A sample of 325 mg of neon occupies 2 dm^3 at 20.0 c. use the perfect gas law to determine the pressure of the gas.

Answers

Answer 1

Answer:

0.1935atm is the pressure of the gas.

Explanation:

In the problem you have the mass of Neon (You can obtain moles using molar mass), you have volume, and temperature. You can use general ideal gas law to solve pressure:

PV = nRT

P = nRT / V

Where P is pressure in atm, n are moles, R is gas constant (0.082atmL/molK), T is absolute temperature (20.0°C + 273.15 = 293.15K), and V is volume in L = Volume in dm³

Moles gas:

Molar mass Ne = 20.1797g/mol

0.325g * (1mol / 20.1797g) = 0.0161 moles

Replacing:

P = nRT / V

P = 0.0161 moles * 0.082atmL/molK * 293.15K / 2L

P = 0.1935atm is the pressure of the gas


Related Questions

Significant Figures
Tell me what's wrong. When adding and subtracting, you line up decimal and look for the least decimal places.
0.87
4.061
+10.4
--------------
15.331
My answer is 15 but teacher said I'm wrong

Thank You

Answers

You show the work right but your addition was wrong
4.061
+10.4
————
14.461

Rhodium has an atomic radius of 0.1345 nm and density of 12.41 gm/cm3 . Determine whether it has an FCC or BCC crystal structure.

Answers

Answer:

FCC.

Explanation:

Hello,

In this case, since the density is defined as:

[tex]\rho =\frac{n*M}{Vc*N_A}[/tex]

Whereas n accounts for the number of atoms per units cell (2 for BCC and 4 for FCC), M the atomic mass of the element, Vc the volume of the cell and NA the Avogadro's number. Thus, for both BCC and FCC, the volume of the cell is:

[tex]Vc_{BCC}=(\frac{4r}{\sqrt{3} } )^3=(\frac{4*0.1345x10^{-7}cm}{\sqrt{3} } )^3=2.997x10^{-23}cm^3\\\\Vc_{FCC}=(2\sqrt{2}r)^{3} =(2\sqrt{2} *0.1345x10^{-7}cm)^3=5.506x10^{-23}cm^3[/tex]

Hence, we compute the density for each crystal structure:

[tex]\rho _{BCC}=\frac{n_{BCC}*M}{Vc_{BCC}*N_A}=\frac{2*102.9g/mol}{2.337x10^{-23}cm^3*6.022x10^{23}/mol} =14.62g/cm^3\\\\\rho _{FCC}=\frac{n_{FCC}*M}{Vc_{FCC}*N_A}=\frac{4*102.9g/mol}{5.506x10^{-23}cm^3*6.022x10^{23}/mol} =12.41g/cm^3[/tex]

Therefore, since the density computed as a FCC crystal structure matches with the actual density, we conclude rhodium has a FCC crystal structure.

Regards.

calculate the ph of the solution resulting from the addition of 20 ml of 0.100 m naoh to 30 ml of 0.100 m hno 3

Answers

Answer:1.6989

Explanation:

The pH of the solution prepared by mixing NaOH and [tex]\rm HNO_3[/tex] has been 1.69.

The moles of NaOH can be given by:

Moles = Molarity × Volume (L)

Moles of NaOH = 0.1 M × 0.02 L

Moles of NaOH = 0.002 mol

Moles of [tex]\rm HNO_3[/tex] = 0.1 M × 0.03 L

Moles of [tex]\rm HNO_3[/tex] = 0.003 mol

Since,  [tex]\rm HNO_3[/tex] has been remaining in the solution after neutralization, the moles of [tex]\rm HNO_3[/tex] has been :

[tex]\rm HNO_3[/tex] = 0.003 - 0.002 mol

[tex]\rm HNO_3[/tex] = 0.001 mol.

The moles of [tex]\rm HNO_3[/tex] has been equivalent to the moles of hydrogen ion.

The hydrogen ion has been 0.001 mol.

