A 6.13 g sample of an unknown salt (MM = 116.82g/mol) is dissolved in 150.00 g water in a coffee cup
calorimeter. Before placing the sample in the water, the
temperature of the salt and water is 23.72°C. After the
salt has completely dissolved, the temperature of the
solution is 28.54°C.


If 3.15 x 10J of heat was gained by the solution, what
is the total heat for the dissolution reaction of the 6.13 g
of salt?

Answers

Answer 1
Answer:

According to law of conservation of energy, if 31.5 J of heat is gained than same amount of heat is lost .

What is law of conservation of energy?

According to law of conservation of energy, it is evident that energy is neither created nor destroyed rather it is restored at the end of a chemical reaction .

Law of conservation of mass and energy are related as mass and energy are directly proportional which is indicated by the equation E=mc².Concept of conservation of mass is widely used in field of chemistry, fluid dynamics.

Law needs to be modified in accordance with laws of quantum mechanics under the principle of mass and energy equivalence.This law was proposed by Julius Robert Mayer in the year 1812.

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Answer 2
Answer:

Answer:

-3.19x10³ J

Explanation:

Since the surroundings absorbed 3.19 × 10³ J (or 3190 J) of heat, the system, or the dissolution reaction, must have lost the same amount of heat. The heat for the system, then, is -3.19 × 10³ J (or -3190 J). We know this is true because of the first law of thermodynamics, "heat is a form of energy, and thermodynamic processes are therefore subject to the principle of conservation of energy".


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The flask contains 10.0 mL of HCl and a few drops of phenolphthalein indicator. The buret contains 0.160 M NaOH . It requires 11.9 mL of the NaOH solution to reach the end point of the titration. A buret filled with a titrant is held above a graduated cylinder containing an analyte solution. What is the initial concentration of HCl

Answers

Answer:

0.190 M

Explanation:

Let's consider the neutralization reaction between HCl and NaOH.

HCl + NaOH = NaCl + H2O

11.9 mL of 0.160 M NaOH were used. The reacting moles of NaOH were:

0.0119 L × 0.160 mol/L = 1.90 × 10⁻³ mol

The molar ratio of HCl to NaOH is 1:1. The reacting moles of HCl are 1.90 × 10⁻³ moles.

1.90 × 10⁻³ moles of HCl are in 10.0 mL of solution. The molarity of HCl is:

M = 1.90 × 10⁻³ mol / 10.0 × 10⁻³ L = 0.190 M

Answer:

The initial concentration of HCl was 0.1904 M

Explanation:

Step 1: Data given

Volume of HCl solution = 10.0 mL = 0.010 L

Volume of a NaOH solution = 11.9 mL = 0.0119 L

Molarity of NaOH solution = 0.160 M

Step 2: The balanced equation

HCl + NaOH → NaCl + H2O

Step 3: Calculate the concentration of HCl

C1*V1 = C2*V2

⇒with C1 = the concentration HCl = TO BE DETERMINED

⇒with V1 = the volume of HCl = 0.010 L

⇒with C2 = the concentration of NaOH = 0.160 M

⇒with V2 = the volume of NaOH = 0.0119 L

C1 * 0.010 L = 0.160 M * 0.0119 L

C1 = (0.160 M * 0.0119 L) / 0.010 L

C1 = 0.1904 M

The initial concentration of HCl was 0.1904 M

Question 51 pts A breathalyzer is a device used to estimate the blood alcohol content of a suspected drunk driver by measuring the amount of alcohol in one's breath. The fuel cell breathalyzer employs the reaction below: CH3CH2OH(g)+O2(g)→HC2H3O2(g)+H2O(g) When a suspected drunk driver blows his or her breath through the fuel-cell breathalyzer, the device measures the current produced by the reaction and calculates the percent alcohol in the breath. How many moles of electrons are transferred per mole of ethanol, CH3CH2OH, in the reaction?

Answers

Answer:

Four moles of electrons

Explanation:

The reactions in a breathalyzer are redox reactions. Fuel cell breathalyzers consists of fuel cells with platinum electrodes. The current produced depends on the amount of alcohol in the breath. Detection of alcohol involves the oxidation of ethanol to ethanoic acid. The two half cells set in the process are;

Anode;

C2H5OH(aq) + 4OH^-(aq) ----------> CH3COOOH(aq) + 3H2O(l) + 4e

Cathode;

O2(g) + 2H2O(l) +4e--------> 4OH^-(aq)

Hence four electrons are transferred in the process.

