:

Amount of compound J = 10 gram

Mass ratio[lithium (Li): Hydrogen (H)] = 8:2

Find:

Mass of hydrogen in 42 gram of J

Computation:

Mass of hydrogen in 10 gram of J = [2 / 10] 10

Mass of hydrogen in 10 gram of J = 2 gram

Mass of hydrogen in 42 gram of J = [2 / 10] 42

Mass of hydrogen in 42 gram of J = 8.4

Write the symbol for every chemical element that has atomic number less than 5 and atomic mass greater than 8.3 u.

Answers

Atomic m. Element Atomic no. Symbol

1.0079u Hydrogen 1 H

4.0026u Helium 2 He

6.941u Lithium 3 Li

Additional information:-

★ Atomic number : The number of protons in one atom of an element is known as atomic number.

Atomic number of an element = No. of protons in one atom of element.

The atomic no. of element is denoted by the letter z .

The symbol for every chemical element that has an atomic number less than 5 and atomic mass greater than 8.3 u are hydrogen H, helium He, lithium Li and atomic masses 1,4 and 6.94 respectively in the periodic table.

What is a periodic table?

A periodic table is an arrangement of the elements in which they are classified on the basis of the number of electrons, neutrons, and protons present in their nucleus and shell of them.

Hydrogen has one electron in it and atomic mass is also one, helium comes in the second number with 4 amu and lithium comes in the third number containing 6.94 mass with it.

Therefore, hydrogen H, helium He, lithium Li and atomic masses 1,4 and 6.94 respectively in the periodic table contain symbols for every chemical element that has an atomic number less than 5 and atomic mass greater than 8.3 u are hydrogen H.

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Gold (Au) has a density of 19.3 g/ml. A gold filling falls out of your teeth and you bring it to chemistry class. When you drop it into a graduated cylinder the water rises from 9.27 ml to 12.03 ml.
What is the mass of your gold filling?

Answers

Answer:

53.27 g.

Explanation:

From the question given above, the following data were obtained:

Density of gold = 19.3 g/mL

Volume of water = 9.27 mL

Volume of water + gold = 12.03 mL

Mass of gold =.?

Next, we shall determine the volume of the gold filling. This can be obtained as follow:

Volume of water = 9.27 mL

Volume of water + gold = 12.03 mL

Volume of gold =?

Volume of gold = (Volume of water + gold) – (Volume of water)

Volume of gold = 12.03 – 9.27

Volume of gold = 2.76 mL

Finally, we shall determine the mass of the gold filling as follow:

Density of gold = 19.3 g/mL

Volume of gold = 2.76 mL

Mass of gold =.?

Density = mass/volume

19.3 = mass of gold /2.76

Cross multiply

Mass of gold = 19.3 × 2.76

Mass of gold = 53.27 g

The, the mass of the gold filling is 53.27 g.

Calculate the wavelength of light produced if an electron moves from n=6 state to n=5 state of an electron in a hydrogen atom.

Answers

The questions asks us to calculate the wavelength of light produced if an electron moves from n = 6 to n = 5 state in a hydrogen atom. The answer would be:

λ = [tex]7.46 * 10^{-6}[/tex]  m

m = [tex]7.46 * 10^{-4}[/tex] cm

Explanation:

Please see my graphic below:

The wavelength of light produced if an electron moves from n=6 state to n=5 state of an electron in a hydrogen atom is 7.59 × [tex]10^-6[/tex] m

Using the Rydberg formula;

1/λ = 1.09737 × [tex]10^-6[/tex] (1/[tex]nf^2[/tex] - 1/[tex]ni^2[/tex])

Where;

λ = wavelength

nf = final state of the electron

ni = initial state of the electron

Substituting values;

1/λ = 1.09737 × [tex]10^7[/tex] ([tex]1/5^2[/tex] - [tex]1/6^2[/tex])

1/λ = 1.09737 × [tex]10^7[/tex] (1/25 - 1/36)

1/λ = 1.09737 × [tex]10^7[/tex] (0.04 - 0.028)

1/λ = 1.09737 × [tex]10^7[/tex] (0.012)

λ = 7.59 × [tex]10^-6[/tex] m

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An iceberg has a volume of 783 m3. If the density of ice = 0.917 g/cm3, how much mass does the iceberg have in kg?

Answers

Answer:

7.18 × 10⁵ kg

Explanation:

Step 1: Given data

Volume of the iceberg (V): 783 m³

Density of ice (ρ): 0.917 g/cm³

Step 2: Convert "V" to cm³

We will use the relationship 1 m³ = 10⁶ cm³.

