To find pH of hydrochloric acid.
Lab data: on the attached file
Acid used to titrate 25.0mL of 0.150M potassium hydroxide.

A scientific question may lead to a(n)

The answer is may lead to a hypothesis.

Explanation:

Answer: observations

Hypothesis

Experimentation

Explanation:in order and right on edge

Answer:

0.0075 milliliters (7.5 microliters) will be taken from the stock solution and then diluted with 0.0249925 milliliters (24.9925 microliters) of water to make 0.025 milliliters (25 microliters) of NaOH solution.

Explanation:

This is a problem of dilution using the equation:

initial concentration x initial volume = final concentration x final volume.

The final volume to be prepared is 25 microliters.

The final concentration to be prepared is 3 M.

The initial volume to be taken is not known yet.

The initial concentration is 10 M.

Now, let's substitute these parameters into the the equation above.

10 x initial volume = 3 x 25

Initial volume = 3 x 25/10

     = 7.5 microliters

Note that: 1 microliter = 0.001 milliliters

Hence,

7.5 microliters = 0.0075 milliliters

This means that an initial volume of 0.0075 milliliters (7.5 microliters) will be taken from the stock solution. This amount will then be diluted with 0.0249925 milliliters (24.9925 microliters) of water to make 0.025 milliliters (25 microliters) of NaOH solution.

Answer:

Explanation:

The chemical equation for the reaction is :

[tex]C_6H_{12}O_6_{(s)} \to 2C_2H_5OH _{(l)} + 2CO_{2(g)}[/tex]

The standard enthalpy of formation [tex]\Delta H ^0_f[/tex]  of the above equation is as follows:

[tex]\Delta H^0_{f, C_6H_{12}O_6}[/tex] = -1274.4 kJ/mol

[tex]\Delta H ^0_{f, C_2H_{5}OH[/tex] = -277.7 kJ/mol

[tex]\Delta H ^0_{f, CO_2}[/tex] = -393.5 kJ/mol

[tex]\Delta H ^0_{rxn }= \sum n_p \Delta H ^0_{f,p} - \sum n_r \Delta H ^0_{f,r}[/tex]

where ;

[tex]n_p[/tex] = stochiometric coefficients of products

[tex]n_r=[/tex] stochiometric coefficients of reactants

[tex]\Delta H^0_{f.p}[/tex] = formation standard enthalpy of products

[tex]\Delta H^0_{f.r}[/tex] = formation standard enthalpy of reactants

[tex]\Delta H ^0_{rxn }= (2 \ mol* -2777.7 \ kJ/mol + 2 \ mol * - 393.5 \ kJ/mol) - (1\ mol *(-1274.4 \ kJ/mol)[/tex]

[tex]\Delta H ^0_{rxn }= -68 \ kJ[/tex]

For [tex]\Delta S ^0[/tex] ;

The standard enthalpy of formation of [tex]\Delta S ^0_f[/tex] of the reactant and the products are :

[tex]\Delta S ^0 _{f \ C_6H_{12}O_6} = 212 \ \ J/Kmol[/tex]

[tex]\Delta S ^0 _{f \ C_2H_5O_H} = 160.7 \ \ J/Kmol[/tex]

[tex]\Delta S ^0 _{f \ CO_2} = 213.63 \ \ J/Kmol[/tex]

The [tex]\Delta S ^0_{rxn}[/tex] is as follows:

[tex]\Delta S ^0_{rxn} = \sum n_p \Delta S^0_{f.p} - \sum n_r \Delta S^0_{f.r}[/tex]

[tex]\Delta S ^0_{rxn} = (2 \ mol *160.7 \ \ J/Kmol + 2 \ mol *213.63 \ \ J/Kmol) -(1*212 J/Kmol)[/tex]

[tex]\Delta S ^0_{rxn} =536 . 7 \ J/K[/tex]   (to kJ/K)

[tex]\Delta S ^0_{rxn} =536 . 7 \ J/K * \frac{1 \ kJ}{1000 \ J}[/tex]

[tex]\Delta S ^0_{rxn} =0.5367 \frac{kJ}{K}[/tex]

Given that;

at T = 25°C = ( 25 + 273) K = 298 K

[tex]\Delta G^0 _{rxn} = \Delta H^0_{rxn} - T \Delta S^0_{rxn}[/tex]

[tex]\Delta G^0 _{rxn} = -68 \ kJ - 298 * 0.5367 \frac{kJ}{K}[/tex]

[tex]\Delta G^0 _{rxn} = -227. 9 \ \ \ kJ[/tex]

As [tex]\Delta G^0 _{rxn}[/tex] is negative; the reaction is spontaneous

[tex]\Delta H^0 _{rxn}[/tex] = negative

[tex]\Delta S^0 _{rxn}[/tex] = positive; Therefore , the reaction is spontaneous at all temperature , We can then say that the spontaneity of this reaction [tex]\Delta G^0 _{rxn} = \Delta H^0_{rxn} - T \Delta S^0_{rxn}[/tex]  is dependent on temperature.

