vessel a contains co(g) at 0 oc and 1 atm. vessel b contains so2(g) at 20 oc and 0.5 atm. the two vessels have same volume. a. which vessel contains more molecules? b. which contains more mass? c. in which vessel is the average kinetic energy of molecules higher? d. in which vessel is the rms speed of molecules higher?

Answer: 50%

Explanation:

The best we get from the moon is a full moon, which is just half of the moon shining from the earth while the other half is in complete darkness.

Answer:

Can cause sunburn, premature aging of the skin and damage molecules in skin cells

Explanation:

Answer:

pueden afectar directo al adn de las celulas de la piel y son causantes de las quemaduras del sol

Explanation:

The major product formed when the given compound reacts with RCO3H is an ester. This reaction, known as an esterification reaction, involves the conversion of a carboxylic acid to an ester. The specific ester formed will depend on the structure of the given compound and the nature of the R group in RCO3H.

The given compound, when reacted with RCO3H, undergoes an esterification reaction. In this reaction, the carboxylic acid group (-COOH) of the compound reacts with the RCO3H, which is a reagent known as an acid anhydride. The reaction proceeds through an intermediate where the oxygen of the carboxylic acid attacks the carbon of the acid anhydride, leading to the formation of an acyl-oxygen bond. Simultaneously, one of the oxygens in the anhydride group is cleaved off, forming a leaving group. The acyl-oxygen bond is then broken, and the resulting oxygen forms a bond with the carbon, leading to the formation of the ester. The specific ester formed will depend on the structure of the given compound and the nature of the R group in RCO3H. The R group can vary and may contain different substituents, such as alkyl or aryl groups. The structure of the given compound will determine which functional groups participate in the reaction and influence the regioselectivity and stereochemistry of the product. Thus, without knowing the exact structure of the compound and the R group in RCO3H, it is not possible to determine the specific major product formed in this reaction.

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Answer:

25N

Explanation:

The force calculated to be 25kgm/s² has a value of 25N.

To solve this problem, we need to consider the unit of force. The unit of force is expressed in N, newtons.

  To find force;

           Force  = mass x acceleration

            Unit of mass  = kg

                      acceleration  = m/s²

 So;

        1N  = 1kgm/s²

Therefore 25kgm/s² is the same as 25N

Answer:  The atomic mass in the periodic table is a weighted average of all naturally occuring isotopes of zinc.

Explanation:  The elements consist of protons, neutrons, and electrons.  Protons and nuetrons are assigned a mass of 1 amu (atomic mass unit).  Electrons are too small to make a big difference in the mass, so they are assign a 0 for their amu.  Since the elements consist of just these three particles, any one atom must add up to a whole number atomic mass.  Zinc consists of 30 protons, by definition.  A neutral atom has 30 electrons.  The balance of a zinc atom is made up of neutrons.  A Zn-65 atom would have 30 protons and 35 neutrons:

30 Protons = 30 amu

35 Neutrons = 35 amu

30 Electrons = 0 amu

Total AMU = 65

Any ONE Zn atom should have a whole number atomic mass.  But zinc in the environment has isotopes of zinc - atoms that contain differing numbers of neutrons.  A Zn-66 atom would have 36, instead of 35 neutrons.  (P:30, N:36, E:0 = 66 amu).  The atomic mass in the Periodic Table reflects the average of these isotopes as found in the environment.  If the 65 and 66 isotopes were the only ones found naturally, an atomic mass of 65.38 would suggest around 60% of the atoms are Zn-65 and 40% are Zn-66.  The average weight atomic mass would would be 65.40

The atomic mass of zinc which is listed in periodic table is 65.38 but there is no zinc atom with that mass this is due to average atomic mass of naturally occurring isotopes of zinc.

Atomic mass is defined as the mass of an atom which has units of Dalton which is a non-SI unit.One Dalton is defined as approximately equal 1/12 of mass of carbon-12 atom.

