What volume of O2 is produced when 28. 5 g of hydrogen peroxide (H2O2) decomposes to form water and oxygen at 150°C and 2. 0 atm?

Answer:

liquid

Explanation:

Gas has very low density but spreads fast. Solid has high density.

The oxygen consumption rate of a 200 kg seal is 10 ml/(g h). If it has oxygen stores of 2 liters, then minutes can it dive before running completely out of oxygen is 0.06min.

Oxygen is one of the most vital element of the organism which is required by almost all the cellular and for all metabolic activities to perform body function.

It is consumed by the respiratory system and passed ro each cell by circulatory system of the body.

Given,

Mass of seal = 200kg = 200000 g

Rate of oxygen consumption = 10 ml/gh

The volume of stored O2 = 2000mL

The time of consumption is calculated as

t = V/mr

t = 2000/200000×10

t = 0.001 hrs

In minutes

t = 0.06min.

Thus 0.06min is required to dive out the oxygen stores of 2 litre.

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The number of electrons, which are present in the nonbonding π molecular orbital of the allyl anion is "0".

Anions, cations, but also allylic radicals have always been frequently mentioned as reaction intermediates. Each one has three adjacent [tex]sp^{2}[/tex]-hybridized carbon centers, and they all rely on resonance for stability. Two resonance structures would be used to present each species, with the charge as well as unpaired electron scattered across both the 1,3 and 0 positions.

The Aufbau principle states that these orbitals would fill up based on the order of stability, therefore a typical pi bond, will have 2 electrons in the Pi orbital as well as zero in the Pi* orbital.

Therefore, the correct answer will be option (d)

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Answer:

1.0535×10(exponent24)

Explanation:

N (number of molecules)

n(number of moles)

L( Avogadro's constant )

N=n×L

N=1.75×6.02×10(exponent 23)

Explanation:

In one mole we always have:

1 mol = 6.02 X 10²³ molecules

So emulates simple rule of 3 we will have;

1 mol ----------> 6.02 X 10²³

1.75 moles ----> X

X = 1.0535 X 10²⁴ molecules

Hope this helps, Good studies!

Carbon dioxide released by automobiles is an example of a(n) non-point source of pollution.

Nonpoint source pollution is typically caused by hydrologic alteration, drainage, seepage, precipitation, especially land runoff. Contrary to pollution from industries and sewage treatment facilities, nonpoint source (NPS) pollution originates from a variety of diffuse sources.

Point-source pollution can be defined as pollution coming from a single, distinguishable source, such as a factory and sewage plant output pipe. Nonpoint-source pollution would be pollution that doesn't come from a single source or point.

Therefore, Carbon dioxide released by automobiles is an example of a(n) non-point source of pollution.

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At  -230 °C temperature will He atoms have the same rms value as [tex]N_{2}[/tex] molecules at 25°C .

Calculation ,

Given temperature =  25°C  =  25+273  = 298 K

[tex]V_{rms}[/tex] =√3RT/M

[tex]V_{rms}[/tex] =√3R[tex]T_{1}[/tex]/[tex]M_{He}[/tex]      ......( i )

[tex]V_{rms}[/tex] =√3R[tex]T_{2}[/tex]/[tex]M_{N_{2}[/tex]      ......( ii )

Equalize equation  ( i ) and ( ii )

√3R[tex]T_{1}[/tex]/[tex]M_{He}[/tex] = √3R[tex]T_{2}[/tex]/[tex]M_{N_{2}[/tex]  

[tex]T_{1}[/tex]/[tex]M_{He}[/tex]  = [tex]T_{2}[/tex]/[tex]M_{N_{2}[/tex]  

[tex]T_{1}[/tex] = [tex]M_{He}[/tex] × [tex]T_{2}[/tex]/[tex]M_{N_{2}[/tex]  =  4 ×298 K/28 = 42.57 K

Temperature in°C = 42.57 - 273  = -230 °C

Therefore , at  -230 °C temperature will He atoms have the same rms value as [tex]N_{2}[/tex] molecules at 25°C .

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The total heat of hydration of 1. 00 mol of gas phase [tex]K^{+}[/tex] ions and [tex]Cl^{-}[/tex]ions is - 684 kJ/mol.

Calculation ,

Given data ,

Heat of solution =  17. 2 kJ/mol

lattice energy of KCl(s) = 701. 2 kJ/mol.

heat of hydration = ?