783 m³ × (10⁶ cm³/1 m³) = 7.83 × 10⁸ cm³

Step 3: Calculate the mass (m) of the iceberg

We will use the following expression.

ρ = m/V

m = ρ × V

m = 0.917 g/cm³ × 7.83 × 10⁸ cm³

m = 7.18 × 10⁸ g

Step 4: Convert "m" to kg

We will use the relationship 1 kg = 10³ g.

7.18 × 10⁸ g × (1 kg/10³ g) = 7.18 × 10⁵ kg

Which of the following is not one of the four groups of organic compounds

Answers

where are the options ?

The following that is not one of the four groups of organic compounds is a monosaccharide. The correct option is D.

What are organic compounds?

Any of a vast group of chemical compounds known as organic compounds contain one or more carbon atoms that are covalently connected to atoms. Organic compounds are made up of carbon and hydrogen, and oxygen.

Carbohydrates, lipids, proteins, nucleic acids, and nucleotides are organic compounds.  Due to their presence of both carbon and hydrogen, these substances are referred to as organic.

Monosaccharides are the single unit of carbohydrates. They join to form polymers of carbohydrates. They are glucose, fructose, etc. It is a single unit of carbohydrates. It is not one of the organic compound.

Thus, the correct option is D) monosaccharides.

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The question is incomplete. Your most probably complete question is given below:

A) carbohydrates D) monosaccharides B) lipids E) proteins C) nucleic acids

Which of the following describes freezing? A)A substance changing from the liquid state to the solid state. B) A substance changing from the solid state to the liquid state. C) A substance changing from the liquid to the gaseous state. D) A substance changing from the gaseous state to the liquid state.

Answers

Answer:

A

Explanation:

A because water turns into ice

437 mm converted to m

Answers

Answer:

0.437

Explanation:

Divide the length value by 1000 to get the answer.

1. For one atom of lithium -6 I will need_____ atom of lithium -7 for an average atomic mass of 6.944? 2. For one atom of lithium -6 needed _____ Adams of lithium -7 for an average atomic mass of 6.944.

Answers

Answer:

For one atom of lithium -6 I will need_16.85____ atom of lithium -7 for an average atomic mass of 6.944 .

Explanation:

1 ) Let required atom be m .

[tex]\frac{6\times 1 + 7\times m}{1 + m} = 6.944[/tex]

6 + 7m = 6.944 + 6.944m

7 m - 6.944 m = .944

.056 m = .944

m = 16.85

Here, we are required to find how many atoms of lithium -7 is to combine with 1 atom of lithium -6 for an average atomic mass of 6.944.

For one atom of lithium -6 I will need 16.86 atoms of lithium -7 for an average atomic mass of 6.944.

Let the total number of lithium -6 and lithium -7 required be; x.

Therefore, the number of lithium -7 required is;

(x-1)

Therefore;

{(1/x) × 6 + ((x-1)/x)× 7} = 6.944

Therefore, (6 + 7x - 7)/x = 6.944

And, 7x -1 = 6.944x

0.056x = 1

And, x = 1÷0.056 = 17.86 atoms.

Therefore, the number of lithium -7 atoms needed is;

x - 1 = 17.86 - 1 = 16.85 atoms of lithium -7.

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One possible source of error in this experiment is not completely drying the NaCl. Effect of Percent yield ? And other questions

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Answer:

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12.9 Theoretical Yield and Percent Yield

FlexBooks® 2.0  >  CK-12 Chemistry For High School  >  Theoretical Yield and Percent Yield

It is best to have high yields for chemical reactions

Can we save some money?

The world of pharmaceutical production is an expensive one. Many drugs have several steps in their synthesis and use costly chemicals. A great deal of research takes place to develop better ways to make drugs faster and more efficiently. Studying how much of a compound is produced in any given reaction is an important part of cost control.

Percent Yield

Chemical reactions in the real world don’t always go exactly as planned on paper. In the course of an experiment, many things will contribute to the formation of less product than would be predicted. Besides spills and other experimental errors, there are usually losses due to an incomplete reaction, undesirable side reactions, etc. Chemists need a measurement that indicates how successful a reaction has been. This measurement is called the percent yield.

To compute the percent yield, it is first necessary to determine how much of the product should be formed based on stoichiometry. This is called the theoretical yield, the maximum amount of product that could be formed from the given amounts of reactants. The actual yield is the amount of product that is actually formed when the reaction is carried out in the laboratory. The percent yield is the ratio of the actual yield to the theoretical yield, expressed as a percentage.

