Based on the diagram below, which would have the most inertia?


The sphere on the left, because it has more mass

The sphere on the right because it has less mass

The sphere on the left because it is a bigger diameter

The sphere on the right because it is a smaller diameter

Answer: Glucose and oxygen

Explanation:

The basic products of photosynthesis are glucose and oxygen. Photosynthesis is a process in which green plants produce there own food from carbon dioxide (CO2) and water (H2O) and in the presence of energy from the sun to produce glucose(food) and oxygen

After series of reaction, the overall equation for photosynthesis is

6CO2 + 6H20 + (sunlight ) → C6H12O6 + 6O2

Carbon dioxide + water + energy from sunlight to produces glucose and oxygen.

Answer is glucose and oxygen

Answer:

b) O2, because it has weaker intermolecular forces

Explanation:

The preassure is produced by the collisions of the gas molecules with the walls of its container.

When the intermolecular forces between the gas molecules increase, those molecules start to "slow down" by effect of the interactions. The collisions decrease in frequency and intensity producing a smaller preassure in the container.

Both O2 and Cl2 are non-polar gases and the only intermolecular forces they have are the London ones. Given that the O2 molecules are smaller than the Cl2, the last ones attract each other with more strengh.

Being all that said, the container with the oxygen is expected to have a higher preassure.

Oxygen molecule exert a higer pressure because it has weaker intermolecular forces.

The pressure is produced by the collisions of the gas molecules with the walls of its container. If the intermolecular forces between the gas molecules increase, those molecules start to slow down which leads to lowering of pressure on the wall of the container.

So we can conclude that decreasing intermolecular force, increase the pressure on the wall of the container.

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The balanced chemical equations for the reactions are

(a) K₂O + H₂O  → 2KOH

(b) P₂O₃ + 3H₂O → 2H₃PO₃

(c) Cr₂O₃ + 6HCl → 2CrCl₃ + 3H₂O

(d) SeO₂ + 2KOH → K₂SeO₃ + H₂O

From the question

We are to write a balanced equations for the given reactions

The reaction between potassium oxide and water gives potassium hydroxide

The balanced chemical equation for the reaction between potassium oxide and water is

K₂O + H₂O  → 2KOH

The reaction between diphosphorus trioxide with water gives phosphorous acid

The balanced chemical equation for the reaction between diphosphorus trioxide with water is

P₂O₃ + 3H₂O → 2H₃PO₃

The reaction between chromium(III) oxide with dilute hydrochloric acid produces chromium(II) chloride and water

The balanced chemical equation for the reaction between chromium(III) oxide with dilute hydrochloric acid is

Cr₂O₃ + 6HCl → 2CrCl₃ + 3H₂O

The reaction between selenium dioxide with aqueous potassium hydroxide gives potassium selenite and water

The balanced chemical equation for the reaction between selenium dioxide with aqueous potassium hydroxide

SeO₂ + 2KOH → K₂SeO₃ + H₂O

Hence, the balanced chemical equations for the reactions are

(a) K₂O + H₂O  → 2KOH

(b) P₂O₃ + 3H₂O → 2H₃PO₃

(c) Cr₂O₃ + 6HCl → 2CrCl₃ + 3H₂O

(d) SeO₂ + 2KOH → K₂SeO₃ + H₂O

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A balanced reaction equation contains the same number of atoms of each element on both sides of the reaction equation.

To write a balanced chemical reaction equation, the number of atoms of each element on the right hand side must be the same as the number of atoms of the same element on the left hand side.

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Explanation:

The given data is as follows.

     mass = 0.158 g,      volume = 100 ml

     Molarity = 1.0 M,      [tex]\Delta T[/tex] = [tex](32.8 - 25.6)^{o}C = 7.2^{o}C[/tex]  

The given reaction is as follows.

         [tex]Mg(s) + 2HCl(aq) \rightarrow MgCl_{2}(aq) + H_{2}(g)[/tex]  

So, moles of magnesium will be calculated as follows.

         No. of moles = [tex]\frac{mass}{\text{molar mass}}[/tex]                            

                               = [tex]\frac{0.158 g}{24.305 g/mol}[/tex]

                               = [tex]6.5 \times 10^{-3}[/tex]

                               = 0.0065 mol

Now, formula for heat released is as follows.

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

                  = [tex]\text{volume} \times \text{density} \times C \times \Delta T[/tex]

                  = [tex]100 ml \times 1.0 g/ml \times 4.184 \times 7.2^{o}C[/tex]

                  = 3010.32 J

Hence, heat of reaction will be calculated as follows.