What is the empirical formula of a compound that is 64.3 % c, 7.2 % h, and 28.5 % o by mass?

Answers

Asnwer : Empirical formula of a compound is : C_(3)H_(4)O

Given information : C = 64.3 % , H = 7.2 % , O = 28.5 %

Step 1 : Convert the given percentage (%) to grams.

Explanation : Let the total mass of the compound be 100 grams.

Mass of C = 64.3 g

(100g)* ((64.3percent))/((100percent)) = 64.3g

Mass of H = 7.2 g

(100g)* ((7.2percent))/((100percent)) = 7.2g

Mass of O = 28.5 g

(100g)* ((28.5percent))/((100percent)) = 28.5g

Step 2 : Convert the grams of each compound to moles.

Moles = (Grams)/(Molar mass)

Molar mass of C = 12.0g/mol  

Molar mass of H = 1.0 g/mol

Molar mass of O = 16.0g/mol

Moles of C = (64.3g)/(12.0(g)/(mol))

Moles of C = 5.36 mol

Moles of H = (7.2g)/(1.0(g)/(mol))

Moles of H = 7.2 mol

Moles of O = (28.5g)/(16.0(g)/(mol))

Moles of O = 1.78 mol

Step 3 : Find the mole ratio of C , H and O

Mole ratio is calculated by dividing the mole values by the smallest value.

Mole of C = 5.36 mol , Mole of H = 7.2 mol , Mol of O = 1.78 mol

Out of the three mole values , mole value of O that is 1.78 mol is less , so we divide all the mole values by 1.78 mol.

Mole of C = (5.36mol)/(1.78mol) = 3.0

Mole of H = (7.2mol)/(1.78mol) = 4.0

Mole of O = (1.78mol)/(1.78mol) = 1.0

C : H : O = 3 : 4 : 1

So empirical formula of the compound is C_(3)H_(4)O_(1) or C_(3)H_(4)O

The amount of heat required to melt 2 lbs of ice is twice the amount of heat required to melt 1 ib of ice. is this observation a macroscopic or microscopic description of chemical behavior? Explain your answer.

Answers

The observation in this instance relates to the quantity of heat needed to melt ice, and it is expressed in terms of weights (2 lbs and 1 lb) and a comparison (twice the amount).

Without going into detail into the different molecules or their interactions, it concentrates on the general behaviour and characteristics of the substance (ice) as a whole.

A microscopic description, on the other hand, would describe the behaviour in terms of the molecular or atomic interactions that take place at the particle level. It would go into ideas such as the amount of heat required to dissolve the intermolecular interactions between water molecules.

Therefore, the observation regarding how much heat is needed to melt ice is a macroscopic description since it ignores the underlying molecular interactions in favour of the substance's general behaviour and qualities.        

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The observation that melting 2 lbs of ice requires twice the heat of melting 1 lb is a macroscopic description, focusing on observable properties and behavior without exploring microscopic details.

This observation is a macroscopic description of chemical behavior. Macroscopic descriptions involve the properties and behavior of substances on a large scale that can be observed directly, without delving into the molecular or atomic details. In this case, the statement refers to the amount of heat required to melt a certain quantity of ice, and it is expressed in terms of macroscopic, measurable quantities (pounds of ice and the associated heat).

The macroscopic observation does not provide insight into the molecular or atomic interactions within the ice but rather focuses on the overall behavior of the substance. The concept that the amount of heat required to melt 2 lbs of ice is twice that needed for 1 lb of ice is a statement about the material's behavior at a larger scale.

This observation aligns with the macroscopic principles of heat and phase transitions, where the heat required for a phase change is directly proportional to the mass of the substance undergoing the transition. The macroscopic perspective is concerned with observable properties and measurements, making it a practical and accessible way to describe chemical behavior without delving into microscopic details.