Percent yield is very important in the manufacture of products. Much time and money is spent improving the percent yield for chemical production. When complex chemicals are synthesized by many different reactions, one step with a low percent yield can quickly cause a large waste of reactants and unnecessary expense.

Typically, percent yields are understandably less than 100% because of the reasons indicated earlier. However, percent yields greater than 100% are possible if the measured product of the reaction contains impurities that cause its mass to be greater than it actually would be if the product was pure. When a chemist synthesizes a desired chemical, he or she is always careful to purify the products of the reaction.

Sample Problem: Calculating the Theoretical Yield and the Percent Yield

Potassium chlorate decomposes upon slight heating in the presence of a catalyst according to the reaction below:

In a certain experiment, 40.0 g KClO3 is heated until it completely decomposes. What is the theoretical yield of oxygen gas? The experiment is performed and the oxygen gas is collected and its mass is found to be 14.9 g. What is the percent yield for the reaction?

First, we will calculate the theoretical yield based on the stoichiometry.

Step 1: List the known quantities and plan the problem.

Known

given: mass of KClO3 = 40.0 g

molar mass KClO3 = 122.55 g/mol

molar mass O2 = 32.00 g/mol

Unknown

theoretical yield O2 = ? g

Apply stoichiometry to convert from the mass of a reactant to the mass of a product:

Step 2: Solve.

The theoretical yield of O2 is 15.7 g.

Step 3: Think about your result.

The mass of oxygen gas must be less than the 40.0 g of potassium chlorate that was decomposed.

Now, we use the actual yield and the theoretical yield to calculate the percent yield.

Step 1: List the known quantities and plan the problem.

Known

Actual yield = 14.9 g

Theoretical yield = 15.7 g (from Part 12.11A)

Unknown

Percent yield = ? %

Use the percent yield equation above.

Step 2: Solve.

Step 3: Think about your result.

Since the actual yield is slightly less than the theoretical yield, the percent yield is just under 100%.

 

 

Summary

Theoretical yield is calculated based on the stoichiometry of the chemical equation.

The actual yield is experimentally determined.

The percent yield is determined by calculating the ratio of actual yield/theoretical yield.

Review

What do we need in order to calculate theoretical yield?

If I spill some of the product before I weigh it, how will that affect the actual yield?

How will spilling some of the product affect the percent yield?

I make a product and weigh it before it is dry. How will that affect the actual yield?

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Explanation:

What is the Temperature of the suns most middle layer the chromosphere

Answers

The temperature in the chromosphere varies between about 4000 K at the bottom (the so-called temperature minimum) and 8000 K at the top (6700 and 14,000 degrees F, 3700 and 7700 degrees C), so in this layer (and higher layers) it actually gets hotter if you go further away from the Sun

If u burn paper in the jar, is it the same mass as before u burn it or less or more. And why.

Answers

Answer:

the mass would not be the same because of the ashes in the jar

Explanation:

Answer: The total mass before and after burning is the same

Explanation: When the carbon and hydrogen of the hydrocarbon-containing substance (the paper) chemically combine with the oxygen, the remaining materials may appear as ash, the solid remains of a fire.

hope this helps

plz mark brainliest

Draw all of the constitutional isomers of the molecule with formula C3H5Br. Ignore geometric and stereoisomers.

Answers

Answer:

-) 3-bromoprop-1-ene

-) 2-bromoprop-1-ene

-) 1-bromoprop-1-ene

-) bromocyclopropane

Explanation:

In this question, we can start with the I.D.H (hydrogen deficiency index):

[tex]I.D.H~=~\frac{(2C)+2+(N)-(H)-(X)}{2}[/tex]

In the formula we have 3 carbons, 5 hydrogens, and 1 Br, so:

[tex]I.D.H~=~\frac{(2*3)+2+(0)-(5)-(1)}{2}~=~1[/tex]

We have an I.D.H value of one. This indicates that we can have a cyclic structure or a double bond.

We can start with a linear structure with 3 carbon with a double bond in the first carbon and the Br atom also in the first carbon (1-bromoprop-1-ene). In the second structure, we can move the Br atom to the second carbon (2-bromoprop-1-ene), in the third structure we can move the Br to carbon 3 (3-bromoprop-1-ene). Finally, we can have a cyclic structure with a Br atom (bromocyclopropane).

See figure 1

I hope it helps!