            [tex]\Delta H_{rxn} = \frac{\text{-heat released}}{\text{moles of Mg}}[/tex]

                      = [tex]\frac{3010.32 J}{0.0065 mol}[/tex]

                      = -4.63 J/mol

or,                  = [tex]-463 \times 10^{-5} kJ/mol[/tex]                (as 1 kJ = 1000 J)

Thus, we can conclude that heat of given reaction is [tex]-463 \times 10^{-5}[/tex] kJ/mol.    

Answer:

24.44 atm

Explanation:

Considering that this gas mixture behaves like an ideal gass, and that all component gases are ideal gases, we can use:

PV=nRT

Then:

P=nRT/V

Where:

n= N° of moles

R= gas constant= 0.082 Lt*atm/K*mol

T= temperature (in Kelvin)

V = volume (in Lt)

Finally, statement says:

T = 25°C = 298 K

V = 5 Lt

n = 8 moles (for O₂)

P = [8 molx(0.082 Lt*atm/K*mol)x298 K]/5 Lt

P = 24.44 atm would be the pressure due to O₂ (partial pressure of the oxygen)

Explanation:

Endothermic animals are also known as warm-blooded, they have the capacity to regulate their body temperature independent of the environment. They have mechanisms to compensate if heat loss exceeds heat generation (shivers) Or if heat generation exceeds the heat loss (panting, sweating).

On the other hand, ectothermal animals are known as cold blooded organisms and depend on external sources, like sunlight, to regulate their body temperature, reptiles are ectothermals.

To determine if the animal of interest is endo or ectothermal you’ll have to consider that is a reptile, you’ll also observe that it consumes less food and finally it’ll have more difficulties to adapt to sudden temperature changes.

I hope you find this information useful and interesting! Good luck!

To determine if a tropical reptile is an endotherm or an ectotherm, observe its behavior in response to environmental temperature changes and test its metabolic rate and body temperature in various conditions.

To determine whether a large tropical reptile with a high and relatively stable body temperature is an endotherm or an ectotherm, one should observe the animal's behavior and response to changes in environmental temperature. If the reptile is an endotherm, it would generate its own heat through metabolic processes and maintain a stable body temperature regardless of the environment.

Endotherms demonstrate behaviors like shivering to generate heat when cold. Ectotherms, on the other hand, rely on the environment to regulate their body temperature, and this can be observed if the animal basks in the sun to warm up or seeks shade to cool down.

Most reptiles are known to be ectotherms and can exhibit behaviors such as brumation in response to cold temperatures. This state is similar to hibernation but doesn't rely on fat reserves. Therefore, if the reptile is less active in colder conditions and depends on external heat sources like sunlight, it is likely an ectotherm. However, if the reptile maintains an active metabolism and body temperature in colder environments without relying on external heat sources, it may be an endotherm.

Specific tests could include monitoring the reptile's body temperature in controlled environments with varying temperatures and measuring the metabolic rate under these conditions to see if there is internal heat production.

Answer:

[tex]\large \boxed {\text{24.3 g Cl}_{2}; 82.3 \%}[/tex]

Explanation:

We are given the masses of two reactants and asked to determine the mass of the product.

This looks like a limiting reactant problem.

1. Assemble the information

We will need a balanced equation with masses and molar masses, so let’s gather all the information in one place.  

MM:          86.94    36.46                 70.91

                MnO₂ + 4HCl ⟶ MnCl₂ + Cl₂ + 2H₂O

Mass/g:     86.0      50.0                  20.00

2. Calculate the moles of each reactant

[tex]\text{Moles of MnO}_{2} = \text{86.0 g MnO}_{2} \times \dfrac{\text{1 mol MnO}_{2}}{\text{86.94 g MnO}_{2}} = \text{0.9892 mol MnO}_{2}\\\\\text{Moles of HCl} = \text{50.0 g HCl} \times \dfrac{\text{1mol HCl }}{\text{36.46 g HCl }} = \text{1.371 mol HCl}[/tex]

3. Calculate the moles of Cl₂ formed from each reactant

From MnO₂:

[tex]\text{Moles of Cl$_{2}$} = \text{0.9892 mol MnO}_{2} \times \dfrac{\text{1 mol Cl$_{2}$}}{\text{1 mol MnO}_{2}} = \text{0.9892 mol Cl}_{2}[/tex]

From HCl:

[tex]\text{Moles of Cl$_{2}$} = \text{1.371 mol HCl} \times \dfrac{\text{1 mol Cl$_{2}$}}{\text{4 mol HCl}} = \text{0.3428 mol Cl}_{2}[/tex]

4. Identify the limiting reactant

The limiting reactant is HCl, because it forms fewer moles of Cl₂.

