The density of water is about 1.0 g/mL at room temperature. Briefly explain how the density of an aqueous solution at room temperature can be significantly less than 1.0 g/mL. Give an example of such a solution.

Answers

Answer 1

Answer:

The density of water is about 1.0 g/mL at room temperature.

Briefly explain how the density of an aqueous solution at room temperature can be significantly less than 1.0 g/mL.

Give an example of such a solution.

Explanation:

That means 1.0mL of water weighs ---- 1.0g

If any other aqueous solution which has mass less than 1.0g will have density less than 1.0g.

For example aqueous solution of sulfuric acid has density 0.98g/mL.

That means 1mL of sulfuric acid has mass 0.98g.

Answer 2

Dissolving gases or liquids in water whose density are less than that of water will result in an aqueous solution whose density is less than that of water, for example, sulfuric acid which has a density 0.98g/mL.

What is density of a substance?

The density of a substance is the ratio of the mass to the volume of that substance.

Density measures compactness.

More compact substances are more dense.

The density of water is about 1.0 g/mL at room temperature.

This means that 1.0mL of water weighs 1.0g

1.0 mL of an aqueous solution whose mass is less than 1.0g will have density less than that of water.

Dissolving gases or liquids in water whose density are less than that of water will result in an aqueous solution whose density is less than that of water.

An example of an aqueous solution whose density is less than that of water is sulfuric acid which has a density 0.98g/mL.

Sulfuric acid is prepared by dissolving sulphur (vi) oxide, a gas, in water.

This means 1mL of sulfuric acid has mass 0.98g.

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Related Questions

Guys I don't know science, if you are intelligent tell me what is science ​

Answers

Explanation:

Science is the pursuit and application of knowledge and understanding of the natural and social world following a systematic methodology based on evidence

identify the organ system pictured below and state two functions of this system in the body​

Answers

Answer:

skeletal system

Explanation:

to create and fliter blood and provide frame-work to the human body and support

given two equations representing reactions: which type of reaction is represented by each of these equations?

Answers

Answer:

Equation 1 - nuclear fission

Equation 2 - nuclear fusion

Explanation:

Nuclear fission is a reaction in which a large nucleus is split into smaller nuclei when it is bombarded by neutrons. The process produces more neutrons to continue the chain reaction. This is clearly depicted in equation 1 as shown in the question.

Nuclear fusion is a reaction in which two light nuclei combine in order to form a larger nuclei. This is clearly depicted in equation 2 as shown in the question.

In the first reaction, a neutron is released, and in the second a helium atom is released. The given two equations represent nuclear fission and fusion.

What are nuclear reactions?

A nuclear reaction is a reaction that involves the nuclei of the atom and the absorption and release of energy. In the first reaction, a big nucleus is split by the neutron bombardment into smaller nuclei.

In the second reaction the process of nuclear fusion, two nuclei combine into a single larger nucleus that is shown as:

₁¹H+ ²₁H → ³₂He

Therefore, nuclear fission and fusion are represented by each of these equations.

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Can someone please help with these 2?

Answers

Equilibrium shifts to the right.

OPTION A


State three natural conditions under which plasmolysis could occur

Answers

Answer:

When the cell is placed in salt solutionWhen leave celss dryWhen cells are placed in sugar solution

Explanation:

Hope this helps

Según la cinética química para que una reacción ocurra, los átomos o moléculas deben

I. Chocar con la suficiente energía. II. Chocar con una concentración adecuada. III. Ser choques efectivos

A) Solo I.
B) Solo I y II.
C) Solo I y III.
D) Solo II y III.
E) I, II, III.

Answers

Answer:

solo I

Explanation:

Según esta teoría para que se produzca una reacción deben cumplirse tres condiciones: Las moléculas de los reactivos tienen que chocar entre sí. Estos choques deben de producirse con energía suficiente de forma que se puedan romper y formar enlaces químicos.

How much energy does an X-ray with an 8 nm (8 x 10-9m) wavelength have?
A. 1.99 x 10-25 J
B. 3.33 x 1016 J
C. 2.48 x 10-17 j
D. 8.28 x 10-26 J

Answers

Answer:

it would be option C

Explanation:

Speed of light = 3×10^8m/s

Planck's constant = 6.626×10^-34 Js

Wavelength = 8 x 10^-9 m

Energy = [(3×10^8) * (6.626×10^-34)] / 8 x 10^-9

Energy = [19.878×10^(8-34)] / 8 x 10^-9

Energy = 2.48475 × 10^(-26+9)

Energy = 2.48×10^-17 J

You are an intermediate product of an industrial process which intends to separate iron from its ore. A well known iron ore is hematite. Which of these ores does not contain iron?
Goethite
Malachite
Siderite
Limonite

Answers

Answer:

Malachite

Explanation:

Malachite is a copper carbonate hydroxide mineral, with the equation Cu2CO3(OH)2. This dark, green-joined mineral solidifies in the monoclinic precious stone framework, and frequently shapes botryoidal, sinewy, or stalagmitic masses, in cracks and profound, underground spaces, where the water table and aqueous liquids give the way to synthetic precipitation. So, the answer is malachite. Best of Luck!

sound waves? like what they do.