The concentration of Hydrogen ions can be given by:

Hydrogen ion concentration = [tex]\rm \dfrac{Moles}{Volume(L)}[/tex]

Hydrogen ion concentration = [tex]\rm \dfrac{0.001}{0.03\;+\;0.02}[/tex]

Hydrogen ion concentration = 0.02 M.

pH can be defined as the negative log of the hydrogen ion concentration.

pH = - log ([tex]\rm H^+[/tex])

pH = - log (0.02)

pH = 1.69

The pH of the solution prepared by mixing NaOH and [tex]\rm HNO_3[/tex] has been 1.69.

For more information about the pH of the solution, refer to the link:

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What is the mass (g) of a 17.4 mL piece of material, if the density is 3.4 g/mL?

Answers

Answer:

59.16 g

Explanation:

Mass = Density × Volume

= 3.4 g/mL × 17.4 mL

= 59.16 g or

59.2 g (rounded to three significant figures) or

59 g (rounded to two significant figures)

Hope that helps.

The mass of a 17.4 mL piece of material, if the density is 3.4 g/mL is 59.16 grams.

What is density?

Density is defined as the degree to which a material is packed together.

It can also be defined as a material is defined as its total mass (m) divided by its total volume (V).

Density is an important topic because it tells us which compounds will float and which will sink when placed in a liquid. Usually substances float as long as their density is less than that of the liquid in which they are immersed.

Density can be calculated in kilogram per cubic meter or gram per cubic meter.

Density can be expressed as

Density = mass / volume

Given Volume = 17.4 ml

           Density = 3.4 g / ml

Mass = density x volume

Mass = 3.4 x 17.4

Mass = 59.16 grams

Thus, the mass of a 17.4 mL piece of material, if the density is 3.4 g/mL is 59.16 grams.

To learn more about density, refer to the link below:

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How much energy is transferred when 30.0 g of water is cooled from 25.0’C to 12.7’C

Answers

Answer:

1549.8 Joules.

Explanation

Energy transferred =  mass ( in kg) * specific heat of water * change in temperature in ( celsius).

= 0.03 * 4200 * (25 - 12.7)

= 1549.8 Joules.

According to specific heat capacity, 1549.8 joules of  energy is transferred when 30.0 g of water is cooled from 25.0°C to 12.7°C.

What is specific heat capacity?

Specific heat capacity is defined as the amount of energy required to raise the temperature of one gram of substance by one degree Celsius. It has units of calories or joules per gram per degree Celsius.

It varies with temperature and is different for each state of matter. Water in the liquid form has the highest specific heat capacity among all common substances .Specific heat capacity of a substance is infinite as it undergoes phase transition ,it is highest for gases and can rise if the gas is allowed to expand.

It is given by the formula ,

Q=mcΔT, substitution in formula gives, Q=30×4.2×12.3=1549.8 joules.

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A 50 ml graduated cylinder is filled with 23.0 mL of water. Water is then added to the cyl- inder dropwise until 36.0 mL of water are in the cylinder. If it took 265 drops to increase the volume in the cylinder to 36.0 mL. How many drops are in 1 ml?

Answers

Answer:

20 drops

Explanation:

Step 1: Given data

Initial volume of water: 23.0 mLFinal volume of water: 36.0 mLDrops of water added: 265 drops

Step 2: Calculate the volume of water added

We calculate the volume of water added by subtracting the initial volume to the final volume.

36.0 mL - 23.0 mL = 13.0 mL

Step 3: Calculate the number of drops in 1 mL

We will divide the number of drops of water added by the volume they represent.

265 drops / 13.0 mL = 20.4 drops/1 mL ≈ 20 drops/1 mL

Based on the bond energies for the reaction below, what is the enthalpy of the reaction?HC≡CH (g) + 5/2 O₂ (g) → 2 CO₂ (g) + H₂O (g)

Answers

Answer:

1219.5 kj/mol

Explanation:

To reach this result, you must use the formula:

ΔHºrxn = Σn * (BE reactant) - Σn * (BE product)

ΔHºrxn = [1 * (BE C = C) + 2 * (BE C-H) + 5/2 * (BE O = O)] - [4 * (BE C = O) + 2 * (BE O-H).