The KCl is formed by[tex]K^{+}[/tex] ions and [tex]Cl^{-}[/tex]– ions

Δ[tex]H_{solution}[/tex] = U° + Δ[tex]H_{hydration}[/tex]

Δ[tex]H_{hydration}[/tex] = Δ[tex]H_{solution}[/tex] - U° =  17. 2 kJ/mol - 701. 2 kJ/mol = - 684 kJ/mol

Hence, heat of hydration of 1. 00 mol of gas phase [tex]K^{+}[/tex] ions and [tex]Cl^{-}[/tex] ions is - 684 kJ/mol.

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8.62%

8.62% is the percent error of the experiment.

Rounding to the nearest hundredth value will be 8.620.

(-655 - -603)/(-603) = 0.0862

Multiply by 100%,

(0.0862) x 100% = 8.62%

The percent error is the difference between the estimated value and the actual value in relation to the actual value.

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Answer: The process is diffusion.

Explanation:

a dark liquid will have the highest concentration of dark purple atoms/molecules so it gives a concentrated colour.

when water is added to it, the colourless water molecules fill up the gaps between the purple particles and so their colour fades and becomes lighter and lighter as we add more water. see the image attached where imagine the red particles are water and the blue particles are purple particles. thats why the colour fades.

Answer:

Explanation:

The final answer has a different set of units. In particular, kilograms (kg) changes to grams (g) and meters (m) changes to centimeters (cm). To make this change, you need to multiply the first value by proportions.

When writing these proportions, it is important that they are arranged in a way that allows for the cancellation of units. For instance, since both kg and m are located in the numerator, they must be located in the denominators of the conversions.

Proportions:

1 kg = 1,000 g

1 m = 100 cm

The full expression:

As you can see, the old units cancel out and you are left with g and cm in the numerator.

The pH at the equivalence point for the titration of 0. 22 m HCN with 0. 22 m NaOH is 11.17

Calculation,

Concentration of NaCN = 0. 22 m/ 2 = 0.11 M ( at equal volumes of acid and base will be used).

The equilibrium is ,

HCN +[tex]H_{2} O[/tex]  → [tex]H^{+} + CN^{-}[/tex]

C(1-x)              Cx      Cx

Where x , is the degree of hydrolysis and

[tex]K_{h}[/tex] = C[tex]x^{2}[/tex]/(1-x)

We know that [tex]K_{h}[/tex] = [tex]K_{w}/K_{a}[/tex] = 1 ×[tex]10^{-14}[/tex]/4. 9 ×[tex]10^{-10}[/tex] =  2.04×[tex]10^{-5}[/tex]

[tex]K_{h}[/tex] =  C[tex]x^{2}[/tex] =   2.04×[tex]10^{-5}[/tex]  = 0.11 M×[tex]x^{2}[/tex]

[tex]x^{2}[/tex] =  2.04×[tex]10^{-5}[/tex]/0.11 M

x = 1.36×[tex]10^{-2}[/tex]

[tex][OH^{-} ][/tex] = Cx =  1.36×[tex]10^{-2}[/tex] × 0.11 M = 0.15×[tex]10^{-2}[/tex]

[tex][H^{+} ][/tex] =  1 ×[tex]10^{-14}[/tex]/ 0.15×[tex]10^{-2}[/tex] = 6.66×[tex]10^{-12}[/tex]

pH = -㏒[tex][H^{+} ][/tex] = -㏒6.66×[tex]10^{-12}[/tex] = 11.17

The pH at the equivalence point for the titration is  11.17.

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Theoretical yield = 2.397

The product could be sodium carbonate

percent yield = 98.456%

When nahco3 completely decomposes, it can follow this balanced chemical equation:

2nahco3 → na2co3 h2co3

If the mass of the NaHCO3 sample is 3.80 g, we must use stoichiometry to calculate the theoretical yields of each of the products.

mass of NaHCO₃ = 3.80 g

molar mass of NaHCO₃ = 84 g/mol

so the no of moles of NaHCO₃ = 3.80/84 =  0.0452 mol

You see, one mole of sodium carbonate and one mole of hydrogen carbonate are produced from two moles of sodium bicarbonate.

so, the no of moles of sodium carbonate = 0.0452/2 = 0.0226 mol

∴ mass of sodium carbonate ( Na₂CO₃) = no of moles of Na₂CO₃ × molar mass of Na₂CO₃

=  0.0226 × 106 ≈ 2.397 g

no of moles of hydrogen carbonate = 0.0452/2 = 0.0226 mol

mass of the hydrogen carbonate ( H₂CO₃) = no of moles of H₂CO₃ × molar mass of H₂CO₃