The protons and neutrons account for the mass of an atom as they reside inside the nucleus . Electrons do not contribute to the atomic mas as they have negligible mass.

The atomic mass of atoms,ions or nuclei is slightly less than the sum of masses of constituent protons , neutrons and electrons due to the loss in binding energy.Addition of atomic masses of atoms leads to a quantity called as molecular mass.

The observed atomic mass is slightly less than the sum of masses of protons, neutrons  which is called mass defect which arises due to difference in binding energy.

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Answer:

D) 14 Electrons

Explanation:

I took the test and got it right

The total number of electrons that can occupy the f sublevel is 14 electrons. Thus option D is correct.

Electrons can be defined as a negatively charged subatomic particle that together with protons and neutrons forms an atom's nucleus. It is the lightest subatomic particles.

s orbital has 2 electrons in pair

p orbital has 6 electrons in pair

d orbital has 10 electrons in pair

f orbital has 14 electrons in pair.

Thus, the total number of electrons that can occupy the f sublevel is 14 electrons. Thus option D is correct.

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The molar mass of the nondissociating, nonvolatile sugar is estimated to be 342.73 g/mol.

When 39.8 g of a nondissociating, nonvolatile sugar is dissolved in 200.0 g of water, the boiling point of the water is raised by 0.30°C. To estimate the molar mass of the sugar, we can use the formula:

∆Tb = (Kb x W2)/W1

Here, Kb is the boiling point elevation constant with a value of 0.512 °C/m. W1 represents the mass of the solvent in kilograms, which is 0.200 kg. W2 is the mass of the solute in kilograms.

Given that the boiling point elevation (∆Tb) is 0.30°C and the mass of the solute (sugar) W2 is 39.8 g (0.0398 kg), we can calculate the molar mass using the formula:

Molar mass = (mass × (R × ∆Tb)) / (Kb × W2)

Substituting the known values:

Molar mass = (0.0398 × (8.31 × 0.30)) / (0.512 × 0.200)

After calculation, we find that the molar mass of the nondissociating, nonvolatile sugar is estimated to be 342.73 g/mol.

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A) Solution B of 100 mM is more concentrated than Solution A of 0.5 mM.

B) Solution B of 3.7mg/dL is more concentrated than Solution A of 3.5mg/dL.

C) Solution A = 0% (w/v) and Solution B = 0% (w/v) equal concentrations.

D) Solution A of 10⁻³ M is more concentrated than Solution A of 10⁻⁵M.

E) Solution B of 12.3% (w/v) is more concentrated than Solution A of 12.23% (w/v)

In chemistry, concentration can be described as the abundance of components divided by the total volume of a solution. Several kinds of mathematical descriptions can be distinguished: mass concentration, number concentration, molar concentration, and volume concentration.

The concentration is defined as the kind of chemical mixture, but most frequently refers to solutes and solvents in solutions. The molar concentration has variants, such as osmotic concentration and normal concentration.

Molarity, molality, Normality, and weight/ volume percentage are also used to define the concentration of the solution.

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Answer:

The correct answer is - butanoic acid

Explanation:

An empirical formula is the general formula or base formula that is the simplest positive integer ratio of atoms present in a compound. The empirical formula is specific to specific compounds or types of compounds.

Among the given options, the empirical formula C2H4O is butanoic acid as the molecular formula of butanoic acid is - C4H8O2 which is multiple of C2H4O. Other compounds have different than this empirical formula.

Explanation:

basically, you'll just have to convert the given grams into moles, convert that into moles of Br, and then convert it again into grams of Br

The amount of gram of \(Br_{2}\) will be  65.1 grams in equation \(2AlCl_{3} + 3Br_{2}\)→ \(2AlBr_{3} + 3Cl_{2}\).

The definition of a gram was given as the mass of one cubic centimeter of pure water at 4 degrees Celsius.

Calculation of amount of  \(Br_{2}\)

The given equation is \(2AlCl_{3} + 3Br_{2}\)→ \(2AlBr_{3} + 3Cl_{2}\).