The BE values are:

BE C = C: 839 kj / mol

BE C-H: 413 Kj / mol

BE O = O: 495 kj / mol

BE C = O = 799 Kj / mol

BE O-H = 463 kj / mol

Now you must replace the values in the above equation, the result of which will be:

ΔHºrxn = [1 * 839 + 2 * (413) + 5/2 * (495)] - [4 * (799) + 2 * (463) = 1219.5 kj/mol

Based on the bond energies, the reaction

HC≡CH(g) + 5/2 O₂(g) → 2 CO₂(g) + H₂O(g) has a standard enthalpy of reaction of - 1222 kJ/mol.

The given reaction is: HC≡CH(g) + 5/2 O₂(g) → 2 CO₂(g) + H₂O(g)

The bond energies (E) of the reactions can help to calculate the standard enthalpy of the reaction (ΔH°rxn) using the following expression.

[tex]\triangle H_ \degrees {rxn} = \sum E_{reactants} - \sum E_{products}[/tex]  

The broken bonds are:   HC≡CH  1 triple C-C bond 2 single C-H bonds  O₂  5/2 double O-O bonds The formed bonds are:    CO₂  4 double C-O bonds H₂O  2 single O-H bonds

Solution:

bond energies in the formula:

[tex]\triangle H_ \degrees {rxn} = 835+ 2(411) + 2.5(494) - 4(799) - 2(459)[/tex]

= 1222 kJ/mol  

Thus, Based on the bond energies, the reaction:

HC≡CH(g) + 5/2 O₂(g) → 2 CO₂(g) + H₂O(g) has a standard enthalpy of reaction of -1222 kJ/mol.  

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Which of the changes are chemical changes?

Sugar is dissolved in water.

A warm soda bottle fizzes when opened.

The copper roof turns green over time.

Baking soda is dissolved in vinegar and bubbles form.

A fog appears on the windshield in cold weather.

Answers

Answer:

two of them are chemical changes

Explanation:

1.) the copper roof turns green over time: it forms a new substance which is the green deposit known as verdigris .

2.) baking soda is dissolved in vinegar and bubbles form: it forms a new substance which is carbon dioxide gas.

   I hope it helps:)

evidence and reasning on when puppies are going to be born

Answers

Answer:

Keep a close eye on her in case she starts giving birth to the next pup at the same time. A greenish/brown discharge may suggest a placenta has separated. If you see this, a puppy should be born within the next 2-4 hours.

You are the manager of a chemical stockroom, and find a bottle containing approximately one liter of a clear and colorless solution of unknown identity and concentration. Your only clue to its identity is that it was found between bottles of silver fluoride and sodium fluoride, so it is likely an aqueous solution of one of those two compounds. You will need to develop a procedure to determine the following: a) The identity of the unknown solution b) The concentration of the unknown solution

Answers

Answer:

Explanation:

Given that for Part A,

the arrangement can be either AgF or NaF.

On the off chance that the arrangement is AgF ,

For this situation AgF conc. can be controlled by titration with standard choride arrangement with potassium chromate as marker.

Get ready 1 M NaCl standard arrangement by dissolving 58.5 gm of NaCl in one liter refined water. Take 25 mL of 1 N NaCl arrangement in a funnel shaped carafe include hardly any drops of potassium chromate marker.

Take the example in the burette and titrate until a perpetual ruddy earthy colored hasten is seen.

[AgF responds with NaCl as follows

AgF + NaCl \small \rightarrow AgCl ( white precipitate)+ NaF

For whatever length of time that CaCl is available in the arrangement Ag structures AgCl hasten as above. When all the chloride particles expelled this way, the overabundance silver structures Ag2CrO4 ( rosy earthy colored encourage ) with the marker.

The dissolvability of AgCl is a lot of lower than the solvency of Ag2CrO4. Consequently the later won't hasten until all chloride particles exhausted.]

