Calculate the molar enthalpy change of H2O (l) when liquid water is heated from 41.7 C to 76.2 C

Final answer:

The percentage yield of aspirin is calculated by comparing the actual yield from the experiment to the theoretical yield determined by stoichiometry. Calculations are based on the molar masses of salicylic acid and aspirin and the stoichiometry of the reaction between them.

Explanation:

To calculate the percentage yield of aspirin from salicylic acid, first determine the theoretical yield and then compare it to the actual yield. The reaction between salicylic acid and ethanoic anhydride is given as:

C7H6O3 + (CH3CO)2O → C9H8O4 + CH3COOH

The molar mass of salicylic acid (C7H6O3) is 138.12 g/mol, and the molar mass of aspirin (C9H8O4) is 180.16 g/mol. If all 50.05 g of salicylic acid reacted, the maximum amount of aspirin that could be formed is calculated using stoichiometry:

Theoretical yield = (50.05 g of salicylic acid * 1 mol salicylic acid / 138.12 g salicylic acid) * (180.16 g aspirin / 1 mol aspirin)

From the actual experiment, 55.45 g of aspirin was obtained. Now, calculate the percentage yield:

Percentage yield = (Actual yield / Theoretical yield) * 100%

By inserting the values into this equation, the percentage yield can be found. Remember that a percentage yield above 100% is not practically possible, indicating that there may have been an error in the experiment or in calculating the yields.

Answer:

1. Oxidation is when a molecule, atom or ion losses an Electron. 2. Reduction is when a molecule, atom or ion gains an electron. 3. endergonic means absorbing energy in the form of work whereas Exergonic means releasing energy in the form of work.

Explanation:

Answer is: 2.0 M NaCl.

Change in freezing point from pure solvent (water) to solution (sodium chloride solution): ΔT = i · Kf · c.

Kf - molal freezing-point depression constant for water is 1.86°C/m.

c-  molarity of the solution.

i - Van't Hoff factor.  

Because molal freezing-point depression constant and Van't Hoff factor are the same for all five solutions, freezing point depends on molarity of the solution.

The higher is the molarity, the lower is freezing point.

The temperature at which the phase transition solid-liquid occurs is the melting point or the freezing point.

Answer : The correct option is, 2.0 M NaCl

Explanation :

Formula used for lowering in freezing point :

[tex]\Delta T_f=i\times k_f\times m[/tex]

where,

[tex]\DeltaT_f[/tex] = change in freezing point  or freezing point depression

[tex]k_f[/tex] = freezing point constant

m = molality

i = Van't Hoff factor

As we know that the Van't Hoff factor for NaCl will be same for all given concentrations of NaCl and [tex]k_f[/tex] is the constant term. So, freezing point depression directly depends only on the molality of the solution.

That means the more the value of molality, the lower will be the freezing point and vice-versa.

From the given options, 2.0 M NaCl has the lowest freezing point.

Hence, the correct option is, 2.0 M NaCl

Final answer:

Iron (Fe) has the highest binding energy per nucleon, making it the most stable element and unable to release energy by fusion or fission.

Explanation:

The element that has the lowest mass per nuclear particle and cannot release energy by either fusion or fission is iron (Fe). Iron is the most stable element due to its highest binding energy per nucleon. Energy can be extracted by fusing elements lighter than iron, but once iron is formed, fusing heavier elements requires the addition of energy instead of releasing it. Similarly, fission reactions release energy with heavy, unstable nuclei that have low binding energies, such as uranium-235 or uranium-238. Since iron has a high binding energy and lies at the peak of the binding energy curve, it neither releases energy via fusion past this point, nor can it effectively release energy through fission.

Final answer:

The concentration of the HBr solution is 0.001888 M.