The mass of  \(Br_{2}\)  = Count of bromine molecule × molar mass

The mass of  \(Br_{2}\)   = 0.4072 × 159.8

The mass of  \(Br_{2}\)   = 65.08 mol.

The amount of gram of  \(Br_{2}\)  will be  65.1 grams in equation  \(2AlCl_{3} + 3Br_{2}\)→ \(2AlBr_{3} + 3Cl_{2}\).

Therefore, the answer will be option (a).

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Answer:

ok

Explanation:

Answer:

Yes it is balanced

Explanation:

You can check by finding the total of each element on both sides so:

Left Hand Side :

Carbons: 6

Oxygens: 6x2 (carbon dioxide) and 6 (water) so there are 18

Hydrogens: 6x2 (carbon dioxide) so 12

Right Hand Side:

Carbons: 6

Oxygens: 6 (glucose) then 6x2 (oxygen) so there are 18

Hydrogens: 12

Answer:

The density of the mystery metal is 2.7g/cm^3.  It is likely aluminum, Al.

Explanation:

Density is defined as (mass/unit volume).  We know the volume of the metal sample by the fact it displaced 6.2 ml of water.  Since the mateal sample had a mass of 16.74 grams, we find the density by dividing the mass by the volume:

Density = (mass)/(volume) = (16.74g)/(6.2ml)

Density = 2.7 g/ml

Note that the ubntis for density can vary greatly. kg/cm^3 is a possible unit.

since 1 ml = 1 cm^3, we can also say the density of the metal sample is 2.7 g/cm^3.  [This is a more common unit for this type of measurement]

A reference book of densities can be search to find what metals have this density.  See the attached excerpt form one such table.

While not definitive, it can be seen that aluminum, Al, is a good candidate for the ID of this metal.  It has a density of 2.7 g/cm^3, although different forms may deviate slightly.  The metal is most likely aluninum.

The mass of Nickel(II) phosphate = 5508.646 g

Given

9.06 x 10²⁴ of Nickel(II) phosphate

Required

The mass

Solution

1 mol = 6.02 x 10²³ particles(molecules,ions, atoms)⇒Avogadro's number

So mol Nickel(II) phosphate :

n = N : No

n = 9.06 x 10²⁴ : 6.02 x 10²³

n = 15.05

mass of Nickel(II) phosphate(Ni₃(PO₄)₂ , MW=366.023 g/mol) :

mass = mol x MW

mass = 15.05 x 366.023

mass = 5508.646 g

The average reservoir pressure just before the initiation of a flame event, as well as the greatest normal operating pressure at a specific point in the system.

What is a good instance of pressure?

Holding a knife against a piece of fruit will demonstrate pressure in an easy way. It won't cut the surface of the fruit if you press the flat part of the knife against it. A sizable area is affected by the force (low pressure)

What are the many sorts of pressure?

The force applied physically to an object is referred to as pressure. Per unit area, the force is applied perpendicularly to the surfaces of the objects. The fundamental pressure formula is F/A. (Force per unit area). measure of pressure

Briefing:

W = 352 joule & n = 0.325 & T by kelvin = 22+273 = 295 K & Pf= 1.76

352 = (0.325 x 8.314 x 295)㏑(Pi/Pf)

∴㏑(Pi/Pf) = -352 / (0.325 x 8.314 x 295)

㏑(Pi/Pf) = - 0.44

Pi/Pf = 0.644

∴ Pi =1.76 x 0.644 = 1.13 atm

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Answer: C. Seal

Explanation: because a seal has alose characteristics

Answer:

The rate reduces by a factor cf /2.

Answer:

malleable

Explanation:

The image represent in malleable property of metal.

The image possibly represents the photoelectric effect of a metal, which is when it emits electrons after being exposed to electromagnetic radiation. Metals are also characterized by physical properties such as conductivity, malleability, metallic luster, and metallic bonding.

