The heat of vaporization of water at 100°c is 40.66 kj/mol. Calculate the quantity of heat that is absorbed/released when 5.00 g of steam condenses to liquid water at 100°c.

Answer:

Metalloids have properties of metals and non-metals.

Explanation:

A metalloid is a type of chemical element which has properties in between, or that are a mixture of, those of metals and nonmetals.

The six commonly recognized metalloids are boron, silicon, germanium, arsenic, antimony, and tellurium.

Metalloids have a metallic appearance, but they are brittle (nonmetal property) and only fair conductors of electricity. Chemically, they behave mostly as nonmetals. They can form alloys with metal.

Most of their other physical properties and chemical properties are intermediate in nature.

That means Metalloids  have properties of both metal and nonmetal.

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:

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:

1 billion molecules O₂

Explanation:

From my research, a human red blood cell contains approximately 270 million hemoglobin molecules.    

A hemoglobin molecule contains four heme groups, each of which has an iron ion forming a coordination complex that carries every dioxygen molecule. Therefore for each hemoglobin molecule, we will have 4 dioxygen molecules. The heme groups are responsible for the transport of every dioxygen and other diatomic gases.                    

Hence, the number of O₂ molecules in a red blood cell saturated with 100% will be:                

[tex] \frac{270 \cdot10^{6} hemoglobine molecules}{1 red blood cell} \cdot \frac{4 heme group}{1 hemoglobine molecule} \cdot \frac{1 O_{2} molecules}{1 heme group} = 1 \cdot 10^{9} O_{2} \frac{molecules}{red blood cell} [/tex]

So, the correct answer is 1 billion of O₂ molecules.  

Have a nice day!

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:

-2312 kJ

Explanation:

By the Hess Law, when a reaction is formed by several steps, the enthalpy change (ΔH) of the global reaction can be calculated by the sum of the enthalpy changes of the steps reactions.

Besides that, when a reaction is multiplied by a constant, ΔH must be multiplied by the same constant, and when the reaction is inverted, the signal of ΔH must be inverted.

B₂O₃(s) + 3H₂O(g) → 3O₂(g) + B₂H₆(g) ΔH = +2035 kJ (multiply by 2 and inverted)

2B(s) + 3H₂(g) → B₂H₆(g) ΔH = +36 kJ (multiply by 2)

H₂(g) + 1/2O₂(g) → H₂O(l) ΔH = -285 kJ (multiply by 6 and inverted)

H₂O(l) → H₂O(g) ΔH = +4 kJ (multiply by 6 and inverted)

-------------------------------------------------------

6O₂(g) + 2B₂H₆(g) → 2B₂O₃(s) + 6H₂O(g) ΔH = -4070 kJ

4B(s) + 6H₂(g) → 2B₂H₆(g) ΔH = +72 kJ

6H₂O(l) → 6H₂(g) + 3O₂(g) ΔH = +1710 kJ

6H₂O(g) → 6H₂O(l) ΔH = -24 kJ

Summing the equation, simplifying the equal terms in both sides:

4B(s) + 3O₂(g) → 2B₂O₃(s)

ΔH = -4070 + 72 + 1710 - 24

ΔH = -2312 kJ

Answer:

The answer to your question is P = 17.73 atm

Explanation:

V = 250 ml = 0.25 liters (L)

mass = 3.4 g

Temperature = 45°C = 45 + 273 = 318°K

Pressure = ?

R = 0.082 atm L / mol°K

Atomic mass Ne = 20.18 g

Process

1.- Calcule moles of Neon

                           20.18 g of Ne ------------------ 1 mol

                              3.4 g of Ne -------------------  x

                                            x = (3.4 x 1)/20.18

                                            x = 0.17 moles of Neon

2.- Find pressure

Ideal gases formula             PV = nRT

                                            [tex]P = \frac{nRT}{V}[/tex]

                                            [tex]P = \frac{(0.17)(0.082)(318)}{0.25}[/tex]

                                            [tex]P = \frac{4.433}{0.25}[/tex]        

                                                  P = 17.73 atm                                                                          

To calculate the pressure of the neon gas inside the flask, you can use the ideal gas law equation PV = nRT.