= 0.0226 × 62 g = 1.401 g

mass of one of the products was measured to be 2.36 g , from above data, we can say it must be sodium carbonate because value is the nearest of 2.397 g.

percentage yield = experimental yield/theoretical yield × 100

here experimental yield of Na₂CO₃ = 2.36 g

and theoretical yield of Na₂CO₃ = 2.397 g

∴ % yield = 2.36/2.397 × 100 ≈ 98.456%

Therefore the percentage yield of the product is 98.456%

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THEIR ANSWER IS RIGTH, ALL THREE PARTS

Theoretical yeild is 2.397, sodium bicarbonate is the possible product I chose, and percent yeild is 98.456

The reactant [tex]O_{2}[/tex] is limiting if we begin with 9. 50 grams of al and 9. 50 grams of  [tex]O_{2}[/tex]   .

The reactant whose number of moles is present in less quantity is called limiting reactant .

Calculation,

Given mass of Aluminium and oxygen

Mass of Aluminium = 9. 50 grams

and Mass of oxygen = 9. 50 grams

Number of moles of aluminium = given mass / molar mass of aluminium

Number of moles of aluminium = 9. 50 grams/ 26.99 = 0.351 mole

Number of moles of oxygen = given mass / molar mass of oxygen

Number of moles of oxygen = 9. 50 grams/ 31.999  = 0.296 mole

The reactant [tex]O_{2}[/tex] is limiting because oxygen present in less quantity or in limiting quantity and consume rapidly.

Therefor , the reactant [tex]O_{2}[/tex] is limiting if we begin with 9. 50 grams of al and 9. 50 grams of  [tex]O_{2}[/tex]   .

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The kf of Ag(NH₃)₂ for the given equation is 2.786 x 10⁷.

kf value can be calculated by dividing the molal concentration with the freezing point of depression.

The reaction are given:

Ag+(aq) + e- ⇋ Ag(s) E0= 0.81 V

Ag(NH₃)₂ +(aq) + e- ⇋ Ag(s) + 2NH₃(aq) E0= 0.37 V

To get Kf of Ag(NH₃)₂  we will rearrange the above written equations

a. Ag +(aq) + e- ⇋ Ag(s) E0= 0.81 V

b. Ag(s) + 2NH₃(aq) ⇋ Ag(NH₃)₂ +(aq) + e- E0= - 0.37 V

Adding equation (a) and equation (b) we will get

Ag +(aq) + 2NH₃(aq) ⇋ Ag(NH₃)₂  + (aq) Eo = 0.81 -0.37 = 0.44 V

By the Ernst equation,

E = E° - 0.0591 logkf

Now Kf is calculated at equilibrium and at equilibrium E = 0,

The number of transfer of electron is 1 ( 1 e-) , n=1

E° = 0.44 V

Substituting all the values in Ernst equation

0 - 0.44 - 0.0591/ 1 logkf

logkf = 0.44/ 0.0591

kf = 10 (0.44/ 0.0591)

Kf = 2.786 x 10⁷

Hence, Kf of Ag(NH₃)₂ for the given equation is 2.786 x 10⁷.

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The hydroxide ion concentration [OH-] in an aqueous solution is 6.66 x 10⁻⁹ M

Experimental measurements of the pKw of water at 25ºC yielded a result of 14.0. The same method used to determine the pKa of all other water-soluble compounds that can function as acids in aqueous solution can also be used to determine this value from the examination of thermodynamic or electrochemical data for these aqueous solutions.

Water has a pkw=14

So it can be represented as,

[H+] [OH-] = 1×10^-14

Given [H+] = 1. 5×10^−6

So, [OH-] = (1*10^-14) / ( 1.5*10^-6)

               = 6.66 x 10⁻⁹ M

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The percentage of volume of ethanol is 9.5%.

Volume/volume percentage (v/v% or percent v/v) is a unit used to express how much of a material is present in a solution. It is defined as the volume of the solute divided by the sum of the volumes of the solution, multiplied by one hundred. Examples: The average alcohol concentration (v/v%) of wine is 12 percent.

The formula used to get the volume of ethanol in percentages is

Percent of volume of ethanol = (volume of solute / total volume) *  100

Percent volume of Ethanol = (volume of ethanol  / total volume)  * 100

                                            =  95mL  / 1000mL  *  100

                                            = 9.5%

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The reaction with H₂ & Raney Ni (Catalytic Hydrogenation) can be used to reduce aldehydes and ketones to alcohols.