Let V2 be the volume of test utilized. At that point grouping of AgF in the example is given by

C2 = V1C1/V2 = 25 x 1/V2.

Rehash the titration by making legitimate weakening if any required.

On the off chance that the arrangement is NaF (Indirect strategy)

Here the Fluoride particle in a referred to volume of test is hastened as leadchlorofluoride with NaCl and PbNO3.

NaF + NaCl + Pb(NO3)2 \small \rightarrow PbClF (hasten) + 2 NaNO3

The above hasten is separated and washed and the washed encourage is taken in a measuring utencil and re-broke up in nitric corrosive to deliver the chloride particles.

PbClF + 2 HNO3 \small \rightarrow Pb(NO3)2 + HCl + HF

Presently include a known abundance of Silver nitrate answer for the above blend to encourage the chloride as silver chloride. Channel wash and evacuate the AgCl encourage.. The Filtrate contains the overabundance silver nitrate is to be dictated by titration with Standard chloride arrangement as in the past technique.

From the first amount of silver nitrate taken and the overabundance silver nitrate decided from the titration, we can decide the amount of silver nitrate responded.

The moles of silver nitrate responded = the moles of Cl delivered = the moles of F in the example.

In this way the convergence of sodium fluoride in the obscure example might be resolved.

For a chemical reaction to be spontaneous at all temperatures, which of the following conditions must be met?
A. ?S° > 0, ?H° > 0.
B. ?S° > 0, ?H° < 0.
C. ?G° > 0.
D. ?S° < 0, ?H° > 0.
E. ?S° < 0, ?H° < 0.

Answers

Answer:

B. S° > 0, ?H° < 0.

Explanation:

Expression for change in free energy is

     ΔG = ΔH - TΔS

For a reaction to be spontaneous , ΔG should be negative . When we watch the relation above , we find that for  ΔG will be negative at any temperature , if ΔH is negative and  ΔS is positive . Then both the terms on the right hand side will be negative and then   ΔG will become negative.

So option B is correct .

What is the mass of an object if its height is 10 meters above ground and its gravitational potential energy (GPE) is 3920 Joules? Remember Gravity = 9.8 meters per seconds squared

Answers

Answer:

40

Explanation:

Your trying to find out the meters so your going to divide 3920J by 10 and 9.8

3920/10/9.8

How many signals does the aldehyde (CH3)3CCH2CHO have in 1H NMR and 13C NMR spectra?
a. five 'H signals and six 13C signals.
b. three 'H signals and four 13C signals.
c. five fH signals and four 13C signals.
d. three 'H signals and six 13C signals.
e. Impossible to know without knowing which isomer was under study.

Answers

Answer:

B- three 1H signals and four 13C signals

Explanation:

Protons in similar environment have the same chemical shift; deshielding by electro-negative elements is also a factor that plays in signal emission in spectroscopy.

All C- atoms in methyl-3 group would count as a single peak, as well as the hydrogen atoms present there on the 13C and 1H spectra respectively.

Let me know if you have any further questions.

what is 8.000 mol / 2.00 L

Answers

Answer:

[tex]M=4.00\frac{mol}{L}=4M[/tex]

Explanation:

Hello.

In this case, we should remember that a molar concentration is defined in terms of the moles and volume in liters as shown below:

[tex]M=\frac{mol}{L}[/tex]

Thus, for the given information we obtain:

[tex]M=\frac{8.000mol}{2.00L}\\ \\M=4.00\frac{mol}{L}=4M[/tex]

Best regards.