Answers

Answer:

A sound wave is the pattern of disturbance caused by the movement of energy traveling through a medium (such as air, water, or any other liquid or solid matter) as it propagates away from the source of the sound. The source is some object that causes a vibration, such as a ringing telephone, or a person's vocal chords.

HEY. HOPE THIS HELPS♡

What is its density in kilograms per cubic meter?

Answers

Copper has a density of 8950 kg/m3 = 8.95 kg/dm3 = 8.95 g/cm3. Water has a density of 1000 kg/m3 = 1000 g/L = 1 kg/dm3 = 1 kg/L = 1 g/cm3 = 1 g/mL.

HOPE IT HELPS❤️

Use the following information to calculate the concentration, Ka and pka for an unknown monoprotic weak acid. (8 pts.) 20.00 mL
Volume of unknown weak acid used : 20.00 mL Total volume of 0.20 M NaOH required to reach the equivalence point: 18.50 mL Initial pH of the weak acid 2.87

Answers

Answer:

Concentration: 0.185M HX

Ka = 9.836x10⁻⁶

pKa = 5.01

Explanation:

A weak acid, HX, reacts with NaOH as follows:

HX + NaOH → NaX + H2O

Where 1 mole of HX reacts with 1 mole of NaOH

To solve this question we need to find the moles of NaOH at equivalence point (Were moles HX = Moles NaOH).

18.50mL = 0.01850L * (0.20mol / L) = 0.00370 moles NaOH = Moles HX

In 20.0mL = 0.0200L =

0.00370 moles HX / 0.0200L = 0.185M HX

The equilibrium of HX is:

HX(aq) ⇄ H⁺(aq) + X⁻(aq)

And Ka is defined as:

Ka = [H⁺] [X⁻] / [HX]

Where [H⁺] = [X⁻] because comes from the same equilibrium

As pH = 2.87, [H+] = 10^-pH = 1.349x10⁻³M

Replacing:

Ka = [H⁺] [H⁺] / [HX]

Ka = [1.349x10⁻³M]² / [0.185M]

Ka = 9.836x10⁻⁶

pKa = -log Ka

pKa = 5.01

Air bags are activated when a severe impact causes a steel ball to compress a spring and electrically ignite a detonator cap. This causes sodium azide (NaN3) to decompose explosively according to the following reaction. 2 NaN3(s) --> 2 Na(s) 3 N2(g) What mass in grams of NaN3(s) must be reacted in order to inflate an air bag to 79.5 L at STP

Answers

Answer:

154 g

Explanation:

Step 1: Write the balanced decomposition equation

2 NaN₃(s) ⇒ 2 Na(s) + 3 N₂(g)

Step 2: Calculate the moles corresponding to 79.5 L of N₂ at STP

At STP, 1 mole of N₂ occupies 22.4 L.

79.5 L × 1 mol/22.4 L = 3.55 mol

Step 3: Calculate the number of moles of NaN₃ needed to form 3.55 moles of N₂

The molar ratio of NaN₃ to N₂ is 2:3. The moles of NaN₃ needed are 2/3 × 3.55 mol = 2.37 mol.

Step 4: Calculate the mass corresponding to 2.37 moles of NaN₃

The molar mass of NaN₃ is 65.01 g/mol.

2.37 mol × 65.01 g/mol = 154 g

A 15.0 mL urine from a dehydrated patient has a density of 1.019g/mL. What is the mass of the sample, reported in mg?

Answers

Answer:

Mass of sample in mg = 15,285 mg

Explanation:

Given:

Volume of urine sample = 15 ml

Density of sample = 1.019 g/ml

FInd:

Mass of sample in mg

Computation:

Mass = density x volume

Mass of sample in mg = Volume of urine sample x Density of sample

Mass of sample in mg = 1.019 x 15

Mass of sample in mg = 15.285 gram

Mass of sample in mg = 15.285 x 1,000

Mass of sample in mg = 15,285 mg

A diver exhales a bubble with volume of 250 mL at pressure of 2.4 atm and temperature of 15 C. How many gas particulate in this bubble?