Explanation:

To determine the concentration of the HBr solution, we need to use the equation:

HBr(aq) + NaOH(aq) → NaBr(aq) + H2O(l)

From the balanced equation, we can see that the mole ratio of HBr to NaOH is 1:1. Since the volume of NaOH required to reach the equivalence point is 18.88 mL, we can calculate the number of moles of NaOH used:

Moles of NaOH = concentration of NaOH (M) × volume of NaOH (L)

Moles of NaOH= 0.100 M × 0.01888 L = 0.001888 mol of NaOH

Since HBr and NaOH have a 1:1 mole ratio, the concentration of the HBr solution is also 0.001888 M.

To find the final volume of a gas when the temperature increases at constant pressure, apply Charles's Law. The final volume of the 500 ml gas sample that is heated from 150 K to 350 K is calculated to be 1166.67 ml.

The problem relates to Charles's Law, which states that for a fixed amount of gas at constant pressure, the volume is directly proportional to its temperature in Kelvin. If we know the initial volume and temperature of a gas sample and the temperature changes, we can use Charles's Law (V1/T1 = V2/T2) to find the final volume of the gas.

To calculate the final volume of the 500 ml sample of gas that increases in temperature from 150 K to 350 K while keeping the pressure constant, we use the formula:

V1/T1 = V2/T2

Substituting in the given values:

500 ml / 150 K = V2 / 350 K

We solve for V2:

V2 = 500 ml × (350 K / 150 K)

V2 = 500 ml × (7/3)

V2 = 1166.67 ml

The final volume of the gas will be 1166.67 ml.

To produce aluminum from aluminum hydroxide and carbon, the balanced chemical equation is: 2 Al(OH)₃(s) + 3 C(s) → 2 Al(s) + 3 CO(g) + 3 H₂O(g). This ensures all elements are balanced.

To write the net chemical equation for the production of aluminum from aluminum hydroxide and carbon, we must consider the typical reaction: Al(OH)₃(s) + C(s) → Al(s) + CO(g) + H₂O(g)

Now let’s balance it:

Therefore, the balanced equation is: 2 Al(OH)₃(s) + 3 C(s) → 2 Al(s) + 3 CO(g) + 3 H₂O(g)

The dipole moment of HF can be calculated through the charge of an electron and bond length. Upon conversion into Debye, it's found to be approximately 4.4 D assuming the HF bond to be completely ionic.

The dipole moment of a molecule depends on the difference in electronegativity between the atoms and the distance between the atoms' nuclei. In the case of HF (Hydrogen Fluoride), assuming the bond is completely ionic, we can calculate the dipole moment using the charge of an electron (1.60218 × 10-¹⁹ C) and the bond length (converted to meters).

The bond length of HF given is 0.917 angstroms, which is equivalent to 0.917 x 10-10m or 9.17 x 10-11m. This allows us to calculate the dipole moment in coulomb-meters (C.m) as (1.60218 × 10-¹⁹ C) * (9.17 x 10⁻¹¹m) = 14.69 x 10⁻³⁰ Cm.

We usually express dipole moments in Debyes. To convert C.m to Debye, we use the conversion factor where 1 Debye (D) = 3.336 x 10⁻³⁰ C.m, which gives us a dipole moment of approximately 4.4 D for HF.

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Answer:

its not electrons are negatively charged

Explanation:

Answer: The correct statements are the concentration of one or more of the products is small, the reaction will not proceed very far to the right and the reaction will generally form more reactants than products.

Explanation:

[tex]K_{eq}[/tex] is defined as the equilibrium constant of the reaction. It is basically the ratio of concentration of products to the concentration of reactants, each raised to the power their stoichiometric coefficients.

For a reaction:

[tex]aA+bB\rightarrow cC+dD[/tex]

The expression for [tex]K_{eq}[/tex] is:

[tex]K_{eq}=\frac{[C]^c[D]^d}{[A]^a[B]^b}[/tex]

When [tex]K>1[/tex], forward reaction is favored and when [tex]K<1[/tex], backward reaction is favored.

When K < 1, the expected possibilities are:

Hence, the correct statements are the concentration of one or more of the products is small, the reaction will not proceed very far to the right and the reaction will generally form more reactants than products.