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Part D 2ClO2(g)+2I−(aq)→2ClO−2(aq)+I2(s) Drag the appropriate labels to their respective targets. ResetHelp e−→e Superscript- rightarrow ←e−leftarrow e Superscript- Cathode Cathode Anode Anode II Superscript- ClO−2C l O Subscript 2 Superscript- Request Answer Part E Indicate the half-reaction occurring at Anode. Express your answer as a chemical equation. Identify all of the phases in your answer. nothing

Answers

The half reaction occurring at anode is:

2I^-(aq)---- > I_2(s)+2e^-

Half reaction for the cell:

The substance having highest positive  potential will always get reduced and will undergo reduction reaction.

Balanced chemical equation:

2ClO_2(g)+2I^-(aq)----- > 2ClO^(2-)(aq)+I_2(s)

The half reaction follows:

Oxidation half reaction:  2I^-(aq)---- > I_2(s)+2e^- , Reduction potential is 0.53V

Reduction half reaction:  ClO_2(g)+e^----- > ClO_2^-   ( × 2 ), Oxidation potential is +0.954 V

Oxidation reaction occurs at anode and reduction reaction occurs at cathode.

Hence, the half reaction occurring at anode is :

2I^-(aq)---- > I_2(s)+2e^-

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Answer: The half reaction occurring at anode is 2I^-(aq.)\rightarrow I_2(s)+2e^-

Explanation:

The substance having highest positive E^o potential will always get reduced and will undergo reduction reaction.

For the given chemical equation:

2ClO_2(g)+2I^-(aq)\rightarrow 2ClO^(-2)(aq.)+I_2(s)

The half reaction follows:

Oxidation half reaction:  2I^-(aq.)\rightarrow I_2(s)+2e^-;E^o_(I_2/I^-)=0.53V

Reduction half reaction:  ClO_2(g)+e^-\rightarrow ClO_2^-(aq.);E^o_(ClO_2/ClO_2^-)=+0.954V    ( × 2 )

Oxidation reaction occurs at anode and reduction reaction occurs at cathode.

Hence, the half reaction occurring at anode is 2I^-(aq.)\rightarrow I_2(s)+2e^-

Identify the true statement(s) about the valence bond theory. 1. The strength of a bond depends on the amount of overlap between the two orbitals of the bonding atoms. 2. The greater the overlap between two bonding atoms, the lesser the bond strength. 3. Orbitals bond in the directions in which they protrude or point to obtain maximum overlap

Answers

Answer:

The strength of a bond depends on the amount of overlap between the two orbitals of the bonding atoms

Orbitals bond in the directions in which they protrude or point to obtain maximum overlap

Explanation:

The valence bond theory was proposed by Linus Pauling. Compounds are firmed by overlap of atomic orbitals to attain a favourable overlap integral. The better the overlap integral (extent of overlap) the better or stringer the covalent bond.

Orbitals overlap in directions which ensure a maximum overlap of atomic orbitals in the covalent bond.

Answer:

THE STRENGTH OF THE BOND DEPENDS ON THE AMOUNT OF OVERLAP BETWEEN THE TWO ORBITALS OF THE BONDING ATOMS

ORBITALS BOND IN THE DIRECTION OR POINT IN WHICH THEY PROTRUDE OR POINT TO OBTAIN MAXIMUM OVERLAP.

Explanation:

Valence bond theory describes the covalent bond as the overlap of half-filled atomic orbital yields a pair of electrons shared between the two bonded atoms. Overlapping of orbitals occurs when a portion of one orbital and the other occur in the same region of space. The strength of a bond is determined by the amount of overlap between the two orbitals of the bonding atoms. In other words, orbitals that overlap more and in the right orientation of maximum overlapping form stronger bonds that those with less overlap and right orientation for maximum overlap. The bonding occurs at a varying distance in different atoms from which it obtains its stable energy caused by the increase in the attraction of nuclei for the electrons.

Orbitals also bond in the direction to obtain maximum overlap as orientation of the atoms also affect overlap. The greater overlap occurs when atoms are oriented on a direct line mostly end to end or side by side between the two nuclei depending on the type of bond formed. A sigma bond is formed when atoms overlap end to end in which a straight line exists between the two atoms that is the internuclear axis indicating the concentrated energy  density in that region. Pi bond exits in when overlap occurs in the side -to -side orientation and the energy density is concentrated opposite the internuclear axis.