Answer:

See explaination

Explanation:

Please see the attached file for the structural diagram of the molecular formation.

It is detailed and properly presented at the attachment.

A slice or part of a homogeneous substance has the same density as the whole piece because the density is uniform throughout. In a heterogeneous substance, the density can vary and depends on the composition and structure of the part being measured. Density is found by dividing mass by volume and is useful for identifying substances.

The density of a substance is defined as its mass divided by its volume. In a homogeneous material, the density is consistent throughout, which means a slice or part of the substance would have the same density as the whole piece. For example, a solid iron bar is homogeneous, and therefore any part of it would have the same density as the entire bar. However, in a heterogeneous material, the density can vary from one part to another; an instance of this is Swiss cheese, which contains air pockets resulting in variable local densities.

To determine the density of a substance, you would divide the mass by the volume of the substance. This method applies to both whole objects and parts thereof. In the case of a cube, the volume is found by cubing the edge length. In practice, density is often used to help identify substances by comparison with known values, as different substances have characteristic densities.

Answer:

Check the explanation

Explanation:

Here, Nitrogen (N) undergoes oxidation and Chlorine (Cl) undergoes reduction.

To answer your question:

N is oxidized from an oxidation number of -3 to an oxidation number of -1.

Cl is reduced from oxidation number of +1 to an oxidation number of -1.

Now,

Borneol should have a lower Rf because of boiling point.

Option b (O=N−F and N=O−F) represents a pair of resonance structures.

The correct choice representing a pair of resonance structures is option b. O=N−F and N=O−F.

Resonance structures are different representations of a molecule or ion that show the delocalization of electrons. In option b, the double bond can be moved between the nitrogen and oxygen atoms, resulting in two resonance structures.

Resonance occurs when there are multiple valid Lewis structures that can be drawn for a molecule or ion. It indicates that the actual structure is a combination or hybrid of all the resonance structures.

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Answer:

Yes, a hydrocarbon is any compound that contains hydrogen and oxygen.

Explanation:

Answer:

A) No , because Qsp<Ksp for calcium fluoride.

Check attachment for calculation

Without specific calculations, to determine if a precipitate forms when mixing calcium chlorate with sodium fluoride, one assesses if the product of the ion concentrations exceeds the Ksp for calcium fluoride; if Qsp > Ksp, precipitation occurs.

To determine if a precipitate will form when calcium chlorate is mixed with sodium fluoride, one must consider the ionic products formed and compare the reaction quotient (Qsp) with the solubility product constant (Ksp) of the relevant compound, in this case, calcium fluoride (CaF2). The Ksp for calcium fluoride is 3.45 × 10∑11. In a mixture of calcium chlorate and sodium fluoride, the calcium ions (Ca2+) from calcium chlorate can react with the fluoride ions (F∑) from sodium fluoride to potentially form solid calcium fluoride. However, without the exact concentrations of Ca2+ and F∑ after mixing, one cannot directly calculate Qsp. The principle, however, is that if Qsp > Ksp, a precipitate of calcium fluoride will form, whereas if Qsp < Ksp, no precipitate forms because the solution is not supersaturated with respect to CaF2.

The concentration of each ion in the final solution depends on the dilution that occurs when the two solutions are mixed. Given the initial conditions and the principle that the formation of a precipitate requires Qsp to exceed Ksp, we generally assess the likelihood of precipitation by considering the concentrations of the reactive ions and the solubility product of the insoluble compound they may form. In this scenario, without performing a specific calculation, the correct answer would be based on understanding whether the mixing leads to a situation where the product of the concentrations of calcium and fluoride ions exceeds the Ksp of calcium fluoride.