5. Calculate the theoretical yield of Cl₂

[tex]\text{ Mass of Cl$_{2}$} = \text{0.3428 mol Cl$_{2}$} \times \dfrac{\text{70.91 g Cl$_{2}$}}{\text{1 mol Cl$_{2}$}} = \textbf{24.3 g Cl}_\mathbf{{2}}\\\\\text{The theoretical yield is $\large \boxed{\textbf{24.3 g Cl}_\mathbf{{2}}}$}[/tex]

6. Calculate the percentage yield of Cl₂

[tex]\text{Percentage yield} = \dfrac{\text{Actual yield}}{\text{Theoretical yield}} \times 100 \, \%= \dfrac{\text{20.0 g}}{\text{24.3 g}} \times 100 \, \% = 82.3 \, \%\\\text{The percentage yield is $\large \boxed{\mathbf{82.3 \, \%}}$}[/tex]

In the substitution reaction of P-fluoroanisole with sulfur trioxide and sulfuric acid, the major product is P-fluoroanisole-sulfonic acid. This is an example of an Electrophilic Aromatic Substitution reaction.

The question involves a reaction of P-fluoroanisole with sulfur trioxide and sulfuric acid. This is a substitution reaction that happens under the influence of a strong acid like sulfuric acid.

In this reaction, sulfur trioxide (SO3) reacts with P-fluoroanisole to replace a hydrogen atom with a sulfonic acid group (–SO3H), forming P-fluoroanisole-sulfonic acid as the major product of the reaction. This type of reaction is a part of a broader class of reactions known as Electrophilic Aromatic Substitution reactions.

However, it is important to note that drawing the resulting chemical structure requires knowledge of organic chemistry and its conventions. Given the complexity of this information, a written description may not fully capture the details, and it is recommended to refer to a textbook or online resource where visualizations of such substitutions are possible.

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Answer:

The correct answer is a A molecule that brings about a cellular response to a signal.

Explanation:

Effector molecule is a small molecule that act as ligand to generate various cellular response by binding with target proteins(receptors).

  Effector molecules can regulate catalytic activity of enzymes,gene expression and various signaling processes that are very much important for proper functioning of the cell that the effect molecule target.

 Examples Allosteric effectors which can be either modulators or inhibitors depending on the nature of the effector molecule.

Answer : The molar mass of catalase is, [tex]2.40\times 10^5g/mol[/tex]

Explanation :

Formula used :

[tex]\pi =CRT\\\\\pi=\frac{w}{M\times V}RT[/tex]

where,

[tex]\pi[/tex] = osmotic pressure  = 0.745 torr = 0.000980 atm   (1 atm = 760 torr)

C = concentration

R = solution constant  = 0.0821 L.atm/mol.K

T = temperature  = [tex]27^oC=273+27=300K[/tex]

w = mass of catalase = 10.03 g

M = molar mass of catalase = ?

V = volume of solution  = 1.05 L

Now put all the given values in the above formula, we get:

[tex]0.000980atm=\frac{10.03g}{M\times 1.05L}\times (0.0821L.atm/mole.K)\times (300K)[/tex]

[tex]M=2.40\times 10^5g/mol[/tex]

Therefore, the molar mass of catalase is, [tex]2.40\times 10^5g/mol[/tex]

The molar mass of catalase is 326.99 g/mol.

To calculate the molar mass of catalase, we can use the formula for osmotic pressure, II = MRT, where M is the molar concentration of the solution, R is the ideal gas constant, and T is the temperature in Kelvin. Rearranging the formula to solve for M, we get M = II / RT. Plugging in the given values, II = 0.745 torr, R = 0.08206 L atm/mol K, and T = 27°C = 300 K, we can solve for M. M = (0.745 torr) / (0.08206 L atm/mol K) / 300 K = 0.0306 mol/L. Since the solution has a volume of 1.05 L and contains 10.03 g of catalase, we can use the formula M = mass / (volume * molar mass) to solve for the molar mass. Rearranging the formula to solve for molar mass, we get molar mass = mass / (volume * M). Plugging in the given values, molar mass = 10.03 g / (1.05 L * 0.0306 mol/L) = 326.99 g/mol.

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Answer:

The reaction that will occur are:

[tex]Mn(s)+NiCl_{2}(aq)->MnCl_{2}(aq)+Ni(s)[/tex]

Explanation:

In the activity series, a higher metal can displace a lower metal from its salt.

Mn is placed in a higher position than Ni in the activity series. Hence, it is able to displace Ni from its salt  [tex]NiCl_{2}[/tex].

All the other reactions are not feasible based on the activity series.