Based on your question, the image possibly represents the photoelectric effect, a key property of metals. This phenomenon occurs when a metal surface exposed to electromagnetic waves of a certain frequency absorbs radiation and emits electrons. These emitted electrons are called photoelectrons. Metals can also exhibit free electron model behavior, where electrons freely roam within the metal structure.

Metals possess unique physical properties like conductivity, malleability, and metallic luster. Malleability refers to the metal's ability to deform without breaking, while conductivity refers to the metal's ability to transfer heat or electricity. A metallic luster gives metals their characteristic shiny appearance.

Finally, metals are also known for their metallic bonding—a unique force that holds together the atoms within a metallic solid. Metallic bonding gives rise to many useful and varied bulk properties of metals.

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Answer:

evaluation

Explanation: Big brain

Answer:

3.Test for unsaturation is carried out by either Baeyer's test or bromine water test.

(a) What is Baeyer's reagent?

Ans:

Baeyer's reagent is an alkaline solution of Potassium Permanganate \(\bold{KMnO_4}\) .

(b) How does it detect presence of double bond in organic compound? Write the reaction involved.

Ans:

When a dilute alkaline solution of \(\bold{KMnO_4}\) is added to the given unsaturated organic compound, if the purple color of permanganate solution disappears, the presence of double bond is indicated.

For example:

2\(\bold{KMnO_4}\) + 2KOH \(\longrightarrow\)\(\bold{2K_2MnO_4 + H_2O+[O]}\)

\(\bold{CH_2=CH_2(ethene) +H_2O+[O]\longrightarrow CH_2OH-CH_2OH (ethylene \:glycol :Colourless)}\)

(c) Write the product when ethyne is treated with Baeyer's reagent.

Ans:

Product will not produce as ethyne doesn't contain double bond.

(d) Benzene (an aromatic hydrocarbon) does not give Baeyer's test though it has double bonds. Why?

Ans:

As Benzene is an unsaturated compound. The structure of benzene contains 3 alternate single double bonds.

(e) Write the reaction for bromine water test for ethyne.

Ans:

Reaction:

\(\bold{CH_2CH_2(ethene)+Br_2( red \: color) \longrightarrow CH_2Br-CH_2Br}\)  

(1,2-dibromoethane (colorless}

Kolbe electrolytic method can be used to prepare alkane, alkene and alkyne.

(a) At which electrode, alkane, alkene or alkyne is obtained?

Ans:

alkane, alkene or alkyne is obtained at anode.

(b) Which electrolyte should be chosen for alkane, alkene or alkyne for the method? Suggest an example of electrolyte for each.

Ans:

alkane: Potassium or sodium salt of Carboxylic acid

alkene: Potassium or sodium salt of dicarboxylic acid

alkyne: Potassium or sodium maleate

(c) If potassium salt of dicarboxylic acid is taken for the method, which hydrocarbon is prepared?

Ans: If potassium salt of dicarboxylic acid is taken for the method, Alkene hydrocarbon is prepared.

(d) Write the detail process for the electrolysis of aqueous potassium maleate.

Ans: Attachment

(e) Suggest a demerit of the method.

Ans:

Demerits:

Answer:

By just counting sigma bond and lone pair in central atom , one can easily find hybridization.

CO2 sp

SO2 sp2

NH3 sp3

BCl3 sp2

The following equation can be used to determine the equilibrium constant (Keq) for this reaction: Keq is equal to [SO3]2/[SO2][O2].

Since the concentration of SO3 in this situation is 0.250M at equilibrium, the Keq value is calculated as 0.2502 / (0.500 x 0.400) = 0.4.

Given that the Keq value is more than 1, this indicates that the reaction is marginally biassed in favour of the products.

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Answer:1. The sound waves arrive at the pinna (auricle), the only visible part of the ear.

2. Once the sound waves have passed the pinna, they move into the auditory canal (external

acoustic meatus) before hitting the tympanic membrane (eardrum).