In a chemical reaction between sulfur and oxygen that produces sulfur dioxide, there is a 1:1 ratio between sulfur and sulfur dioxide. Using the molar mass of sulfur and sulfur dioxide, it is calculated that to produce 8g of sulfur dioxide, 4g of sulfur needs to be burned.

To calculate the mass of sulfur that must be burned to give 8g of sulfur dioxide, we first need to understand mole ratio and molar mass, since the question is about a chemical reaction described by a balanced chemical equation. The balanced equation is: S + O2 --> SO2.

The molar masses are: S (32.06 g/mol), O2 (32 g/mol, because there are 2 O atoms each 16 g/mol), and SO2 (64.06 g/mol, which sum the molar masses of S and O2).

From the balanced chemical equation, we see a 1:1 ratio exists between sulfur and sulfur dioxide. So, the number of moles in 8g of SO2 would be the mass divided by the molar mass: 8g SO2 * (1 mol SO2/64.06 g SO2) = 0.125 mol SO2.

Because of the 1:1 ratio between sulfur and sulfur dioxide, 0.125 mole of sulfur is needed. Convert this to grams, we have: 0.125 mol S * (32.06 g S/1 mol S) = 4 g of sulfur

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Answer:

Check the explanation

Explanation:

Kindly check the attached image below for the step by step explanation to the question above.

Answer: The molarity of [tex]H^+[/tex] in this solution is [tex]2.2\times 10^{-11}M[/tex]

Explanation:

pH or pOH is the measure of acidity or alkalinity of a solution.

pH is calculated by taking negative logarithm of hydrogen ion concentration.

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

Given : pH = 10.66

Putting in the values:

[tex]10.66=-\log[H^+][/tex]

[tex][H^+]=10^{-10.66}[/tex]

[tex][H^+]=2.2\times 10^{-11}[/tex]

Thus the molarity of [tex]H^+[/tex] in this solution is [tex]2.2\times 10^{-11}M[/tex]

The molarity of hydrogen ions, H⁺ in this solution is  2.2 * 10⁻¹¹

The molarity of a solution is a measure of the concentration in moles of a substance present in a liter of solution of that substance.

The pH of a solution is the negative logarithm in base 10 of the hydrogen ion concentration of the solution.

pH = -log[H⁺]

-pH = log[H⁺]

pH of solution = 10.66

-10.66 = log[H⁺]

taking antilogarithm of both sides

[H⁺] = 10⁻¹⁰°⁶⁶

[H⁺] = 2.2 * 10⁻¹¹

Therefore, the molarity of hydrogen ions, H⁺ in this solution is  2.2 * 10⁻¹¹

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Answer:

Zn(s) + Fe^2+(aq) ---------> Zn^2+(aq) + Fe(s)

∆G= 61760J

K= 6.5 ×10^10

Explanation:

Zn(s) + Fe^2+(aq) ---------> Zn^2+(aq) + Fe(s)

E°reaction= E°Fe - E°Zn

E°reaction= (-0.44)-(-0.76)

E°reaction= 0.32V

We know that

∆G= -nFE°

Since n= 2 from the reaction equation and F= 96500C

∆G= -(2×96500×0.32)

∆G= 61760J

From

E°= 0.0592/n log K

0.32= 0.0592/2 log K

log K= 0.32/0.0296

log K= 10.81

K= Antilog(10.81)

K= 6.5 ×10^10

Answer: 0.225, x, 0.0788

Explanation:

Answer:

1670 ml

Explanation:

molarity x Volume (Liters) = moles => Volume (Liters) = moles/Molarity

Volume needed = 2.50mol/1.50M = 1.67 Liters = 1670 ml.

Answer:

hes right

Explanation:

ik all

Answer:1.587J / g degC.

Explanation:SOLUTION

Specific heat of aluminum = 0.897 J / g degC.

Heat gained by calorimeter = 100 * 10 * 0.897 J = 897 J

Specific heat of water = 4.18 J / g degC.

Heat gained by water = 250 * 10 * 4.18 J = 10450 J

Specific heat of copper = 0.385 J / g degC.

Heat loss by copper block = 50 * (80-10) * 0.385 J =1347.5  J

Let specific heat of unknown block be x J / g degC.

Heat loss by unknown block = 70 * (100-10) * x J = 6300x J

Heat gain = Heat loss

897 + 10450 = 1347.5 + 6300x

6300x = 9999.5

x =1.58722= 1.587

Specific heat of unknown substance is 1.587J / g degC.