ILL GIVE YOU BRAINLIST !!

Which are the quantitative data in the example above? Explain.

Answers

Answer:

helicona

throated carib

eulampis jugularis

Answer:

The percentages

Explanation:

Quantitative data are numbers, while qualitative data are characteristics. So I think that the numbers are your answer.

Hope that helps!

What is the volume of an object that has a mass of 18 g and a density of 0.5 g/cm3 ?

Answers

The volume is 36 g/cm^3.

Explanation: For this particular equation, you would use the volume formula of p=m/V. You would then rearrange this formula to look like V=m/p so that you can solve for the volume.
p=density
m=mass
V=volume

Once you have set up your equation, all that is left to do is plug in the values to your formula and solve, here is the work so that your understand.
V=m/p
V=18/0.5
V=36
Therefore, the volume is 36 g/cm^3.

Hope this helped!

Calculate the mole fraction of nitric acid of a(n) 15.7% (by mass) aqueous solution of nitric acid. Calculate the mole fraction of nitric acid of a(n) 15.7% (by mass) aqueous solution of nitric acid. 2.56×10−2 0.102 5.33×10−2 5.11×10−2 The density of the solution is needed to solve the problem.

Answers

Answer:

0.0505

Explanation:

Step 1: Calculate the masses of solute and solvent

We have a 15.7% by mass nitric acid solution, that is, there are 15.7 g of nitric acid (solute) per 100 g of solution. The mass of water (solvent) is:

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

m(solvent) = m(solution) - m(solute)

m(solvent) = 100 g - 15.7 g = 84.3 g

Step 2: Calculate the moles of nitric acid

The molar mass of nitric acid is 63.01 g/mol.

15.7 g × (1 mol/63.01 g) = 0.249 mol

Step 3: Calculate the moles of water

The molar mass of water is 18.02 g/mol.

84.3 g × (1 mol/18.02 g) = 4.68 mol

Step 4: Calculate the mole fraction of nitric acid

[tex]X(HNO_3) = \frac{nHNO_3}{nHNO_3+nH_2O} = \frac{0.249mol}{0.249mol+4.68mol} = 0.0505[/tex]

An unknown substance is measured. It has a mass of 0.221 g and a volume of 2.25 mL. What is its density?

Answers

Answer:

d ≈ 0.098 g/mL

Explanation:

The density of a substance can be found by dividing the mass by the volume.

d=m/v

The mass of the substance is 0.221 grams and the volume is 2.25 milliliters.

m= 0.221 g

v= 2.25 mL

Substitute the values into the formula.

d= 0.221 g / 2.25 mL

Divide

d= 0.098222222 g/mL

Let’s round to the nearest thousandth. The 2 in the ten thousandths tells us to keep the 8 in the thousandth place.

d ≈ 0.098 g/mL

The density of the substance is about 0.098 grams per milliliter.

Answer:

Explanation:5.70

Which molecule does not exhibit hydrogen bonding?a. HFb. CH3NH2c. CH2F2d. HOCH2CH2OH

Answers

Answer:

(c) CH₂F₂

Explanation:

Hydrogen bonds are weak intermolecular forces. They are the strongest kind of intermolecular forces, although they are weaker than the covalent bonds.

Hydrogen bonds arise from molecules which contain a hydrogen atom which is bonded to one of the most electronegative elements such as N, O or F.

(a) HF,  → has H-F bond

(b) CH₃NH₂,   → has N-H bond

(c) CH₂F₂,  → has no H-F bond ( F- C- F)

(d) HOCH₂CH₂OH, → has O-H bond

Therefore, only CH₂F₂ does not exhibit hydrogen bonding.

For which reaction, carried out at standard conditions, wouldboth the enthalpy and entropy changes drive the reaction in thesame direction? Please EXPLAIN.

A. 2H2(g) + O2(g) --->2H2O(l) ΔH = -571.1 kJ

B. 2Na(s) + Cl2(g)---->2NaCl(s) ΔH = -822.0kJ

C. N2(g) + 2O2(g) ---->2NO2(g) ΔH= +67.7 kJ

D. 2NH3(g) ----> N2(g) +3H2(g) ΔH= +92.4 kJ

Answers

Answer:

2NH3(g) ----> N2(g) +3H2(g) ΔH= +92.4 kJ

Explanation:

Entropy increases with increase in the number of particles from left to right in a reaction. Hence, the reaction; 2NH3(g) ----> N2(g) +3H2(g) ΔH= +92.4 kJ is favoured by increase in entropy.