3. Once the sound waves reach the tympanic membrane, it begins to vibrate and they enter

into the middle ear.

4. The vibrations are transmitted further into the ear via three bones (ossicles): malleus

(hammer), incus (anvil), and the stapes (stirrup). These three bones form a bridge from

the tympanic membrane to the oval window.

5. Once sound passes through the oval window, it enters into the cochlea in the inner ear.

6. Hair cells in the organ of Corti (within the cochlea) are stimulated which in turn

stimulates the cochlear branch of the vestibulocochlear nerve.

7. The cochlear nerve then transmits electrical impulses to the auditory region of the brain

in the temporal lobe.

Explanation:

ΔH for this reaction : -2668.4 (exothermic)

The change in enthalpy in the formation of 1 mole of the elements is called enthalpy of formation  

The enthalpy of formation measured in standard conditions (25 ° C, 1 atm) is called the standard enthalpy of formation (ΔHf °)  

(ΔH) can be positive (endothermic = requires heat) or negative (exothermic = releasing heat)  

The value of ° H ° can be calculated from the change in enthalpy of standard formation:  

∆H ° rxn = ∑n ∆Hf ° (product) - ∑n ∆Hf ° (reactants)  

you can search the value of ΔHf on the internet

∆H ° rxn =( ∆H Al₂O₃+∆H AlCl₃+3.∆H NO + 6.∆H H₂O)-(∆H Al+∆H NH₄ClO₄)

\(\tt \Delta H^o~rxn=(-1669.8+-704+(3\times 90.4)+(6.\times -241.8)-(3\times 0+3\times -295)\\\\\Delta H6o~rxn=-3553.4+885=-2668.4~kJ/mol\)

Answer:

40km/hr

Explanation:

Speed = Distance ÷ Time and since they both have different units we have to change one of them. Therefore if we change 30 minutes to half an hour we say 20km ÷ ¹/2

Answer:

5.12x10¹¹ millimeters

Explanation:

Milli is a prefix used in science and engineering to decribe the number as the exponent x10⁻³. In the prefix kilo, the number is at the exponent x10³.

5.12x10⁵ kilometers are:

5.12x10⁵ kilometers * (1000m / 1km) = 5.12x10⁸ meters

5.12x10² meters * (1m / 1000millimeters) = 5.12x10¹¹ millimeters

The half-life of the Uranium-235, given that 25% of it is remaining in the rock is 713 million years

We know that half life is the time taken for half of a substance to decay.

Thus, to obtain the half life of the Uranium-235, we shall calculate the number of half lives that has elapsed while in the rock. This can be obtained as follow:

2ⁿ = N₀ / N

2ⁿ = 100 / 25

2ⁿ = 4

2ⁿ = 2²

n = 2

Knowing that 2 half lives has already elapsed, we shall determine the half-life of the Uranium-235. This can be obtained as follow:

t = n × t½

Divide both sides by n

t½ = t / n

t½ = 1.426 / 2

t½ = 713 million years

Therefore, we can conclude that the half-life is 713 million years

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The molarity of the given solution of HCl is 1.184 M.

The molarity of a solution is defined as the number of moles of solute dissolved in one liter (1000 mL) of the solution.

To calculate the molarity of HCl in the given solution, we need to first determine the number of moles of HCl present in the solution:

Calculate the molecular weight of HCl:

HCl = 1(atomic mass of hydrogen) + 35.5(atomic mass of chlorine) = 36.5 g/mol

Convert the given mass of HCl to moles:

3.24 g HCl / 36.5 g/mol = 0.0888 mol HCl

Calculate the molarity of the solution:

Molarity = moles of solute / volume of solution in liters

Volume of solution = 75 mL = 75/1000 L = 0.075 L

Molarity = 0.0888 mol / 0.075 L = 1.184 M

Therefore, the molarity of the given solution of HCl is 1.184 M.

What is molarity ?