Answers

Answer:

1.5x10²² particulates

Explanation:

Assuming ideal behaviour, we can solve this problem by using the PV=nRT formula, where:

P = 2.4 atmV = 250 mL ⇒ 250 / 1000 = 0.250 Ln = ?R = 0.082 atm·L·mol⁻¹·K⁻¹T = 15 °C ⇒ 15 + 273 = 288 K

We input the given data:

2.4 atm * 0.250 L = n * 0.082 atm·L·mol⁻¹·K⁻¹ * 288 K

And solve for n:

n = 0.025 mol

Finally we calculate how many particulates are there in 0.025 moles, using Avogadro's number:

0.025 mol * 6.023x10²³ particulates/mol = 1.5x10²² particulates

Emily spills concentrated sodium hydroxide solution on her lab bench. What she should do first?

Answers

Answer:

Explanation: hell noo

Emily must notify the Instructor of the Lab/Classroom.

Classify each of the following as a Strong acid (sa) or a Weak acid (wa) and indicate how each should be written in aqueous solution.
1. hydrobromic acid
2. hydrochloric acid
3. carbonic acid
Write a net ionic equation for the reaction that occurs when aqueous solutions of hydrochloric acid and barium hydroxide are combined.

Answers

Answer:

A. 1. Strong acid (sa): Hydrobromic acid: HBr (aq)

2. Strong acid (sa); Hydrochloric acid: HCl (aq)

3. Weak acid (wa); Carbonic acid: H₂CO₃ (aq)

B. H+ (aq) + OH- (aq) ----> H₂O (l)

Explanation:

Strong acids are which ionize completely in aqueous solution into hydrogen ions and the corresponding anion. Examples of strong acids include hydrobromic acid, hydrochloric acid, tetraoxosulfate (vi) acid.

The ionization of hdyrobromic and hydrochloric acids in aqueous solution is given below:

1. Hydrobromic acid: HBr (aq) ----> H+ (aq) + Br- (aq)

Hydrobromic acid in aqueous solution ionizes completely into hydrogen ions and bromide ions

2. Hydrochloric acid: HCl (aq) ----> H+ (aq) + Cl- (aq)

Hydrochloric acid in aqueous solution ionizes completely into hydrogen ions and chloride ions

Weak acids are acids which ionizes only partially in aqueous solutions to hydrogen ions and the corresponding anions. Examples of weak acids are carbonic acid and ethanoic acid. The ionization of carbonic acid in aqueous solution is shown below:

3. Carbonic acid: H₂CO₃ (aq) ⇄ 2 H+ (aq) + CO₃²- (aq)

Carbonic acid ionizes partially only to give hydrogen ions and trioxocarbonate (iv) ions. The unionized acid exists in equilibrium with the ions produced by the partial ionization of the acid.

Part B:

The reaction between hydrochloric acid and barium hydroxide is a neutralization reaction producing barium chloride salt and water.

The net ionic equation of the neutralization reaction is given below :

H+ (aq) + OH- (aq) ----> H₂O (l)

Look at pictures and help please

Answers

Answer: In order to increase the rate of reaction between hydrochloric acid and sugar increase the concentration of hydrochloric acid to 2 M because greater concentration results in more collision between the reactants.

Explanation:

More is the concentration of reactant molecules more will be the number of collisions between their molecules. As a result, more readily the products will be formed.

Hence, for the given reaction when concentration of HCl is increased then there will be increase in the number of collisions between reactants.

Thus, we can conclude that in order to increase the rate of reaction between hydrochloric acid and sugar increase the concentration of hydrochloric acid to 2 M because greater concentration results in more collision between the reactants.

1. Determine the volume of SO2 (at STP) formed from the reaction of 96.7 mol FeS2 and 55.0 L of O2 at 358 K and 1.20 atm.
4 FeS2(s) + 11O2(g)  2Fe2O3(s) + 8SO2(g)

Answers

Answer:

40.0L of SO2 are produced

Explanation:

To solve this question we need to find the moles of O2 using PV = nRT in order to find the moles. Thus, we can find the limiting reactant and the moles (And volume) of SO2 produced as follows:

Moles O2:

n = PV/RT

n = 1.20atm*55.0L / 0.082atmL/molK*358K

n = 2.25 moles of O2.

Clearly, limiting reactant is O2.