To calculate the pressure of the neon gas, we can use the ideal gas law equation: PV = nRT. The given volume is 250 mL, which needs to be converted to liters (1 L = 1000 mL), so the volume is 0.25 L. The given temperature is 45°C, which needs to be converted to Kelvin by adding 273, so the temperature is 45 + 273 = 318 K. The given mass is 3.4 g, and we can use the molar mass of neon (20.18 g/mol) to convert grams to moles: 3.4 g / 20.18 g/mol = 0.1682 mol.

Now we can substitute the values into the ideal gas law equation:

P(0.25 L) = (0.1682 mol)(0.0821 L·atm/mol·K)(318 K)

Solving for P gives:

P = (0.1682 mol)(0.0821 L·atm/mol·K)(318 K) / 0.25 L = 7.22 atm

Therefore, the pressure of the neon gas inside the flask is 7.22 atm.

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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:C

Explanation:

A process is called spontaneous if the process takes place on its own without the intervention of external factors.

The spontaneous processes are generally quick with observable rates of reaction.But processes can be spontaneous even with negligible rates of reaction.

Spontaneity of reactions depend on temperature.

This is because spontaneity is measured by gibbs energy or enthalpy.

Both of these measures are dependent on temperature.

So,temperature affects spontaneity.

Answer:

The halogens located in group 17

Explanation:

The change in energy of a neutral atom when an electron is added is known as electron affinity. Fluorine (F), chlorine (Cl), bromine (Br), iodine (I), and astatine (At) are known as halogens and conform the group 17 of the periodic table, the group that presents higher electron affinity. This characteristic happens because of their atomic structures, it is easier for them to gain electrons because they are closer to form octets.

Here I present you halogens' electron affinities values:

The use of negative signs is to indicate the RELEASE of energy.

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

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:answers are in the explanation

Explanation:

(a). pH less than 7 between 1 - 3.5 are strong acid, and between 4.5-6.9 weak acid.

pH greater than 7; between 10-14 is a strong base, and between 7.1 - 9, it is weakly basic.

(b). Equation of reaction;

HBr + KOH ---------> KBr + H2O

One mole of HBr reacts with one mole of KOH to give one Mole of KBr and one mole of H2O

Calculating the mmol, we have;

mmol KOH = 28.0 ml × 0.50 M

mmol KOH= 14 mmol

mmol of HBr= 56 ml × 0.25M

mmol of HBr= 14 mmol

Both HBr and KOH are used up in the reaction, which leaves only the product,KBr and H2O.

The pH here is greater than 7

(C). [NH4^+] = 0.20 mol L^-1 × 50 ml. L^-1 ÷ 50 mL + 50mL

= 0.10 M

Ka=Kw/kb

10^-14/ 1.8× 10^-5

Ka= 5.56 ×10^-10

Therefore, ka= x^2 / 0.20

5.56e-10 = x^2/0.20

x= (0.20 × 5.56e-10)^2

x= 1.05 × 10^-5

pH = -log [H+]

pH= - log[1.05 × 10^-5]

pH = 4.98

Acidic(less than 7)

(c). 0.5 × 20/40

= 0.25 M

Ka= Kw/kb

kb= 10^-14/1.8× 10^-5

Kb = 5.56×10^-10

x= (5.56×10^-10 × 0.5)^2

x= 1.667×10^-5 M

pH will be basic

Final answer:

In titrations, the pH at the equivalence point depends on the type of titration. For part b, the pH after adding 28.0 mL of KOH can be calculated by determining the moles of HBr and KOH and using the stoichiometry of the reaction. Similarly, for part c and part d, we can calculate the pH by determining the moles of the reactants and using stoichiometry.

Explanation:

For the first part of the question, the pH at the equivalence point depends on the type of titration. In a titration between a weak acid and a strong base, the pH at the equivalence point will be greater than 7. In a titration between a strong acid and a strong base, the pH at the equivalence point will be equal to 7. In a titration between a weak base and a strong acid, the pH at the equivalence point will be less than 7.