The link for the activity series I referred is given below:

https://studylib.net/doc/9281468/topic-9.1-activity-series

Final answer:

To determine the time taken for the decay rate to drop from 15 to 4 disintegrations per minute using a half-life of 5730 years, we calculate that it would take two half-lives, or 11460 years, for this reduction.

Explanation:

To calculate the time it would take for a piece of wood with an initial decay rate of 15 disintegrations per minute to decrease to 4 disintegrations per minute, we can use the concept of half-life, which for carbon-14 is 5730 years. By applying the decay constant and the exponential decay formula, we can find the time that corresponds to the rate of decay reaching 4 disintegrations per minute.

The relationship we use is N = N0e-λt, where N is the final number of disintegrations per minute, N0 is the initial number of disintegrations per minute, λ is the decay constant (which can be calculated using λ = 0.693 / 5730 years), and t is the time in years.

Using the concept of half-lives, we can determine that if after t years the disintegration rate is a quarter of its original rate (15 to 4 dpm), the wood must have gone through two half-lives, because each half-life reduces the rate of disintegration by half. Therefore, t would be equal to 2 × 5730 years = 11460 years.

Answer:

3.41 x10⁶ torr

Explanation:

To solve this problem we need to remember the equivalency:

1 torr = 133.322 Pa

Then we can proceed to convert 4.55×10⁸ Pa into torr. To do that we just need to multiply that value by a fraction number, putting the unit that we want to convert from in the denominator, and the value we want to convert to in the numerator:

4.55x10⁸ Pa * [tex]\frac{1torr}{133.322Pa} =[/tex] 3.41 x10⁶ torr

Answer: a contest in which you are eliminated if you fail to spell a word correctly.

Answer:

what are your statements?

Answer:

B

Explanation:

In counting the electron domains around the central atom in VSEPR theory, a Core level electron pair is not included.

Core level electron pair are the electrons other than the electrons in the valency shell of an atom. They in the proximity of the nucleus. They do not participate in the bonding.

VSEPR is the abbreviation for Valence shell Electron Pair repulsion theory. VESPR is a model for predicting molecular geometries based on the reduction in the electrostatic repulsion of the valence electrons of a molecule around a central atom.

Answer:

E) Scandium (Sc)

Hope this helps!

Here 'n' represents the principal quantum number. The element which has n=4 and has three electrons in its valence p orbital is scandium (Sc). The electronic configuration of 'Sc' is [Ar] 3d¹ 4s². The correct option is E.

The quantum number which represents the main energy level or shell in which the electron is present. It also determines the average distance of the orbital or electron from the nucleus. It is denoted by letter 'n' and can have any whole number values like 1, 2, 3, 4, etc. These values represent different energy levels.

The atomic number of the element scandium is 21 and its electronic configuration is [Ar] 3d¹ 4s². The total number of valence electrons present here is 3 and the energy level is 4.

The element scandium is a transition element and it has the highest melting point and low density.

Thus the correct option is E.

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Answer:

1122.5 mL

Explanation:

In the first scenario, by Dalton's law, the total pressure is the sum of the partial pressures of the components. So, the partial pressure of the gas is:

P1 = Ptotal - Pwater = 742 - 36 = 706 torr

By the ideal gas law, the change in a state of a gas can be calculated by:

P1*V1/T1 = P2*V2/T2

Where P is the pressure, V is the volume, T is the temperature in K, 1 is the state 1, and 2 the state 2.

P1 = 706 torr, V1 = 1350 mL, T1 = 32ºC + 273 = 305K

P2 = 760 torr, T2 = 0ºC + 273 = 273 K

706*1350/305 = 760*V2/273

760V2/273 = 3124.92

760V2 = 853102.62

V2 = 1122.5 mL

Answer:

1)50.007 grams of an isotope will left after 10 years.

1)25.007 grams of an isotope will left after 20 years.

3) 23 half-lives will occur in 40 years.

Explanation:

Formula used :

[tex]N=N_o\times e^{-\lambda t}\\\\\lambda =\frac{0.693}{t_{\frac{1}{2}}}[/tex]

where,

[tex]N_o[/tex] = initial mass of isotope

N = mass of the parent isotope left after the time, (t)

[tex]t_{\frac{1}{2}}[/tex] = half life of the isotope

[tex]\lambda[/tex] = rate constant

We have:

[tex][N_o]=100 g[/tex]

[tex]t_{1/2}=10 years[/tex]

[tex]\lambda =\frac{0.693}{t_{1/2}}=\frac{0.693}{10 year}=0.0693 year^{-1}[/tex]

1) mass of isotope left after 10 years:

t = 10 years

[tex]N=N_o\times e^{-\lambda t}[/tex]