The moles of SO2 produced are:

2.25 moles of O2 * (8mol SO2 / 11mol O2) = 1.6351 moles SO2

Volume SO2:

V = nRT/P

V = 1.6351 moles SO2*0.082atmL/molK*358K / 1.20atm

V = 40.0L of SO2 are produced

HELPPP PLEASEEEEE
Name the following alkane molecule:

Answers

Answer:

5–bromo–9–chlorodecane

Explanation:

To name the compound given above, the following must be obtained:

1. The longest continuous carbon chain. This gives the parent name of the compound.

2. The substituent group attached to the compound.

3. Position of the substituent group.

4. Combine the above to obtain the name.

Now, we shall determine the name of the compound as follow:

1. The longest continuous carbon chain is 10. Thus, the parent name of the compound is decane.

2. The substituent groups attached to the compound are:

I. Bromine (Br) => Bromo

II. Chlorine (Cl) => Chloro

3. The position of the substituent groups are:

I. Br => carbon 5

II. Cl => carbon 9

NOTE: numbering is done alphabetically.

4. Therefore, the name of the compound is:

5–bromo–9–chlorodecane

Answer:

A.

Explanation:

I chose this answer and it was correct ‍♀️

What volume in mL of 0.220 M HBr solution is required to produce 0.0130 moles of HBr?

Answers

Answer: A volume of 59 mL of 0.220 M HBr solution is required to produce 0.0130 moles of HBr.

Explanation:

Given: Moles = 0.0130 mol

Molarity = 0.220 M

Molarity is the number of moles of solute present in liter of a solution.

[tex]Molarity = \frac{moles}{volume (in L)}[/tex]

Substitute the values into above formula as follows.

[tex]Molarity = \frac{moles}{volume (in L)}\\0.220 M = \frac{0.0130 mol}{Volume (in L)}\\Volume (in L) = 0.059 L[/tex]

As 1 L = 1000 mL

So, 0.059 L = 59 mL

Thus, we can conclude that a volume of 59 mL of 0.220 M HBr solution is required to produce 0.0130 moles of HBr.

Lee y analiza a detalle el Anexo #1 de este plan de trabajo el cual habla sobre las fuentes alternativas de energía. Posteriormente con la información elabora un cartel o un cuadro sinóptico en tu cuaderno donde organices la información para darla a conocer a los miembros de tu comunidad.

Answers

Answer: el texto no es tan claro

An infant acetaminophen suspension contains 80 mg/0.80 mL suspension. The recommended dose is 15 mg/kg body weight.
How many milliliters of this suspension should be given to an infant weighing 13 lb.

Answers

Answer:

0.8853 mL

Explanation:

First we convert 13 lb to kg, keeping in mind that 1 lb = 0.454 kg:

13 lb * [tex]\frac{0.454kg}{1lb}[/tex] = 5.902 kg

Then we calculate how many mg of acetaminophen should be given, using the recommended dose and infant mass:

15 mg/kg * 5.902 kg = 88.53 mg

Finally we calculate the required mL of suspension, using its concentration:

88.53 mg ÷ (80 mg/0.80 mL) = 0.8853 mL

Why is iodine always Used in a solution containing excess I2

Answers

Answer:

If a standard iodine solution is used as a titrant for an oxidizable analyte, the technique is iodimetry. If an excess of iodide is used to reduce a chemical species while simultaneously forming iodine.

Iodine always used in a solution excess KI is given to aid in the solubilization of free iodine, which would be insoluble in clean water during normal circumstances.

What is Iodine?

Iodine is a kind of element which are mainly used in iodometry titration. It can be represented by I.

What is solution?

A solution would be a homogenous mixture of two components, usually a solute as well as a solvent.

Iodimetry would be a technique that uses a standard iodine solution as a titrant for such an oxidizable analyte. When an excessive amount of iodide is used to decrease a chemical while somehow producing iodine.

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Why can light be treated like a particle?

Answers

Answer:

Light can be treated like particles because it is made of chunks like things called protons.

What is the molar mass of Na2SO4?
O A. 142.04 g/mol
O B. 71.05 g/mol
O c. 238.22 g/mol
O D. 94.04 g/mol

Answers

I believe its D due to my thinking

Answer: A. 142.04 g/mol

Explanation:

Molecule contains carbon, hydrogen and sulfur atoms. When a sample of 0.535g of this compound is burnt in oxygen, 1.119 g of CO2and 0.229 gof H2O and 0.407g of SO2are obtained.
Calculate its empirical formula.