Answer:

Work done on the door is = 0

Mechanical energy of the person is 100%

Mechanical energy of the door is 50%.

Explanation:

Mechanical energy, M.E., is the energy an object possesses due to its position and motion. When we talk of position, the body is at rest = potential energy. In terms of motion, the body is moving = kinetic energy.

This means that M.E = potential + kinetic

Also, Work done = force × distance

For the person pushing the door, the potential of the energy stored in the person and the kinetics of pushing the door handle are all present. Therefore,  M.E of person is 100%.

But despite the doors potential energy present, there is no motion from it, which means the M.E is only half or 50%.

Also, since distance moved by the door is 0, no work is done on the door.

Answer:

42050 J.

Explanation:

Data obtained from the question:

Mass (M) of ethanol = 50g

Heat of vaporisation (ΔHv) of ethanol = 841 J/g

Heat (Q) =.?

The heat required to boil 50g of ethanol can be obtained as follow:

Q = MΔHv

Q = 50 x 841

Q = 42050 J.

Therefore, the heat required to boil 50g of ethanol is 42050 J.

Absorption of 1.0 rad of gamma radiation would likely cause the greatest tissue damage due to its high energy and penetration ability.

The amount of tissue damage caused by radiation depends on various factors, including the type of radiation and its energy level.

Alpha particles are relatively heavy and highly ionizing, but they have low penetration power, so they are generally less damaging to tissues externally but can be more damaging if ingested or inhaled.

Beta particles have less mass and energy compared to alpha particles and can penetrate deeper into tissues, potentially causing more damage.

Gamma rays are electromagnetic radiation with high energy and penetration power. They can penetrate deeply into tissues, causing damage along their path.

Considering these factors, absorption of 1.0 rad of gamma radiation would likely result in the greatest tissue damage due to its high energy and penetration ability.

Answer: [H3O+]= 0.05 M

Explanation:

Message me for extra explanation.

snap- parkguy786

Answer:

CHC13

Explanation:

Answer:

-5.51 kJ/mol

Explanation:

Step 1: Calculate the heat required to heat the water.

We use the following expression.

[tex]Q = c \times m \times \Delta T[/tex]

where,

The average density of water is 1 g/mL, so 75.0 mL ≅ 75.0 g.

[tex]Q = 4.184J/g.\°C \times 75.0g \times (95.00\°C - 25.00\°C) = 2.20 \times 10^{3} J = 2.20 kJ[/tex]

Step 2: Calculate the heat released by the methane

According to the law of conservation of energy, the sum of the heat released by the combustion of methane (Qc) and the heat absorbed by the water (Qw) is zero

Qc + Qw = 0

Qc = -Qw = -22.0 kJ

Step 3: Calculate the molar heat of combustion of methane.

The molar mass of methane is 16.04 g/mol. We use this data to find the molar heat of combustion of methane, considering that 22.0 kJ are released by the combustion of 64.00 g of methane.

[tex]\frac{-22.0kJ}{64.00g} \times \frac{16.04g}{mol} = -5.51 kJ/mol[/tex]

Complete Question

The complete question is shown on the first uploaded image

Answer:

The pH is  [tex]pH = 4.94[/tex]

Explanation:

From the question we are told that

   The average concentration of NaOH is [tex][NaOH] = 0.101 M[/tex]

    The volume of NaOH is  [tex]V__{NaOH}} = 15.00 mL[/tex]

    The average concentration of Acetic acid is [tex][Acetic \ Acid] =0.497 \ M[/tex]

     The volume of Acetic acid is  [tex]V__{Acetic \ Acid}} = 5.00 \ mL[/tex]

The chemical equation for this reaction is  

         [tex]NaOH + CH_3COOH ---> CH_3 COONa + H_2 O[/tex]

The total volume of the solution is  

  [tex]V__{Total}} = V__{NaOH}} + V__{Acetic \ Acid}}[/tex]

Substituting values

      [tex]V__{Total}} = 15 + 5[/tex]

      [tex]V__{Total}} = 20mL = 20 *10^{-3} L[/tex]

The number of moles of NaOH is mathematically represented as

      [tex]n__{NaOH}} = [NaOH] * V__{NaOH}}[/tex]

substituting values

      [tex]n__{NaOH}} = 0.101 * 15*10^{-3}[/tex]

     [tex]n__{NaOH}} = 0.001515 \ moles[/tex]