For part b, we have a titration between a strong acid (HBr) and a strong base (KOH). The initial volume of HBr is 56.0 mL and the volume of KOH added is 28.0 mL. To calculate the pH after the addition of 28.0 mL of KOH, we need to determine the moles of HBr and KOH and use the stoichiometry of the reaction to find the concentration of the resulting solution, which will give us the pH.

For part c, we have a titration between a weak base (NH3) and a strong acid (HNO3). The initial volume of NH3 is 50.0 mL and the volume of HNO3 added is 50.0 mL. Similarly to part b, we need to determine the moles of NH3 and HNO3 and use the stoichiometry of the reaction to find the concentration of the resulting solution, which will give us the pH.

For part d, we have a titration between a weak acid (CH3COOH) and a strong base (NaOH). The initial volume of CH3COOH is 30.0 mL and the volume of NaOH added is also 30.0 mL. Again, we need to determine the moles of CH3COOH and NaOH and use the stoichiometry of the reaction to find the concentration of the resulting solution, which will give us the pH at this point.

Final answer:

Nonmetals generally gain electrons to form anions in reactions with metals, creating ionic compounds, or they share electrons in reactions with each other to form molecular compounds.

Explanation:

When nonmetals react with other elements, they typically undergo a process where they gain electrons to form anions, or negative ions, as they combine with metals to produce ionic compounds. In reactions with each other, nonmetals often form molecular compounds with covalent bonding, where electrons are shared rather than transferred. The most reactive nonmetals, like fluorine, have a particularly high electron affinity and can take part in vigorous redox reactions, changing their oxidation states. Considering the diverse group of nonmetals, these elements can form a variety of different compounds, both ionic and covalent, depending on the elements they react with and the environmental conditions.

Answer:

61.76%

Explanation:

Percent yield = Actual yield/Theoeetical yield × 100%

We need to calculate the theoretical yield for this reaction.

Firstly, we need to know the number of moles of hydrogen reacted = mass of hydrogen ÷ molar mass = 450 ÷ 2 = 225 moles

From the reaction, we can see that 3 moles of hydrogen yielded 2 moles of ammonia. Hence 225 moles of hydrogen will yield:

(225 × 2) ÷ 3 = 150 moles

The actual mass of ammonia yielded = number of moles × molar mass = 150 × 17 = 2550g

Actual yield = 1575g

Theoretical yield = 2550g

Percentage yield = 1575/2550 × 100% = 61.76%

The correct option is (c). The percent yield of this reaction is 62.1%

To calculate the percent yield of the reaction, we need to compare the actual yield to the theoretical yield. The balanced chemical equation for the production of ammonia is:

[tex]\[ N_2 + 3H_2 \rightarrow 2NH_3 \][/tex]

 From the equation, we see that 1 mole of nitrogen reacts with 3 moles of hydrogen to produce 2 moles of ammonia. The molar masses of[tex]\( N_2 \), \( H_2 \), and \( NH_3 \)[/tex] are approximately 28.02 g/mol, 2.016 g/mol, and 17.031 g/mol, respectively.

 First, we calculate the number of moles of hydrogen gas that reacted:

[tex]\[ \text{moles of } H_2 = \frac{\text{mass of } H_2}{\text{molar mass of } H_2} = \frac{450 \text{ g}}{2.016 \text{ g/mol}} \approx 223.2 \text{ moles} \][/tex]

Since the stoichiometric ratio of[tex]\( H_2 \) to \( NH_3 \)[/tex] is 3:2, we can find the theoretical yield of ammonia:

[tex]\[ \text{moles of } NH_3 \text{ produced theoretically} = \frac{2}{3} \times 223.2 \text{ moles} \approx 148.8 \text{ moles} \][/tex]

 Now, we calculate the theoretical mass of ammonia that could be produced:

[tex]\[ \text{mass of } NH_3 \text{ theoretically} \approx 148.8 \text{ moles} \times 17.031 \text{ g/mol} \approx 2533.5 \text{ g} \][/tex]

The actual yield is given as 1575 g of ammonia. The percent yield is calculated by:

 [tex]\[ \text{Percent Yield} = \left( \frac{\text{actual yield}}{\text{theoretical yield}} \right) \times 100\% \][/tex]

[tex]\[ \text{Percent Yield} = \left( \frac{1575 \text{ g}}{2533.5 \text{ g}} \right) \times 100\% \approx 62.1\% \][/tex]

 Therefore, the percent yield of this reaction is 62.1%.