Answers

Answer:

The empirical formula is, C4H4S

Explanation:

Number of moles of carbon = 1.119 g/ 44g/mol = 0.025 moles

Mass of Carbon= 0.025 moles × 12 g/ mole = 0.3 g

Number of moles of hydrogen = 0.229/18g/mol × 2 = 0.025 moles

Mass of hydrogen = 0.025 moles × 1 = 0.025 g

Number of moles of sulphur = 0.407g/ 64 g/mol = 0.0064 moles

Mass of sulphur= 0.0064 moles ×32 = 0.2 g

Now we obtain the mole ratios by dividing through by the lowest ratio.

C- 0.025 moles/ 0.0064 moles, H- 0.025 moles/ 0.0064 moles, S- 0.0064 moles/0.0064 moles

C4H4S

Which of the following mixtures is best separated by the use of a separating funnel?
methane and water
ethyl ethanoate and water
ethanol and water
ethanoic acid and water

Answers

Answer:

ethyl ethanoate and water

Explanation:

At the point when one fluid doesn't blend in with another yet glides on top of it, an isolating pipe can be utilized to isolate the two fluids. Oil glides on water. This combination can be isolated utilizing an isolating channel as demonstrated on the following page.  

Ethyl liquor and water are two miscible fluids.   Refining is a cycle that can be utilized to isolate an unadulterated fluid from a combination of fluids. An isolating channel can be utilized to isolate the parts of the combination of immiscible fluids.

The answer is ethyl ethanoate and water. Hope this helps you!

A transition metal in the fourth period from the following list : Cu, O , Pr, Ag

Answers

Answer:

Cu

Explanation:

Groups 3 - 12 (or groups IIA - IIB) of the periodic table contain transition elements. Transaction elements start from period four (4) of the periodic table. The phrase alludes to the fact that the d sublevel is filling at a lower main energy level than the s sublevel that came before it.

The transition elements' arrangement is inverted from the fill order, with the 4 s filled prior to the actual 3 d begins. The transition elements are commonly referred to as transition metals since they are all metals. They are less reactive than the metals in Groups 1 and 2 and have normal metallic characteristics.

From the options given Cu is the only transition metal in the fourth period on the periodic table.


1. Complete the following chart for the following atoms:
Element
Mass
number
Number of
electrons
Number of
neutrons
Atomic
Number
(number of
protons)
Potassium
16
56
26
What is the atomic symbol (the nuclide) for the isotope with 15 protons and 16
neutrons? Your answer must be in the form shown:
AX
You must show the correct numbers for A and Z
and have the correct symbol (Z). (3 points)

Answers

Explanation:

If there are 15 protons, 15 nuclear particles of unit positive charge, then

Z

=

15

. Now

Z

the atomic number

, and you look at your copy of the Periodic Table, and you find that for

Z

=

15

, the element phosphorus is specified.

But we are not finished. Along with the 15 defining protons, there are also 16 neutrally charged, massive nuclear particles, 16 neutrons, and the protons and neutrons together determine the atomic mass. The isotope is thus

31

P

, which is almost 100% abundant, and an important nucleus for

NMR spectroscopy.

A 1.375 g sample of mannitol, a sugar found in seaweed, is burned completely in oxygen to give 1.993 g of carbon dioxide and 0.9519 g of water. The empirical formula of mannitol is

Answers

Answer:

[tex]C_3H_7O_3[/tex]

Explanation:

Hello there!

In this case, according to the given information, it turns out possible for us to infer that the empirical formula of mannitol contains carbon, hydrogen and oxygen, so that the first step is to calculate the moles of C and H contained in the CO2 and H2O, respectively, as the only sources of these two elements in the formula:

[tex]n_C=1.993gCO_2*\frac{1molCO_2}{44.01gCO_2}*\frac{1molC}{1molCO_2} =0.0453molC\\\\n_H=0.9519gH_2O*\frac{1molH_2O}{18.02gH_2O}*\frac{2molH}{1molH_2O} =0.106molH[/tex]

Next, we calculate the grams and moles of O by subtracting the mass of C and H from the mass of the sample:

[tex]m_O=1.375g-0.0453molC*\frac{12gC}{1molC}-0.106molH*\frac{1.01gH}{1molH}=0.724gO\\\\n_O=0.724gO*\frac{1molO}{16.0gO} =0.0453molO[/tex]

Finally, we divide the moles of C, H and O by 0.0453 as the fewest moles of both C and O to find the mole ratios in the formula:

[tex]C:\frac{0.0453mol}{0.0453mol} =1\\\\H:\frac{0.106mol}{0.0453mol} =2.34\\\\O:\frac{0.0453mol}{0.0453mol} =1[/tex]

To get:

[tex]CH_{2.34}O[/tex]

Which must be multiplied by 3 to get whole numbers for all the subscripts, and therefore obtain:

[tex]C_3H_7O_3[/tex]

Regards!

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