The following reaction is investigated: 2N2O(g) N2H4(g) <---> 3N2(g) 2H2O(g) Initially there are 0.100 mol of N2O and 0.25 mol of N2H4, in a 10.0-L container. If there are 0.059 mol of N2O at equilibrium, how many moles of N2 are present at equilibrium

Answers

Answer:

6.2 × 10^-2

Explanation:

Given the reaction:

------------2N2O(g) N2H4(g) <---> 3N2(g) 2H2O(g)

Initial: N2O = 0.1, N2H4 = 0.25-- 0 - - - - - 0

Then: N20 = -2x, N2H4= - x - - - +3x, +2x

Equ : 0.1 - 2x ; 0.25 - x ; +3x ; +2x

At equilibrium :

Add both:

N2O = 0.1 - 2x ;

N2H4 = 0.25 - x;

3N2 = 3x

2H2O = 2x

Moles of N2O at equilibrium = 0.059

Then;

0.1 - 2x = 0.059

-2x = 0.059 - 0.1

-2x = - 0.041

x = 0.041 / 2

x = 0.0205

Moles of N2 present at equilibrium ;

3N2 = 3x

3N2 = 3(0.0205)

= 0.0615

= 0.062 = 6.2 × 10^-2

What is the PH of sebacic acid ?

Answers

Answer:

Sebacic acid is a naturally occurring dicarboxylic acid with the formula (CH2)8(CO2H)2. It is a white flake or powdered solid. Sebaceus is Latin for tallow candle, sebum is Latin for tallow, and refers to its use in the manufacture of candles.

Explanation:

pls mark as brainliest!!!!!

The solubility of silver sulfate (Ag2SO4), in moles per liter, can be expressed in terms of the resulting ion concentrations. Which relationship is correct?

Answers

Answer:

Solubility = [SO₄²⁻]

Explanation:

Solubility of Ag₂SO₄ can be understood seen its Ksp equilibrium:

Ag₂SO₄(s) ⇄ SO₄²⁻ + 2Ag⁺

Where 1 mole of silver sulfate dissolves producing 1 mole of sulfate ion and 2 moles of silver ion.

Solubility is defined as: "the amount of solid that can be dissolved per liter of solution".

In the dissolution process, the moles of Ag₂SO₄ dissolved are equal to moles of SO₄²⁻.

That means:

Solubility = [SO₄²⁻]

Is HCL a pure substance? Look at me already asking chemistry questions! //!roll eyes emoji/2#/

Answers

Answer:

yes it is as it has a chemical formula which cannot be changed

Please Help Me What is chromatography? Question 2 options: A) A picture taken with a chrome camera. B) The stationary phase. C) The separation of substances in a mixture by differences in their attraction to a substance over which they are passed.

Answers

Explanation:

Chromatography = It is a method in which a drop of mixture is spot is spotted at one of the end of the paper. After some time whenever it gets dried, it is then dipped into solvent without toushing the spot. As a result the mixtures gets separated.

Hence, the correct option is (c) "The separation of substances in a mixture by differences in their attraction to a substance over which they are passed"

In your own words, describe the concept of specific heat capacity and the effects that it has on temperature changes. Give an example of specific heat capacity that you see in your everyday life. Why did you select this example

Answers

Answer:

The specific heat capacity is the heat that a body or a system needs to administer so that it can increase its internal temperature.

Explanation:

The calorific capacity is measured in several units, it varies a lot between products, reactants or the same systems since each one is independent in its composition and this conditions it.

As for its mathematical calculation, it is the quotient, that is, the division between the dose of energy transferred to a body and the change in temperature that it experiences.

I know how to solve it with D=M/V and M1V1 however the answer isn’t correct. Help me please

Answers

Answer:

23.28 g of O2.

Explanation:

We'll begin by calculating the mass of hexane. This can obtain as follow:

Volume of hexane = 10 mL

Density of hexane = 0.66 g/mL

Mass of hexane =?

Density = mass /volume

0.66 = mass of hexane /10

Cross multiply

Mass of hexane = 0.66 x 10

Mass of hexane = 6.6 g

Next, we shall write the balanced equation for the reaction. This is given below:

2C6H14 + 19O2 —> 12CO2 + 14H2O

Next, we shall determine the masses of C6H14 and O2 that reacted from the balanced equation. This can be obtained as follow:

Molar mass of C6H14 = (12.01x6) + (1.008 x 14)

= 72.06 + 14.112

= 86.172 g/mol

Mass of C6H14 from the balanced equation = 2 x 86.172 = 172.344 g

Molar mass of O2 = 16x2 = 32 g/mol

Mass of O2 from the balanced equation = 19 x 32 = 608 g

From the balanced equation above,

172.344 g of C6H14 reacted with 608 g of O2.

