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.

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.

Answer:

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

Equilibrium shifts to the right.

OPTION A

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.

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.

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

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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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

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!

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)

Explanation:

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

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.

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❤️

Answer:

Explanation:

Hope this helps

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.

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!

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 ‍♀️

Answer:

skeletal system

Explanation:

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

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

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.

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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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

Answer:

Explanation: hell noo

Answer:

0.8853 mL

Explanation:

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

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

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

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!

Answer: A. 142.04 g/mol

Explanation:

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

Answer:

1.5x10²² particulates

Explanation:

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

We input the given data:

And solve for n:

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

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

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.

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.

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Answer: el texto no es tan claro

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