An Aldehyde is a functional group which contains the carbonyl group that is -C = O. The structural formula for aldehyde is RCHO.

Reaction of Aldehyde with H₂ & Raney Ni gives Primary alcohol

RCHO + H₂ & Raney Ni → RCH₂OH

A Ketone is a functional group which contains the carbonyl group that is -C = O. The structural formula for ketone is RCOR.

Reaction of Ketone with H₂ & Raney Ni gives Secondary alcohol

RCOR + H₂ & Raney Ni → RCH(OH)R

Thus from the above conclusion we can say that The reaction with H₂ , Raney Ni can be used to reduce aldehydes and ketones to alcohols.

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Answer 6

Explanation: That means a carbon atom has 6 protons, 6 neutrons, and 6 electrons.

The ka of the acid will be-

HA(aq)+OH−(aq)→A−(aq)+H2O(l)

You have the half-equivalence point when

[HA]=[A−]

it suggests that

log([HA][A−])=log(1)=0

As a result, it can be said that the pH of the solution and the pKa of the weak acid are equivalent at the half-equivalence point.

At the halfway point of equivalence: pH=pKa

The acid dissociation constant of the weak acid, Ka, determines the pKa. pKa=log(Ka), which indicates that Ka=10pKa.

Ka=10pH will be present when the two points are half equal.

Enter your value to determine Ka=105.67=2.1106.

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Bohr's model explained the position of the electron, proton, and neutron in the atom of the element. The energy at the n = 2 level of the atom will be - 3.40 eV.

The principal quantum number (n) has been the distance of the electron of that atom in the nucleus and its energy in the structure. It can also be said to define the size of the atomic orbit.

n = 2 is the first excited state whose energy is calculated as:

Eₙ = − 13.6 ÷ n² eV

E₂ = - 13.6 eV ÷ 2²

= -3.40 eV

Therefore, -3.40 eV is the energy of electron at n = 2.

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The best way to differentiate chart elements is by using labels.

So, option B is correct one.

The table should be used when

The column chart is probably the most used chart type. This chart best used to compare different values when specific values are important.

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The the mass of the heavier isotope of chlorine (in amu) is 73.932 amu.

The term isotopes refers atoms that have the same atomic number but different mass number. We must know that all these atoms belong to the same element an differ only in the number of neutrons present in the atom of the element. Hence isotopes are different atoms of the same element hence they have the same chemical properties or reactivity. The mass spectrum is able to show the masses of each of the isotopes according to their order of relative abundance in nature.

Looking at the peaks, we know that the heaviest isotope is the one whose mass will have the highest peak as shown in the mass spectrum of the element as shown. Thus, the the mass of the heavier isotope of chlorine (in amu) is 73.932 amu. The mass spectrum of element also shows this fact.

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Because carbon dioxide dissolves in water and is transported by rain to the surface where it combines with rocks to generate carbonates, the Earth's atmosphere contains significantly less carbon dioxide than Venus's atmosphere.

Since Venus' atmosphere is around 100 times thicker than the earth's and extremely dense. Venus must be sufficiently close to the Sun for the little carbon dioxide in its early atmosphere to have heated the surface and released additional carbon dioxide into the atmosphere. Nitrogen makes up 79% of the atmosphere on Earth, along with oxygen (20%) and a trace amount of other gases including carbon dioxide and water vapor. On Venus and Mars, however, carbon dioxide makes up the majority of the atmosphere.

In Venus's original ocean drained, water vapor molecules were dispersed by ultraviolet light, and hydrogen was released into space. There was no longer any surface water, and the amount of carbon dioxide increased.

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Answer:

i think its -91.6

Explanation:

2 ( - 950.8) - (( - 1130.7) + ( - 393.5) + ( - 285.8))

The standard enthalpy change for the given reaction is approximately 91.9 kJ mol⁻¹.

The heat produced during a chemical reaction under specific conditions is known as the standard enthalpy change, abbreviated as H° (zero delta H). Standard conditions often refer to all reactants and products in their standard states, which are their most stable forms at 298 K and 1 bar (or 1 atm) pressure.

Hess's law, which states that the total enthalpy change of a reaction is equal to the sum of the enthalpy changes of its component phases, is used to calculate the standard enthalpy change of the reaction.