Molarity is a measure of the concentration of a solution, defined as the number of moles of solute dissolved per liter of solution. The unit of molarity is mol/L (or sometimes written as M).

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A) pH after adding 9 mL of NaOH is 6.91 ; B) pH after adding 100 mL of NaOH is 13.18 ; C) pH after adding 400 mL of NaOH is 8.42 ; D) pH after adding 500 mL of NaOH is 12.

pH is used to find out the the basicity and acidity of a solution

HA + NaOH -> NaA + H₂O

Ka = [NaA][H₃O⁺] / [HA]

moles of acid = initial concentration x initial volume = 0.2 M x 0.1 L = 0.02 moles

A.) We add 9 mL of 0.05 M NaOH to acid solution. The total volume of the solution is now 100 mL + 9 mL = 0.109 L.

moles of NaOH added = concentration x volume = 0.05 M x 0.009 L = 4.5 x 10⁻⁴ moles

moles of acid remaining = initial moles - moles of NaOH added = 0.02 moles - 4.5 x 10⁻⁴ moles = 0.01955 moles

moles of NaA formed = moles of NaOH added = 4.5 x 10⁻⁴ moles

moles of H₃O⁺ produced = moles of NaOH added = 4.5 x 10⁻⁴ moles

moles of HA remaining = moles of acid remaining - moles of NaA formed = 0.01905 moles

Ka = [NaA][H₃O] / [HA]

[ H₃O⁺] = Ka x [HA] / [NaA]

[ H₃O⁺] = 2.9 x 10⁻⁶  x 0.01905 / 4.5 x 10⁻⁴ = 1.22 x 10⁻⁷ M

pH = -log[H₃O⁺] = -log(1.22 x 10⁻⁷) = 6.91

Therefore, pH after adding 9 mL of NaOH is 6.91.

B.) 100 mL of 0.05 M NaOH, so all the acid has been neutralized.

moles of NaOH added = concentration x volume = 0.05 M x 0.1 L = 0.005 moles

moles of NaOH remaining = moles of NaOH added - moles of acid = 0.005 moles - 0.02 moles = -0.015 moles (negative value indicates excess NaOH)

[OH-] = moles of excess NaOH / total volume = 0.015 moles / 0.1 L = 0.15 M

pOH = -log[OH-] = -log(0.15) = 0.82

So pH= 14 -pOH = 14 -0.82= 13.18

Therefore, pH after adding 100 mL of NaOH is 13.18.

C.) We have added a total of 400 mL of 0.05 M NaOH.

moles of NaOH added = concentration x volume = 0.05 M x 0.4 L = 0.02 moles

moles of NaOH remaining = moles of NaOH added - moles of acid = 0.02 moles - 0.02 moles = 0 moles

[NaA] = moles of NaA / total volume = 0.02 moles / 0.5 L = 0.04 M

Kb = Kw / Ka = 1.0 x 10⁻¹⁴ / 2.9 x 10⁻⁶ = 3.45 x 10⁻⁹

[OH-] = √(Kb x [NaA]) = √(3.45 x 10⁻⁹ x 0.04) = 2.61 x 10⁻⁶ M

pOH = -log[OH-] = -log(2.61 x 10⁻⁶) = 5.58

So, pH= 14 -pOH = 14 -5.58 = 8.42

Therefore, pH after adding 400 mL of NaOH is 8.42.

D.) We have added a total of 500 mL of 0.05 M NaOH.

moles of NaOH added = concentration x volume = 0.05 M x 0.5 L = 0.025 moles

moles of NaOH remaining = moles of NaOH added - moles of acid = 0.025 moles - 0.02 moles = 0.005 moles

[OH-] = moles of excess NaOH / total volume = 0.005 moles / 0.5 L = 0.01 M

pOH = -log[OH-] = -log(0.01) = 2

So pH =14 -pOH =14 -2=12

Therefore, pH after adding 500 mL of NaOH is 12.

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