Finally, we shall determine the mass of O2 needed to react with 10 mL (i.e 6.6 g) of hexane, C6H14. This can be obtained as follow:

From the balanced equation above,

172.344 g of C6H14 reacted with 608 g of O2.

Therefore, 6.6 g of C6H14 will react with = (6.6 x 608)/172.344 = 23.28 g of O2.

Therefore, 23.28 g of O2 is needed for the reaction.

How much energy has your body used, in joules, if your health indicates the 450 Calories were burned during your workout? How many Calories are in a dinner containing 3.5 x 106 joules?

Answers

Answer:

107.55 J, 88.67 cal

Explanation:

1 Joule = 0.239 Calories

450 cal x 0.239 = 107.55 J

3.5 x 106 = 371

371 J = 371 x 0.239 cal

371 J = 88.67 cal

Perform the following mathematical operations and express the correct answer to the proper number of significant figures or decimal places as required.
7.71 + 4.31 + 1.7 + 4.00141=

7.71 x 4.31 x 1.7 x 4.00141=

Answers

Answer: 7.71 + 4.31 + 1.7 + 4.00141= 17.72141

7.71 x 4.31 x 1.7 x 4.00141=  226.04433255

Explanation:

(i) 7.71 + 4.31 + 1.7 + 4.00141

= 7.71000+4.31000+1.70000+ 4.00141   [make like decimals]

= 17.72141    [By adding the corresponding places ]

(ii)  7.71 x 4.31 x 1.7 x 4.00141 =( 7.71 x 4.31) x 1.7 x 4.00141

= 33.2301 x 1.7 x 4.00141

= (33.2301 x 1.7) x 4.00141

=  56.49117 x 4.00141

= ( 56.49117 x 4.00141 )

=  226.04433255

Hence,  7.71 + 4.31 + 1.7 + 4.00141= 17.72141

7.71 x 4.31 x 1.7 x 4.00141=  226.04433255

A solution of sulfuric acid contains 35.2% by mass of H2SO4 and has a density of 1.27 g/mL. What is the molarity of H2SO4 in this solution?

Answers

Answer:

[H₂SO₄] = 4.56 M

Explanation:

35.2 % by mass of H₂SO₄ means that in 100 g of solution, we have 35.2 grams of solute.

We convert the mass of solute to moles.

35.2 g / 98 g/mol = 0.359 moles

These moles are contained in 100 g of solution, so we use density to determine the volume.

1.27 g/mL = 100 g / volume

Volume = 100 g / 1.27 g/mL = 78.74 mL

Molarity is mmol /mL (either we can say, moles in 1L of solution).

We convert the moles to mmoles → 0.359 mol . 1000 = 359 mmoles

M = 359 mmoles/74.74 mL = 4.56 M

Consider the two gaseous equilibria: The values of the equilibrium constants K 1 and K 2 are related by

Answers

The question is missing parts. The complete question is as follows.

Consider the two gaseous equilibria involving SO2 and the corresponding equilibrium constants at 298K:

[tex]SO_{2}_{(g)} + \frac{1}{2}O_{2}[/tex] ⇔ [tex]SO_{3}_{(g)}[/tex]; [tex]K_{1}[/tex]

[tex]2SO_{3}_{(g)}[/tex] ⇔ [tex]2SO_{2}_{(g)}+O_{2}_{(g)}; K_{2}[/tex]

The values of the equilibrium constants are related by:

a) [tex]K_{1}[/tex] = [tex]K_{2}[/tex]

b) [tex]K_{2} = K_{1}^{2}[/tex]

c) [tex]K_{2} = \frac{1}{K_{1}^{2}}[/tex]

d) [tex]K_{2}=\frac{1}{K_{1}}[/tex]

Answer: c) [tex]K_{2} = \frac{1}{K_{1}^{2}}[/tex]

Explanation: Equilibrium constant is a value in which the rate of the reaction going towards the right is the same rate as the reaction going towards the left. It is represented by letter K and is calculated as:

[tex]K=\frac{[products]^{n}}{[reagents]^{m}}[/tex]

The concentration of each product divided by the concentration of each reagent. The indices, m and n, represent the coefficient of each product and each reagent.