Similarly, the enthalpy change for the reaction is only +92.4 KJ. Hence for this reaction, both enthalpy and entropy changes drive the reaction in the same direction.

What is the relationship between the bases displayed when the arrow is pointed to the left versus when it is pointed to the right?

Answers

Answer:

Options

Explanation:

CHML 1046 general chemistryExperiment "Buffers and Buffer Capacity" 1. What (NH3/ NH4+) ratio is required for a buffer solution that has pH = 7.00? Кь = 1.8x10-5 for NH3 and Ka = 56x10-10 for NH4.2. Why is a mixture of NHs and NH.CI a poor choice for a buffer having pH 7?

Answers

Answer:

Explanation:

For pH of a buffer solution , the equation is

pH = pKa + log [ A⁻ ] / [ HA ]

Putting the values

7 = -  log 56 x 10⁻¹⁰ +  log [ NH₃ ] / [ NH₄⁺]

7 = 10 - log 56 + log [ NH₃ ] / [ NH₄⁺]

- 3 + 1.748 =  log [ NH₃ ] / [ NH₄⁺]

log [ NH₃ ] / [ NH₄⁺] = - 1.25

[ NH₃ ] / [ NH₄⁺] = [tex]10^{-1.25}[/tex]

Required ratio = [tex]10^{-1.25}[/tex]

2 )

In the formation of NH₄⁺ , generally HCl acid is required . A little bit of excess HCl will change the pH required . So , this mixture is a poor choice for getting pH 7 for a buffer solution .

How do the chemical characteristics of carbon affect the characteristics of organic molecules?

Answers

Answer:

The chemical characteristics of carbon affect the characteristics of organic molecules due to its tetravalent nature. It has four valence electrons in which it shares with other elements in order to form an octet configuration.

Carbon atoms are also capable of forming double and triple bonds with other atoms. These properties help determine the functional group present and gives us a knowledge of the chemical features such as polarity, melting and boiling present in the compound.

What can be observed when concentrated nitric acid is added to iron two sulfate solution?

Answers

Answer:

It will produce iron(III) sulphate ,nitrogen dioxide and water

Explanation:

hope it helps you☺️

if I'm correct do mark my answer as brainliest

what is the average speed in miles per hour of a sprinter who who does thehundred yard dash in 11.3 seconds

Answers

Answer:

18.1 mi/h

Explanation:

Step 1: Given data

Distance traveled (d): 100 yd

Time elapsed (t): 11.3 s

Step 2: Convert "d" to miles

We will use the relationship 1 mi = 1,760 yd.

100 yd × (1 mi/1,760 yd) = 0.0568 mi

Step 3: Convert "t" to hours

We will use the relationship 1 h = 3,600 s.

11.3 s × (1 h/3,600 s) = 3.14 × 10⁻³ h

Step 4: Calculate the average speed (s)

We will use the following expression.

s = d / t

s = 0.0568 mi / 3.14 × 10⁻³ h

s = 18.1 mi/h

Thinking about simple organic molecules made with only carbon and hydrogen and
no other elements (hydrocarbons), what are four different ways to make new
molecules? (Hint: Two versions of the same "way" only count as one thing!)

Answers

Answer:

Explanation:

Simple organic molecules having only carbon and hydrogen are alkanes, alkenes and alkynes.

Ways of making these molecules are referred to as synthesis

Synthesis of

(1) alkanes can be achieved by halogenation of an alkene or alkyne as shown in the reaction below

C₂H₄ + H₂ ⇒ C₂H₆

The reaction above shows an alkene (ethene) undergoing halogenation to form an alkane (ethane). This reaction occurs in the presence of a platinum or nickel catalyst

C₂H₂ + 2H₂  ⇒ C₂H₆

The reaction above shows an alkyne (ethyne) undergoing halogenation to form an alkane (ethane). This reaction also occurs in the presence of a nickel or platinum catalyst

(2) Alkanes can be achieved by Wurtz reaction; treating alkyl halides with sodium metal as shown below

CH₃CH₂Br + 2Na + BrCH₂CH₃ ⇒ CH₃CH₂CH₂CH₃ + 2NaBr

This reaction proceed in the presence of a dry ether.