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:

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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Answer:

Atomic radius of sodium = 227 pm

Atomic radius of potassium = 280 pm

Explanation:

Atomic radii trend along group:

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.

Consider the example of sodium and potassium.

Sodium is present above the potassium with in same group i.e, group one.

The atomic number of sodium is 11 and potassium 19.

So potassium will have larger atomic radius as compared to sodium.

Atomic radius of sodium = 227 pm

Atomic radius of potassium = 280 pm

Answer:

Anthracite

Explanation:

Anthracite variety of coal is the coal with the highest energy available per unit burned. Anthracite is hard, brittle, lustrous coal . They also often referred as hard coal. Anthracite has a high percentage of carbon and low percentage of volatile matter. Other varieties of coal  are peat, lignite and bituminous.  

Answer:

B) A series of chemical reactions where enzymes work one at a time to convert a reactant into intermediates and the intermediates into a final product.

Explanation:

In our cells there are multiple chemical reactions happening all the time to convert the  elements we ingest (food, water, air) into the products the organism needs.

This processes can be very complex and offen involves various reactions (steps). Also, many of this reactions need to by catalized by enzymes in order to happen.

Answer: B

Explanation:

A metabolic pathway is series of chemical reactions that are well organized, well co-ordinated, and purposeful manner; all these reactions are collectively called as metabolism.

Metabolic Pathways this is a series of biochemical reaction where enzymes work one at a time to convert a reactant into intermediates and the intermediates into a final product. The product of reactions serve as an a reactants for the next enzymatic reaction.

The reactants, intermediates, products of an enzymatic reaction are known as metabolites.

Types of Metabolic pathway

1. Catabolic

2. Anabolic

3. Amphibolic

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:

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.

Answer:

2Mg  +  O2 →  2MgO

Explanation:

In all conbustion you should know, that reactans are an specific compound and O2, so the products must be CO2 and H2O, or in this case, the corresponding oxide.

The balanced equation for the combustion of Magnesium is 2Mg + O2 -> 2MgO. It's a redox reaction where Magnesium is oxidised and Oxygen is reduced.

The combustion of Magnesium in Oxygen is a well-known reaction in Chemistry. The balanced equation for this reaction is 2Mg + O2 -> 2MgO. In this reaction, two moles of Magnesium react with one mole of Oxygen gas to produce two moles of Magnesium Oxide. This is a redox reaction where Magnesium gets oxidised and Oxygen gets reduced.

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Answer:

Explanation:

It is possible to obtain ΔG of a reaction using:

ΔG = ΔG° + RT ln Q

Where:

ΔG° is standard change in Gibbs free energy (-32,8 kJ/mol)

R is gas constant (8,314472x10⁻³ kJ/molK)

T is temperature (298 K)

And Q is reaction quotient. For the reaction:

N₂(g) + 3H₂(g) → 2NH₃(g)

Q is: [NH₃]² /[H₂]³[N₂]

Replacing:

ΔG = -32,8kJ/mol + 8,314472x10⁻³ kJ/molK×298 K ln [0,6]² /[0,1]³[0,1]

ΔG = -12,5 kJ/mol

I hope it helps!

The change in Gibbs free energy (ΔG) for the reaction N2(g) + 3H2(g) --> 2NH3(g) at 298 K with partial pressures Pn2= 0.1 atm, Ph2=.1 atm and PNH3=.6 atm, is about -29.6 kJ/mol.

To calculate the Gibbs free energy change under nonstandard conditions, we should use the equation ΔG = ΔG° + RTlnQ, where ΔG° is the standard change in Gibbs free energy, R is the universal gas constant (in this case, 8.314 J/(mol·K)), T is the temperature in Kelvin, and Q is the reaction quotient.

Given that ΔG° = -32.8 kJ/mol, R = 8.314 J/(mol·K), and T = 298 K, we find the reaction quotient Q by substituting the partial pressures of the substances into the balanced chemical equation: Q = [NH3]² / ([N₂][H₂]³) = (0.6)² / ((0.1)(0.1)³) = 360.  