The equilibrium constants of each reaction are:

[tex]SO_{2}_{(g)} + \frac{1}{2}O_{2}[/tex] ⇔ [tex]SO_{3}_{(g)}[/tex]

[tex]K_{1}=\frac{[SO_{3}]}{[SO_{2}][O_{2}]^{1/2}}[/tex]

[tex]2SO_{3}_{(g)}[/tex] ⇔ [tex]2SO_{2}_{(g)}+O_{2}_{(g)}[/tex]

[tex]K_{2}=\frac{[SO_{2}]^{2}[O_{2}]}{[SO_{3}]^{2}}[/tex]

Now, analysing each constant, it is easy to see that [tex]K_{1}[/tex] is the inverse of [tex]K_{2}[/tex].

If you doubled the first reaction, it will have the same coefficients of the second reaction. Since coefficients are "transformed" in power for the constant, the relationship is:

[tex]K_{2}=\frac{1}{K_{1}^{2}}[/tex]

. A sample of crude oil has a density of 0.87 g/mL. What volume (in liters) does a 3.6 kg sample of this oil occupy

Answers

Answer:

The volume is 4.13793 L

Explanation:

Density is a quantity that expresses the relationship between the mass and the volume of a body, so it is defined as the quotient between the mass and the volume of a body:

[tex]density=\frac{mass}{volume}[/tex]

Density is a characteristic property of every body or substance.

The most commonly used units of density are [tex]\frac{kg}{m^{3} }[/tex] or [tex]\frac{g}{cm^{3} }[/tex] for solids, and [tex]\frac{kg}{L}[/tex] or [tex]\frac{g}{mL}[/tex] for liquids and gases.

In this case, you know:

density= 0.87 [tex]\frac{g}{mL}[/tex]mass= 3.6 kg= 3,600 g (being 1 kg=1,000 g)volume= ?

Replacing:

[tex]0.87\frac{g}{mL} =\frac{3,600 g}{volume}[/tex]

Solving:

[tex]volume =\frac{3,600 g}{0.87\frac{g}{mL}}[/tex]

volume= 4,137.93 mL

Being 1,000 mL=1 L, then volume= 4,137.93 mL= 4.13793 L

The volume is 4.13793 L

Carl is approaching the Canadian border on a trip from Vermont to Montreal, Canada, and the gas in his tank is low. He wants to determine if it would be more cost efficient to buy gas before or after crossing the border. He consults Gas Buddy and finds that the cost of gas in the US is 2.21 USD per gallon, while just across the border it is $1.40 CAD per liter. The currency exchange rate is $0.750 USD per $1.000 CAD. How much more gas (in gallons) will he get for $20 USD if he buys the gas before crossing the border instead of buying it in Canada? (1.000 liter = 0.264 US gallons).

Answers

Answer:

4.02 gallons

Explanation:

From the illustration:

$0.750 USD = $1.00 CAD

Therefore,

$20 USD =  20 x 1/0.750 = $26.667 CAD

The cost of gas in the US is 2.21 USD per gallon.

20 USD will be able to get 20/2.21 = 9.05 gallons

The cost of gas across the border is 1.40 CAD per liter.

0.264 gallon = 1 liter

1 gallon = 1/0.264 = 3.788 liters

This means that 1 gallon = 3.788 x 1.40 = 5.303 CAD

Hence, 26.667 CAD will get 26.667/5.303 = 5.03 gallons

If he buys a $20 USD gas before crossing the border, Carl will get 9.05 gallons, but if he buys it after crossing the border, he will get 5.03 gallons. Therefore, if he buys the gas before crossing the border instead of buying it in Canada, he will get 9.05 - 5.03 = 4.02 gallons extra.