Answer:

The correct answer is option B, ammonia

Explanation:

A weak base solution like pyridine (C5H5N) can be neutralized in the presence of an acid only. A weak base in no case shall be neutralized by another base. Since ammonia is also a base thus it cannot neutralize another base. In a neutralized mixture, slats are produced by equal contribution from both the acid and base. However, when a weak base is placed into another base their will be more OH- ions but very rare H+ ion. Thus mixture with excess of OH- will again be a base only.

Answer:

35.567mL

Explanation:

Simple, density = mass/volume

Volume = mass/density

Which yields

34.5/0.97 = 35.567milliliters

Answer : To determine the mass of oxygen in the compound produced in the virtual lab, it should be weighed and subtracted from the total weight of the compound. The mass pf each element can be used to determine the empirical formula of the compound by finding out the molar ratios of the individual elements present in the compound. Molar ratios can be obtained by dividing elements by atomic masses of individual elements.

One has to spot the smallest moles of the elements present in the compound and then it has to be divided by rest of the elements in the compound to find the empirical formula of that compound.

Kepler discovered the laws of planetary motion, including orbits being ellipses, equal areas swept out in equal times, and the period of revolution being related to the semi-major axis.

Kepler discovered that:

Each planet moves around the Sun in an orbit that is an ellipse, with the Sun at one focus.

The radius vector from Sun to planet sweeps out equal areas in equal times.

The planet's period of revolution squared is proportional to the cube of the semi-major axis of the ellipse.

The change in reaction rate when doubling the concentration of [b] is dependent on the rate law of the reaction. For one-to-one stoichiometry, the reaction rate should double. However, reaction orders for more complex reactions may alter the rate differently.

In changing the concentration of reactant [b] in any reaction, the change in reaction rate is dependent on what is known as the 'rate law,' and cannot be predicted without knowledge of this law. In simple reactions, typically found in elementary chemistry, if the reaction has a one-to-one stoichiometry, like aA → bB, doubling the concentration of a reactant, in this case [b], will double the rate of reaction. This means, if changes in concentration of [b] only affect the reaction rate, then doubling [b] should double the reaction rate. However, for more complex reactions, we use the concept of reaction orders. The exponent on the concentration term in the rate law, sometimes referred to as 'n' in this context, dictates how a change in [b] alters the reaction rate. If n=1, doubling [b] would indeed double the reaction rate, but if n=2, doubling [b] would quadruple the rate, and so forth.

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The reaction rate of the provided rate law changes by a factor of the square root of two, or approximately 1.41, when the concentration of reactant B is doubled, all else remaining constant.

It refers to the change in reaction rate given a rate equation when the concentration of one reactant (B) is doubled, with the other reactant concentrations held constant. According to the provided rate law,

Rate=k [B][tex]^{\frac{1}{2}}[/tex] [C]²,

when the concentration of B is doubled, the reaction rate changes by the square root of two. Since the order of reaction concerning B is 1/2, the effect of doubling the concentration of B can be represented mathematically as:

New rate when [B] is doubled:

[tex]k(2[B])^{\frac{1}{2}} [C]^{2}[/tex] = 2[tex]^{ \frac{1}{2} }[/tex]R

The factor changes in rate = 1.41R

Therefore, the reaction rate changes by a factor of [tex]\sqrt{2}[/tex] or approximately 1.41, when the concentration of B is doubled.

Answer:

Enthalpy of combustion (for 1 mol of butane)=-2657.4 [tex]\frac{kJ}{mol}[/tex]

Explanation:

Combustion is a rapid oxidation chemical process that is accompanied by low energy shedding in the form of heat and light. Oxygen is the essential element for oxidation to occur and is known as a oxidizer. The material that oxidizes and burns is the fuel, and is generally a hydrocarbon, as in this case butane C4H10 (g)

The balanced reaction is:

2 C4H10   +      13 O2       →      8 CO2   +     10H2O

Note that a balanced equation must have the same amount of each atom in the reagents and in the products, as in the previous reaction.

The heat of formation is the increase in enthalpy that occurs in the formation reaction of one mole of a certain compound from the elements in the normal physical state (under standard conditions: at 1 atmosphere of pressure and at 25 degrees of temperature).