[tex]N=100g\times e^{-0.0693 year^{-1}\times 10}=50.007 g[/tex]

2) mass of isotope left after 20 years:

t = 20 years

[tex]N=N_o\times e^{-\lambda t}[/tex]

[tex]N=100g\times e^{-0.0693 year^{-1}\times 20}=25.007 g[/tex]

3) Half-lives will occur in 40 years

Mass of isotope left after 40 years:

t = 40 years

[tex]N=N_o\times e^{-\lambda t}[/tex]

[tex]N=100g\times e^{-0.0693 year^{-1}\times 40}=6.2537 g[/tex]

number of half lives = n

[tex]N=\frac{N_o}{2^n}[/tex]

[tex]6.2537 g=\frac{100 g}{2^n}[/tex]

[tex]n\ln 2=\frac{100 g}{6.2537 g}[/tex]

n = 23

23 half-lives will occur in 40 years.

Answer:

These properties are basically the inverse of each other.

Explanation:

        Ionization enthalpy, is the energy required to remove an electron from                        an atom.

Ionization energy is the reverse of electronegativity. Ionization energy tells you how easily or how attracted you are to your own electrons so how hard is for somebody to steal them so it's about how much you hold on to your electrons.

Electronegativity is a chemical property and it's about how much can I steal somebody else's electrons. They are like two sides of the same coin. Electronegativity is affected by both its atomic number (number of protons in the core of an atom) and how far the valence electrons (the outermost electrons) are from the core.

Answer:

A solution that is 0.100 M (i.e. 33,12 g of lead (II) nitrate in 1.00 L of water) yields the desired total dissolved ions concentration.

Explanation:

The molecular formula of lead (II) nitrate is Pb(NO[tex]Pb(NO_{3} )_{2}[/tex] and its molecular mass is 331,2 g/mol.

In disolution, the equilibrium will look like this:

Pb(NO[tex]Pb(NO_{3} )_{2}[/tex] -> [tex]Pb^{2+}  + 2(NO)_{3} ^{-1}[/tex]

The equation above means that, one mol of lead (II) nitrate dissolved in 1L will yield one mol of Pb ions and 2 moles of NO3 ions, i.e. 3 moles total.

If we dissolve 0.100 moles of lead (II) nitrate in 1.00 L of water, the stoichiometry of the disolution states that in turn, it will yield 0.100 of Pb ions and 0.200 moles of NO3 ions, i.e. 0.300 M in total dissolved ions.

331,2 g/mol * 0.100 mol/L * 1 L = 33,12 grams of the compound are required.

Answer:

AS we move from bottom to top on periodic table shielding decreased.

Explanation:

As we move from left to right across the periodic table the number of valance electrons in an atom increase. The atomic size tend to decrease in same period of periodic table because the electrons are added with in the same shell. When the electron are added, at the same time protons are also added in the nucleus. The positive charge is going to increase and this charge is greater in effect than the charge of electrons. This effect lead to the greater nuclear attraction.

As we move down the group atomic radii increased with increase of atomic number. The addition of electron in next level cause the atomic radii to increased. The hold of nucleus on valance shell become weaker because of shielding of electrons thus size of atom increased.

As we move from bottom to top the energy level decreased because of decreased in electron thus shielding decreased and atomic size also decreased.

a. [tex]\[ n \approx 0.0711 \, \text{moles} \][/tex]

b. the percentage by mass of magnesium carbonate in the mixture is [tex]\(100\%\)[/tex].

a) To find the total number of moles of carbon dioxide formed, we first need to calculate the number of moles of carbon dioxide using the ideal gas law. Then, we can use stoichiometry to relate the moles of carbon dioxide to the moles of magnesium carbonate reacted.

Given:

- Mass of mixture = 6.53 g

- Volume of carbon dioxide = 1.73 L

- Temperature = 26 °C = 26 + 273.15 K = 299.15 K

- Pressure = 745 torr

First, let's calculate the number of moles of carbon dioxide using the ideal gas law:

[tex]\[ PV = nRT \][/tex]

Where:

- P is the pressure in atm (convert 745 torr to atm),

- V is the volume in liters,

- n is the number of moles,

- R is the gas constant (0.0821 atm L/mol K),

- T is the temperature in Kelvin.