Which of the following is the most effective buffer system for a pH value of 4.45? a) H3PO4/H2PO4 (Kal for H3PO4 is 6.9 times 10-3) b) H_2CO_3/HCO_3^- (K_a1 for H_2CO_3 is 4.8 times 10^-7) c) HCO_3^-/CO_3^2- (K_a2 for H2CO_3 is 4.8 times 10^-11) d) H_2S/HS^- (K_a1 for H_2S is 8.9 times 10^-8) e) HC_2O_4^-/C_2O_4^2- (K_a2 for H_2C_2O_4 is 5.1 times 10^-5)

Answers

Answer:

Option e.

Acid oxalate / oxalate

Explanation:

To determine the most efecttive buffer, we must look the pKas of the system.

If pKa value is similiar to buffer's pH, the variation of pH by  the addition of a small amount of strong acid or base is minimal. It would be a good buffer.

a. H₃PO₄ / H₂PO₄⁻   Ka 1 = 6.9×10⁻³

pKa = - log Ka → - log 6.9×10⁻³ = 2.16

b. H₂CO₃ / HCO₃⁻    Ka1 = 4.8×10⁻⁷

pKa = - log Ka → - log 4.8×10⁻⁷ = 6.32

c. HCO₃⁻ / CO₃⁻²    Ka2 = 4.8×10⁻¹¹

pKa = - log Ka → - log 4.8×10⁻¹¹ =  10.32

d.  H₂S / HS⁻    Ka1 = 8.9×10⁻⁸

pKa = - log Ka → - log 8.9×10⁻⁸ = 7.05

e. HC₂O₄⁻ / C₂O₄⁻²     Ka2 = 5.1×10⁻⁵

pKa = - log Ka → - log 5.1×10⁻⁵ = 4.29

pKa which is so close to the ph's value is e, acid oxalate / oxalate

Speed of light is 3.0x108 m/s. Convert speed of light unit into miles per hour (mi/s). 3.0x108 m/s =

mi/s

Report your answer in the format with correct sig figs:

5x1013
Given:

1 mile = 1.61 km

1 km = 1000 m

Answers

Explanation:

1 mile = 1.61 km and 1 km = 1000 m

The speed of light is [tex]3\times 10^8\ m/s[/tex]. We need to convert it into miles per hour.

1 m = (1/1000) km

[tex]3\times 10^8\ m=\dfrac{3\times 10^8}{1000}\\\\=3\times 10^5\ km[/tex]

1 km = (1/1.61) miles

[tex]3\times 10^5\ km =\dfrac{3\times 10^5}{1.61}\\\\=1.86\times 10^5\ \text{miles}[/tex]

So, [tex]v=1.86\times 10^5\ \text{miles/s}[/tex]

Hence, this is the required solution.

What is the percent yield in a reaction between 42.6 g O2 and 49.2 g Al if 72.4 g of Al2O3 is produced?

Answers

Answer:

229%

Explanation:

The equation of the reaction is;

4Al(s) + 3O2(g) ----> 2Al2O3(s)

We must first determine the limiting reactant;

Number of moles of Al2O3 produced = mass/molar mass = 72.4g/101.96 g/mol = 0.71 moles

For Al

Number of moles reacted = mass/molar mass = 49.2g/27 g/mol = 1.8 moles

If 4 moles of Al yields 0.71 moles of Al2O3

1.8 moles of Al will yield 1.8 × 0.71/4 = 0.32 moles of Al2O3

For O2

Number of moles reacted = mass/molar mass = 42.6g/32g/mol = 1.33 moles

If 3 moles of O2 yields 0.71 moles of Al2O3

1.33 moles of O2 will yield 1.33 × 0.71/3 = 0.31 moles of Al2O3

Oxygen is the limiting reactant.

% yield = actual yield/ theoretical yield × 100/1

% yield = 0.71 moles/0.31 moles × 100

% yield = 229%

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