The number of moles of Acetic acid is mathematically represented as

      [tex]n__{Acetic acid}} = [Acetic \ acid] * V__{Acetic acid}}[/tex]

substituting values

        [tex]n__{Acetic acid}} = 0.497 * 5*10^{-3}[/tex]

      [tex]n__{Acetic acid}} = 0.002485\ moles[/tex]

From the chemical equation

          1 mole of NaOH reacts with   1 mole of Acetic acid  to produce 1 mole of   [tex]CH_3 COONa[/tex] salt and 1 mole of  [tex]H_2 O[/tex]

So

  0.001515 moles  of NaOH reacts with  0.001515 moles of Acetic acid  to produce 0.001515 moles of   [tex]CH_3 COONa[/tex] salt and 0.001515 moles of  [tex]H_2 O[/tex]

This implies the number of moles of NaOH remaining after the react would be

      [tex]\Delta n__{NaOH}} = 0.001515 - 0.001515[/tex]

      [tex]\Delta n__{NaOH}} = 0 \ mole[/tex]

the number of moles of Acetic acid remaining after the react would be

    [tex]\Delta n__{Acetic acid}} = 0.002485 - 0.001515[/tex]

    [tex]\Delta n__{Acetic acid}} = 0.00097 \ moles[/tex]

the number of moles of [tex]CH_3 COONa \ salt[/tex] remaining after the react would be

    [tex]\Delta n__{CH_3 COONa \ salt}} = 0 + 0.001515[/tex]

    [tex]\Delta n__{CH_3 COONa \ salt}} = 0.001515 \ moles[/tex]

the number of moles of [tex]H_2 O[/tex] remaining after the react would be

   [tex]\Delta n__{H_2O}} = 0 + 0.001515[/tex]

   [tex]\Delta n__{H_2O}} = 0.001515 \ moles[/tex]

The expected pH is mathematically evaluated as

      [tex]pH = pK_a + log [\frac{[CH_3 COONa]}{[Acetic \ acid]} ][/tex]

Where [tex]pKa[/tex] is mathematically evaluated as

        [tex]pK_a = - log (K_a)[/tex]

The concentration of [tex]CH_3 COONa \ salt[/tex] is mathematically evaluated a s

[tex][CH_3 COONa] = \frac{\Delta n_CH_3 COONa \ salt }{V__{Total}}}[/tex]

substituting values

   [tex][CH_3 COONa] = \frac{0.001515}{20 *10^{-3}}[/tex]

    [tex][CH_3 COONa] = 0.07575M[/tex]

The concentration of  Acetic acid is mathematically evaluated as

          [tex][Acetic acid] = \frac{\Delta n__Acetic acid}{V__{Total}}}[/tex]

substituting values

        [tex][CH_3 COONa] = \frac{0.00097}{20 *10^{-3}}[/tex]

       [tex][CH_3 COONa] = 0.0485 M[/tex]

Substituting values into the equation for pH

          [tex]pH = - log (1 .8 *10^{-5}) + log [\frac{0.07575}{ 0.0485} ][/tex]

             [tex]pH = log [\frac{0.07575}{ 0.0485} ] - log (1 .8 *10^{-5})[/tex]

            [tex]pH = log [\frac{1.561856}{1.8*10^{-5}} ][/tex]

          [tex]pH = 4.94[/tex]

Answer:

Find the given attachment

Answer:

Check the explanation

Explanation:

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A common addiction to the common water is the elemental chlorine and chlorine ions, the phosphate ions. Recent reports of the higher levels of lead in cities have to lead to the Pb(s) as CI 2 flows in petal pipes some may be oxidized.

They balanced equation of the net ionic reactions.

Learn more about the chloride ions and phosphate ions.

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Answer: The heat of reaction for the combustion of titanium is 15240 kJ/mol

Explanation:

The quantity of heat required to raise the temperature of a substance by one degree Celsius is called the specific heat capacity.

[tex]Q=C\times \Delta T[/tex]

Q = Heat absorbed by calorimeter =?