[tex]\[ P = \frac{745 \, \text{torr}}{760 \, \text{torr/atm}} = 0.980 \, \text{atm} \][/tex]

[tex]\[ n = \frac{PV}{RT} = \frac{(0.980 \, \text{atm})(1.73 \, \text{L})}{(0.0821 \, \text{atm} \cdot \text{L/mol} \cdot \text{K})(299.15 \, \text{K})} \][/tex]

[tex]\[ n \approx 0.0711 \, \text{moles} \][/tex]

b) Now, let's use stoichiometry to relate the moles of carbon dioxide formed to the moles of magnesium carbonate reacted. From the balanced chemical equation:

[tex]\[ \text{MgCO}_3 + 2\text{HCl} \rightarrow \text{MgCl}_2 + \text{CO}_2 + \text{H}_2\text{O} \][/tex]

We see that 1 mole of magnesium carbonate [tex](\( \text{MgCO}_3 \))[/tex] reacts to produce 1 mole of carbon dioxide [tex](\( \text{CO}_2 \))[/tex].

So, the moles of magnesium carbonate reacted is also [tex]\(0.0711 \, \text{moles}\)[/tex].

Now, let's use the mass of magnesium carbonate and the moles reacted to find the percentage:

[tex]\[ \text{mass of MgCO}_3 = 6.53 \, \text{g} \][/tex]

[tex]\[ \text{percentage of MgCO}_3 = \frac{\text{mass of MgCO}_3}{\text{total mass of mixture}} \times 100\% \][/tex]

[tex]\[ \text{percentage of MgCO}_3 = \frac{6.53 \, \text{g}}{\text{total mass of mixture}} \times 100\% \][/tex]

[tex]\[ \text{percentage of MgCO}_3 = \frac{6.53 \, \text{g}}{6.53 \, \text{g}} \times 100\% \][/tex]

[tex]\[ \text{percentage of MgCO}_3 = 100\% \][/tex]

Therefore, the percentage by mass of magnesium carbonate in the mixture is [tex]\(100\%\)[/tex].

(a)The total number of moles of carbon dioxide that forms is [tex]\( 0.0691 \)[/tex] moles. (b) The percentage by mass of magnesium carbonate in the mixture is [tex]\( 30.9\% \)[/tex].

To solve this problem, we need to use the given data to find the total number of moles of carbon dioxide produced and then use stoichiometry to determine the percentage by mass of magnesium carbonate in the mixture.

Part (a): Calculate the total number of moles of carbon dioxide

First, we use the ideal gas law to find the number of moles of carbon dioxide gas produced. The ideal gas law is given by:

[tex]\[ PV = nRT \][/tex]

Where:

- P is the pressure in atmospheres (atm)

- V is the volume in liters (L)

- n is the number of moles

- R is the ideal gas constant [tex](0.0821 LatmK\(^{-1}\)mol\(^{-1}\))[/tex]

- T is the temperature in Kelvin (K)

We need to convert the given pressure from torr to atm and the temperature from Celsius to Kelvin.

[tex]\[ P = 745 \, \text{torr} \times \frac{1 \, \text{atm}}{760 \, \text{torr}} = 0.980 \, \text{atm} \][/tex]

[tex]\[ T = 26^\circ \text{C} + 273 = 299 \, \text{K} \][/tex]

Given [tex]\( V = 1.73 \, \text{L} \)[/tex], we can solve for n:

[tex]\[ n = \frac{PV}{RT} \][/tex]

[tex]\[ n = \frac{(0.980 \, \text{atm})(1.73 \, \text{L})}{(0.0821 \, \text{LatmK}^{-1}\text{mol}^{-1})(299 \, \text{K})} \][/tex]

[tex]\[ n = \frac{1.6954 \, \text{atmL}}{24.5479 \, \text{LatmK}^{-1}\text{mol}^{-1}} \][/tex]

[tex]\[ n = 0.0691 \, \text{mol} \][/tex]

Part (b): Calculate the percentage by mass of magnesium carbonate in the mixture

The reactions are:

[tex]\[ \text{MgCO}_3 (s) + 2\text{HCl} (aq) \rightarrow \text{MgCl}_2 (aq) + \text{CO}_2 (g) + \text{H}_2\text{O} (l) \][/tex]

[tex]\[ \text{CaCO}_3 (s) + 2\text{HCl} (aq) \rightarrow \text{CaCl}_2 (aq) + \text{CO}_2 (g) + \text{H}_2\text{O} (l) \][/tex]

Both reactions produce [tex]\( \text{CO}_2 \)[/tex] gas in a 1:1 molar ratio with their respective carbonates.