The following reaction is given:

[tex]\rm 2 NaHCO_3(s)--- > Na_2CO_3(s)+ CO_2(g) + H_2O(l)[/tex]

The relevant enthalpy changes of formation are:

ΔH°[NaHCO3(s)] = -950.8 kJ mol⁻¹

ΔH°[Na2CO3(s)] = -1130.7 kJ mol⁻¹

ΔH°[CO2(g)] = -393.5 kJ mol⁻¹

ΔH°[H2O(l)] = -285.8 kJ mol⁻¹

By using Hess's Law, the standard enthalpy change (ΔH°) for the reaction can be calculated as follows:

ΔH° = Σ (products) - Σ (reactants)

ΔH° = [ΔH°[[tex]\rm Na_2CO_3[/tex](s)] + ΔH°[[tex]\rm CO_2[/tex](g)] + ΔH°[H2O(l)]] - [2 × ΔH°[[tex]\rm NaHCO_3[/tex](s)]]

ΔH° = [-1130.7 kJ mol⁻¹ + (-393.5 kJ mol⁻¹) + (-285.8 kJ mol⁻¹)] - [2 × (-950.8 kJ mol⁻¹)]

ΔH° = [-1809.7 kJ mol⁻¹] - [-1901.6 kJ mol⁻¹]

ΔH° ≈ -1809.7 kJ mol⁻¹ + 1901.6 kJ mol⁻¹

ΔH° ≈ 91.9 kJ mol⁻¹

Hence, the standard enthalpy change for the given reaction is approximately 91.9 kJ mol⁻¹.

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The corresponding concentration values of OH⁻  for the pH values

of following are -

1- 10⁻¹

3- 10⁻³

5- 10⁻⁵

7- 10⁻⁷

9-  10⁻⁹

11- 10⁻¹¹

pH is a logarithmic measure of the hydrogen ion concentration of an aqueous solution pH = -log[H+]

pH describes how acidic or basic an aqueous solution is, where a pH below 7 is acidic and a pH greater than 7 is basic. pH of 7 is considered neutral .

Functions of OH−    :

OH− functions as a base .

Hydroxide is a diatomic anion with chemical formula OH−. It consists of an oxygen and hydrogen atom held together by a single covalent bond, and carries a negative electric charge.

Hence, this way pH values of these concentrations are mentioned .

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The 6 models of the atom:

1. John Dalton's atomic model: Daltons Billiard Ball (Solid Sphere) Model

2. J.J. Thomson's model: Plum Pudding model

3. Ernest Rutherford's model: Nuclear model

4. Niels Bohr's model: Planetary model

5. Erwin Schrdinger's model: Electron Cloud Model/Quantum Model

6. Wave mechanical model

The smallest component that makes up a chemical element is an atom. Atoms that are neutral or ionized are the building blocks of all solids, liquids, gases, and plasma. A normal atom is 100 picometers across, which is incredibly small. Because of quantum effects, they are so small that it is impossible to predict their behavior with sufficient precision using classical physics, as if they were, say, tennis balls. One or more electrons are attached to the nucleus of every atom, which is made up of a nucleus. Protons and neutrons, in various numbers, make up the nucleus. Neutrons exist only in the most prevalent type of hydrogen. An atom's nucleus makes up more than 99.94% of its mass.

how to build 6 individual atoms?

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Student e accounted for the equivalent weight found for succinic acid by analyzing its titration with [tex]NaOH[/tex] (aq) and concluding that it is diprotic.

One mole of succinic acid required two mole of base to neutralize completely. In this titration phenolphthalein indicator is used to observe the completion of the reaction. In an aqueous solution, succinic acid readily ionize to form succenate. Succinic acid undergoes two successive deprotonation reaction.

The succinic acid results in succinate ( dicarboxylic acid dianion ) which is form by removal of a proton from both carboxy groups of succinic acid.

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Here BA is benzoic acid and B- is conjugate base;

BA is weaker than B- because the Ka < Kb

--> BA is stronger than B- because the Ka > Kb

BA is stronger than B- because the Ka < Kb

BA is weaker than B- because the Ka > Kb

A conjugate acid-base pair, according to the Brnsted-Lowry definition of acids and bases, consists of two substances that are distinct only in that they contain a proton (H+). When a proton is supplied to a base, a conjugate acid is created, and vice versa when a proton is taken away from an acid, a conjugate base is created. Yuki Jung was the creator.

The power of an acid (or base) is inversely proportional to the power of its conjugate base (or conjugate acid): The conjugate base of an acid is weaker the stronger it is. The conjugate base of a weaker acid has a stronger base. A pair of chemical species known as conjugate acid and conjugate base exhibit opposing chemical behaviors.

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