(3) alkanes from Kolbe's electrolytic method; this involves the synthesis of alkanes through the electrolysis of sodium or potassium salt of a carboxylic acid

2CH₃COO⁻Na⁺ + 2H₂O  ⇒ CH₃CH₃ + 2CO₂ + H₂ + 2NaOH

(4) alkenes from dehydration of alcohol, as shown in the reaction below

C₂H₅OH ⇒ C₂H₄    (-H₂O)

The reaction above proceeds by heating the alcohol (ethanol) at high temperature in the presence of strong acid such as sulphuric acid.

Which of the following molecules is achiral? A) (2R,3R)-2,3-Dichloropentane B) (2R,3S)-2,3-Dichloropentane C) (2S,3S)-2,3-Dichlorobutane D) (2R,3S)-2,3-Dichlorobutane E) None of these

Answers

Answer:

E) None of these

Explanation:

For this question, we must remember the definition of a chiral carbon. In the chiral carbons, we have four different groups around the carbon.

So, an achiral molecule would be a molecule that does not have chiral carbons. That is, in an achiral molecule in all carbons we have at least one repeating group.

In all the molecules, which we have in the question, we have absolute configuration. That is, we have chiral carbons. In molecule A we have two chiral carbons (R and R), in molecule b we have two chiral carbons (R and S), in molecule c we have two chiral carbons (S and S) and in molecule d we have two chiral carbons (R and S). Therefore in all molecules, we have chiral carbons.

Now, if we analyze molecules a and c, we have a plane of symmetry. If we have a plane of symmetry, even though there are chiral carbons, there will be no optical activity. That is, these molecules do not have the ability to deflect polarized light despite having chiral carbons. Therefore molecules a and c are meso compounds.

But optical activity is not related to chirality.

See figure 1 to further explanations

I hope it helps!

4. Which statement gives you enough information to say that the atom is electrically neutral?
O A. The atom has 15 neutrons and 15 electrons.
OB. The atom has 4 neutrons and 4 protons.
O C. The atom has 7 protons and 7 electrons.
O D. The atom has 19 electrons and 19 neutrons.

Answers

Answer:

O C. The atom has 7 protons and 7 electrons.

If the proton amd electron of atoms are equal it is said to be electrically nuetral

How is Oxygen -18 different from oxygen -16?

Answers

Answer:

because they are both 2 deferent type of oxygen levels

Which solution below would have the greatest buffering capacity? The solution was prepared from a weak acid and the salt of its conjugate base. 0.160 M C6H5OHCOOH and 0.160 M C6H5OHCOONa, Ka = 1.05e-3 0.0892 M H2NCH2COOH and 0.0892 M H2NCH2COOK, Ka = 4.50e-3 0.0725 M H2NCH2COOH and 0.0725 M H2NCH2COONa, Ka = 4.50e-3 0.1360 M C6H5OHCOOH and 0.1360 M C6H5OHCOOK, Ka = 1.05e-3

Answers

Answer:

0.1360 M C₆H₅OHCOOH and 0.1360 M C₆H₅OHCOOK,

Ka = 1.05x 10⁻³ .

Explanation:

The first mixture

0.160 M C₆H₅OHCOOH and 0.160 M C₆H₅OHCOONa,

Ka = 1.05x 10⁻³ .

The second mixture

0.0892 M H₂NCH₂COOH and 0.0892 M H₂NCH₂COOK,

Ka = 4.5 x 10⁻³ .

The third mixture

.0725 M H₂NCH₂COOH and 0.0725 M H₂NCH₂COONa,

Ka = 4.5 x 10⁻³

fourth mixture

0.1360 M C₆H₅OHCOOH and 0.1360 M C₆H₅OHCOOK,

Ka = 1.05x 10⁻³ .

In all the mixtures the ratio of acid and its salt are same and equal to one so this ratio will not determine their relative buffering capacity .

Now we know that weak acid has more buffering capacity so mixture having acid of less Ka will have more buffering capacity .

Ka is less if

Ka = 1.05 x 10⁻³ .

Dilute acids have greater buffering capacity

So ultimate answer is

0.1360 M C₆H₅OHCOOH and 0.1360 M C₆H₅OHCOOK,

Ka = 1.05x 10⁻³ .

PLEASE HELP ASAP!!!!!!!!!!! WITH PAGE 2 FOR QUESTION # 3 WITH THE GRAPHING THANK YOU. THE QUESTION IS AS FOLLOWED: GRAPH THE VAPOR PRESSURE OF WATER USING THE FOLLOWING TEMPERATURES. Temperature (k) Vapor Pressure (Torr) (dependent variable) 273 4.6 293 17.5 313 55.0 333 149.2 353 355.5 373 760.0

Answers

Answer:

a

Explanation:

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