Let x be the mass of [tex]\( \text{MgCO}_3 \)[/tex] and y be the mass of [tex]\( \text{CaCO}_3 \)[/tex]. We know:

[tex]\[ x + y = 6.53 \, \text{g} \][/tex]

The moles of [tex]\( \text{CO}_2 \)[/tex] produced from each carbonate are:

[tex]\[ \frac{x}{84.31} \, \text{mol} \, (\text{for} \, \text{MgCO}_3) \][/tex]

[tex]\[ \frac{y}{100.09} \, \text{mol} \, (\text{for} \, \text{CaCO}_3) \][/tex]

The total moles of [tex]\( \text{CO}_2 \)[/tex] is the sum of the moles produced by each carbonate:

[tex]\[ \frac{x}{84.31} + \frac{y}{100.09} = 0.0691 \][/tex]

We have two equations:

1. [tex]\( x + y = 6.53 \)[/tex]

2. [tex]\( \frac{x}{84.31} + \frac{y}{100.09} = 0.0691 \)[/tex]

Solve these equations simultaneously. First, solve equation 1 for y:

[tex]\[ y = 6.53 - x \][/tex]

Substitute into equation 2:

[tex]\[ \frac{x}{84.31} + \frac{6.53 - x}{100.09} = 0.0691 \][/tex]

Multiply through by [tex]\( 84.31 \times 100.09 \)[/tex] to clear the denominators:

[tex]\[ 100.09x + 84.31(6.53 - x) = 0.0691 \times 84.31 \times 100.09 \][/tex]

[tex]\[ 100.09x + 84.31 \times 6.53 - 84.31x = 582.5 \][/tex]

Combine like terms:

[tex]\[ 15.78x + 550.56 = 582.5 \][/tex]

Solve for [tex]\( x \)[/tex]:

[tex]\[ 15.78x = 31.94 \][/tex]

[tex]\[ x = 2.02 \, \text{g} \][/tex]

Now, calculate [tex]\( y \)[/tex]:

[tex]\[ y = 6.53 - 2.02 = 4.51 \, \text{g} \][/tex]

Finally, calculate the percentage by mass of magnesium carbonate in the mixture:

[tex]\[ \text{Percentage of MgCO}_3 = \left( \frac{2.02 \, \text{g}}{6.53 \, \text{g}} \right) \times 100\% \][/tex]

[tex]\[ \text{Percentage of MgCO}_3 = 30.9\% \][/tex]

The density of sulfur hexafluoride gas at 15°C and 1 atm is approximately 6.52 g/L. This calculation is done using the ideal gas law and assuming sulfur hexafluoride behaves as an ideal gas under these conditions.

To calculate the density of sulfur hexafluoride gas at 15° C and 1 atm, we can use the ideal gas law. The ideal gas law states PV=nRT, where P is pressure, V is volume, n is the number of moles, R is the gas constant, and T is temperature in Kelvin. First, convert 15°C to Kelvin by adding 273.15, which gives 288.15 K. The molar mass of sulfur hexafluoride is around 146.06 g/mol. R is 0.0821 L·atm/(K·mol), and P is 1 atm. So, you can solve for density (d) which is mass/volume, or essentially (n*M)/V.

By rearranging the ideal gas law, V=nRT/P, and substituting into the density equation, we get d=P*M/(R*T) = (1 atm * 146.06g/mol) / (0.0821 L·atm/mol·K * 288.15 K) = approximately 6.52 g/L to three significant figures.

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The density of sulfur hexafluoride gas at 15°C and 1atm, assuming it behaves as an ideal gas, is approximately 6.52g/L.

The question asks for the density of sulfur hexafluoride gas at exactly 15°C and 1atm. We can calculate this by using the ideal gas law which states PV=nRT, where P is pressure, V is volume, n is number of moles, R is the ideal gas constant, and T is temperature in Kelvin. We rearrange this to n/V=P/RT which gives us the molar density. From the molar density, we can easily determine the density by multiplying it by the molar mass. To convert 15°C to Kelvin, we add 273.15 to give us 288.15K. This gives us a molar density of 1atm/(0.0821 L.atm/K.mol * 288.15K) = 0.0446 mol/L. The molar mass of sulfur hexafluoride (SF6) is about 146g/mol. Multiply this by the molar volume gives us a density of 6.52g/L, to three significant figures.

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Answer:

Sp^3d

Explanation:

Sp^3d hybridization invovles ther combination of 1s, 3p and 1d orbital  to result in the production of sp3d orbital in which three lobes are oriented towards the corners of a triangle and the other lie perpendicular to them to minimize the repulsions. An important and relatively common type of this hybridization is found in Clf3.

In ClF3  (Chlorine Trifluoride), The central atom Cl needs three unpaired electrons to bond with three F-atoms. ClF3 consist of 3 bond-pairs and 2 lone-pairs. Here, one of the paired electrons of Cl in the 3p subshell remains as a lone pair or unpaired. During hybridisation, one 3s, three 3p and one of the 3d orbitals participate in the process which leads to the formation of five sp3d hybrid orbitals. Here two hybrid orbitals will contain a pair of electrons and three hybrids will contain unpaired electrons which will again overlap with the 2p orbital of F to form single bonds.