In literature you can obtain the following heats of formation of each of the molecules involved in the reaction:

Heat of formation of C4H10 = -125.7 kJ/mol

Heat of formation of water = -241.82 kJ/mol

Heat of formation of CO2 = -393.5 kJ/mol

For the formation of one mole of a pure element the heat of formation is 0, in this case we have as a pure compound the oxygen O2

You want to calculate the ∆H (heat of reaction) of the combustion reaction, that is, the heat that accompanies the entire reaction. For that you must make the total sum of all the heats of the products and of the reagents affected by their stoichiometric coefficient (quantity of molecules of each compound that participates in the reaction) and finally subtract them:

Enthalpy of combustion = ΔH = ∑Hproducts - ∑Hreactants

                                             = (-393.5X8) + (-241.82X10) - (-125.7X2)

                                             = -5314.8 kJ/mol

But, if you observe the previous balanced reaction, you can see that 2 moles of butane are necessary in combustion. And the calculation of the heat of reaction previously carried out is based on this reaction. This ultimately means that the energy that would result in the combustion of 2 moles of butane is -5314.8 kJ/mol.

Then, applying a rule of three can calculate energy required for the combustion of one mole of butane: if for the combustion of two moles of butane an enthalpy of -5314.8 kJ / mol is required, how much energy is required for the combustion of one mole of butane?

Enthalpy of combustion (for 1 mol of butane)=[tex]\frac{-5314.8 \frac{kJ}{mol} }{2}[/tex]

Enthalpy of combustion (for 1 mol of butane)=-2657.4 [tex]\frac{kJ}{mol}[/tex]

The enthalpy change for the combustion of ethanol can be calculated using the enthalpies of formation and applying Hess's law. It's determined by the difference in the products and reactants enthalpy of formation. The enthalpy change for the combustion of ethanol is -1366 kJ per mole of ethanol burned.

To determine the enthalpy change for the combustion of ethanol C₂H5OH(l) + 3 O2(g) → 2 CO2(g) + 3 H2O(g), we need to apply the concept known as Hess's law. In simple terms, Hess's law states that the total enthalpy change of a chemical reaction is the sum of the enthalpy changes for the steps of the process.

Here, we know the enthalpies of formation for the reactants and products involved. Enthalpies of formation are defined for one mole of a substance formed from its elements in their standard states. The enthalpies of formation for the substances involved are ethanol (C₂H5OH(l)) -278 kJ/mol, water (H₂O(l)) -286 kJ/mol, and carbon dioxide (CO2(g)) -394 kJ/mol.

The enthalpy change of the reaction ΔH = Σ[ (products moles x products enthalpy of formation) - (reactants moles x reactants enthalpy of formation) ]

On substituting the numbers we get ΔH = [ (2 mol CO₂ x -394 kJ/mol CO₂) + (3 mol H2O x -286 kJ/mol H2O) ] - [ (1 mol C2H5OH x -278 kJ/mol C2H5OH) + (3 mol O2 x 0 kJ/mol O2) ].

So, ΔH = [ -788 kJ + -858 kJ ] - (-278 kJ) = -1366 kJ

Thus, the enthalpy change for the combustion of ethanol is -1366 kJ per mole of ethanol burned.

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The unknown liquid is most likely Propane.

To determine the identity of the unknown liquid, we calculate its density and compare it to the densities of known substances. Density is defined as mass divided by volume. Given that the mass of the unknown liquid is 500 grams and the volume is 1.0 L (which is equivalent to 1000 cm3), we use the formula:

Density = Mass / Volume

This gives us:

Density = 500 g / 1000 cm3 = 0.5 g/cm3

Comparing this value to the given options:

Distilled Water Density = 1.0 g/cm3

Propane Density = 0.494 g/cm3

Salt Water Density = 1.025 g/cm3

Liquid Gold Density = 17.31 g/cm3

The density of the unknown liquid (0.5 g/cm3) is closest to that of Propane (0.494 g/cm3), so the unknown liquid is most likely Propane.