C = heat capacity of calorimeter = 9.84 kJ/K

Initial temperature of the calorimeter  = [tex]T_i[/tex] = [tex]25.00^0C=(25.00+273)=298.00K[/tex]

Final temperature of the calorimeter  = [tex]T_f[/tex]  = [tex]91.60^0C=(91.60+273)K=364.6K[/tex]

Change in temperature ,[tex]\Delta T=T_f-T_i=(364.6-298.0)K=66.60K[/tex]

Putting in the values, we get:

[tex]Q=9.84kJ/K\times 66.60K=655.3kJ[/tex]

As heat absorbed by calorimeter is equal to heat released by combustion of titanium

[tex]Q=q[/tex]

[tex]\text{Moles of titanium}=\frac{\text{given mass}}{\text{Molar Mass}}=\frac{2.060g}{47.8g/mol}=0.0430mol[/tex]  

Heat released by 0.0430 moles of titanium = 655.3 kJ

Heat released by 1 mole of titanium = [tex]\frac{655.3}{0.0430}\times 1=15240kJ[/tex]

The heat of reaction for the combustion of titanium is 15240 kJ/mol

Answer: This reaction is called as Chemiluminescence. A type of reaction in which light is produced due to chemical reaction. While in case of Phosphorescence and Fluorescence light is emitted due to absorption of photons.

Explanation: i looked the answer up but your welcomee ❤

Final answer:

Luminol undergoes a chemiluminescent reaction that produces light when it reacts with hydrogen peroxide and a catalyst, often used in forensic science to detect blood.

Explanation:

Luminol goes through a chemiluminescent reaction, which refers to a process where a chemical reaction produces a molecule in an excited state. In the presence of hydrogen peroxide and a catalyst, such as the iron found in hemoglobin, luminol reacts to generate 3-aminophthalate in an excited state, emitting a bluish light. This reaction is highly valuable in forensic science for the detection of blood.

Chemiluminescence is a phenomenon where the energy from a chemical reaction is released as light. This reaction is possible because certain chemicals, like peroxide and ozone used in chemiluminescent reactions, contain unstable or energetic chemical bonds. The process is a rare occurrence in chemical reactions and is a direct transduction of chemical energy into radiant energy.

Answer:

4.0 moles

Explanation:

The following data were obtained from the question:

Volume (V) = 12L

Pressure = 5.6 atm

Temperature (T) = 205K

Gas constant (R) = 0.08206 atm.L/Kmol

Number of mole (n) =?

Using the ideal gas equation: PV = nRT, the number of mole of the gas can be obtained as follow

PV = nRT

5.6 x 12 = n x 0.08206 x 205

Divide both side by 0.08206 x 205

n = (5.6 x 12)/(0.08206 x 205)

n = 4.0 moles

Therefore, the number of mole of the gas is 4.0 moles

Answer:

C

Explanation:

Well, we can see from the equation that hydrogen gas and ethane are in a 2:1 ratio based on their mole coefficients.

You need 2 mol H2 to make 1 mol C2H6. So that means you need 2*13.78 mol H2 to make 13.78 mol C2H6!

2*13.78 = 27.56 mol

C is correct.

Answer:

4L

Explanation:

To obtain the volume of O2 at stp, first, we need to determine the number of mole of O2.

From the question given above,

Mass of O2 = 5.72g

Molar Mass of O2 = 32g/mol

Number of mole =Mass/Molar Mass

Number of mole of O2 = 5.72/32

Number of mole of O2 = 0.179 mole

Now, we can calculate the volume of O2 at stp as follow:

1 mole of a gas occupy 22.4L at stp.

Therefore, 0.179 mole of O2 will occupy = 0.179 x 22.4 = 4L

Therefore, the volume occupied by the sample of O2 is 4L

The volume of a 5.72 gram sample of oxygen at STP is approximately 4.01 liters, after converting the mass to moles and using the molar volume of a gas at STP.

To calculate the volume of oxygen at Standard Temperature and Pressure (STP) from the given mass, first convert the mass of oxygen to moles. At STP, 1 mole of gas occupies 22.4 liters. Use the molar mass of Oxygen (32 g/mol) to convert grams to moles and then use the volume of 1 mole of gas at STP to find the volume in liters.

First, convert the given mass of oxygen (5.72 g) to moles:

5.72 g [tex]O_{2}[/tex]
x
1 mol [tex]O_{2}[/tex]/ 32 g [tex]O_{2}[/tex]
= 0.179 moles [tex]O_{2}[/tex]

Then calculate the volume at STP (using the fact that 1 mole of any gas at STP occupies 22.4 liters):

0.179 moles [tex]O_{2}[/tex]
x
22.4 liters/mol
= 4.0096 liters

Therefore, the volume of a 5.72 gram sample of [tex]O_{2}[/tex] at STP is approximately 4.01 liters.