Answer:

The sample of indium, has an average atomic mass of 114.818 amu

Explanation:

Step 1: Data given

Indium is made of 95.7 % 115In and 4.3% 113In

113In has an atomic mass of 112.904 amu

115In has an atomic mass of 114.904amu

Step 2: Calculate the average atomic mass

The average atomic mass = X

0.957* 114.904 + 0.043* 112.904 = X

109.963 + 4.855 = X

X = 114.818

The sample of indium, has an average atomic mass of 114.818 amu

Rounded to the nearest tenth, the average atomic mass of indium is approximately 114.9 amu. So, the correct options provided is:C

To calculate the average atomic mass of a sample of indium, you can use the following formula:

Average Atomic Mass = (% abundance of isotope 1 × atomic mass of isotope 1) + (% abundance of isotope 2 × atomic mass of isotope 2) + ...

In this case, you have two isotopes of indium, 115In and 113In, with the following information:

- % abundance of 115In = 95.7%

- % abundance of 113In = 4.3%

- Atomic mass of 115In = 114.904 amu

- Atomic mass of 113In = 112.904 amu

Plug these values into the formula:

Average Atomic Mass = (0.957 × 114.904 amu) + (0.043 × 112.904 amu)

Calculate each part:

Average Atomic Mass = (110.009648 amu) + (4.855672 amu)

Now, add these values together:

Average Atomic Mass = 114.86532 amu

Rounded to the nearest tenth, the average atomic mass of indium is approximately 114.9 amu. So, the correct options provided is:C

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Answer: C) 0.020 m

Explanation:

Molality of a solution is defined as the number of moles of solute dissolved per kg of the solvent.

[tex]Molality=\frac{n\times 1000}{W_s}[/tex]

where,

n = moles of solute  

[tex]W_s[/tex] = weight of solvent in g  

Mole fraction of [tex]CO_2[/tex] is = [tex]3.6\times 10^{-4}[/tex] i.e.[tex]3.6\times 10^{-4}[/tex]  moles of [tex]CO_2[/tex] is present in 1 mole of solution.

Moles of solute [tex](CO_2)[/tex] = [tex]3.6\times 10^{-4}[/tex]

moles of solvent (water) = 1 - [tex]3.6\times 10^{-4}[/tex] = 0.99

weight of solvent =[tex]moles\times {\text {Molar mass}}=0.99\times 18=17.82g[/tex]

Molality =[tex]\frac{3.6\times 10^{-4}\times 1000}{17.82g}=0.020[/tex]

Thus  approximate molality of [tex]CO_2[/tex] in this solution is 0.020 m

Because we do not know if the value of the change in free energy will be positive or negative.  It is not possible to determine without more information.

Option E

Heat Energy is simply defined as an energy that is moved from one body to another using the concept of difference temperature changes.

Generally, the equation for the change in enthalpy  is mathematically given as

ΔG = ΔH - T ΔS

In conclusion, we do not know the magnitude of positive or positive value and hence, we do not know if the value of the change in free energy will be positive or negative.

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Answer:

1. False

2. False

3. True

Explanation:

Identify all statements are true

1. zeroth order rate constants have units of inverse seconds. FALSE.

The rate law for a zeroth order rate reaction has the following form:

r = k

where,

r is the rate of the reaction

k is the rate constant

k has the same units than r, that is, M . s⁻¹.

2. first order reactions have half lives that are dependent on time . FALSE.

Half-life (t1/2) of a first order reaction can be calculated using the following expression.

[tex]t_{1/2}=\frac{ln2}{k}[/tex]

As we can see, half-life does not depend on time.

3. catalysts cannot appear in the rate law for a reaction. TRUE.

The rate law has the following form:

r = k . [A]ᵃ . [B]ᵇ

where,

[A] and [B] are the concentrations of the reactants

a and b are the reaction orders

Catalysts do not appear in the rate law.

Explanation:

Hence, molecules of a solid have low energy.

This means that in kinetic molecular theory molecules of a gas are in rapid, random motion.

                 [tex]P\propto \frac{1}{V}[/tex]     (at constant temperature and number of moles)

                 [tex]V\propto T[/tex]     (at constant pressure and number of moles)

Thus, we can conclude that the given items are matched as follows.

1.  gas - continuously in motion.

2. solid matter - low energy.

3. Kinetic Molecular Theory - rapid, random motion.

4. Boyle's Law - relates pressure and volume.

5. Charles's Law - relates volume and temperature.