Then, you substitute these values into the ΔG equation, making sure to convert ΔG° and the result of RTlnQ to the same units (either J or kJ): ΔG = ΔG° + RTlnQ = -32.8 kJ/mol + 8.314 J/(mol·K) x 298 K x ln(360), which gives ΔG = about -29.6 kJ/mol. The negative value indicates that the reaction is spontaneous under these conditions.

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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]

Answer:

Activation energy is the amount of energy required to convert the reactants into the activated complex (option B)

Explanation:

The activation energy is the minimum needed energy to make a reaction occurs. When the reaction occurs, the molecules of reactants joins together in a determined position, so when they bond all the molecules reactants, they make the activated complex. That's why we say, that the energy needed to make the molecules of reactants form the activated complex is finally the activation energy.

When the activated complex is formed, there is a few moment with all the molecules from reactants bonding together. By the end, they are separated so the products are been formed.

Activation energy is the amount of energy required to convert the reactants into the activated complex.

Activated complex is the specie which is obtain in the transition state of the chemical reaction, and in this complex partial bond of reactants as well as partial bond of products are present.

So, the minimum amount of energy required by the reactant molecules to reach the threshold point where this activated complex is formed, known as the activation energy. After that state complex will finally convert in the product.

Hence, option (B) is correct.

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The pH of the solution after the titration of 0.10M KOH solution with 0.10M HCl is 7. This is due to complete neutralization of the acid and base, leaving water which has a pH of 7 at 25 degrees Celsius.

The question involves a titration process of KOH solution with HCl solution. KOH and HCl are a strong base and a strong acid respectively, and they react in stoichiometrically equal amounts. Since the molarities and amount of KOH and HCl are the same, it means all the KOH and HCl react completely, leaving no excess.

The result of the reaction is water (H2O) and a salt (KCl), and since KCl does not hydrolyze, the only contribution to the pH of the solution will be from the water itself. The pH of pure water is 7 at 25 degrees Celsius because it is neutral (neither acidic nor basic at this temperature).

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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:

In the given chemical reaction:

Species Oxidized: I⁻

Species Reduced: Fe³⁺

Oxidizing agent: Fe³⁺

Reducing agent: I⁻

As the reaction proceeds, electrons are transferred from I⁻ to Fe³⁺

Explanation:

Redox reaction is a chemical reaction involving the simultaneous movement of electrons thereby causing oxidation of one species and reduction of the other species.

The chemical species that gets reduced by gaining electrons is called an oxidizing agent. Whereas, the chemical species that gets oxidized by losing electrons is called a reducing agent.

Given redox reaction: 2Fe³⁺ + 2I⁻ → 2Fe²⁺ + I₂

Oxidation half-reaction: 2 I⁻ +  → I₂ + 2 e⁻                 ....(1)

Reduction half-reaction: [ Fe³⁺ + 1 e⁻ → Fe²⁺ ] × 2

                                   ⇒  2 Fe³⁺ + 2 e⁻ → 2 Fe²⁺       ....(2)

In the given redox reaction, Fe³⁺ (oxidation state +3) accepts electrons and gets reduced to Fe²⁺ (oxidation state +2) and I⁻ (oxidation state -1) loses electrons and gets oxidized to I₂ (oxidation state 0).

Therefore, Fe³⁺ is the oxidizing agent and I⁻ is the reducing agent and the electrons are transferred from I⁻ to Fe³⁺.

In this electron transfer reaction, iodide ions (I-) are oxidized to iodine (I2) and act as the reducing agent, while iron(III) ions (Fe3+) are reduced to iron(II) ions (Fe2+) and act as the oxidizing agent.

In the provided electron transfer reaction, a notable chemical transformation unfolds where the iodide ion (I-) undergoes oxidation to form iodine (I2), while concurrently, the iron(III) ion (Fe3+) experiences reduction to yield iron(II) ion (Fe2+). In terms of the key participants, it is crucial to discern the roles of reducing and oxidizing agents. The reducing agent is the entity that itself undergoes oxidation, and in this specific scenario, the iodide ion (I-) fulfills this role by ceding electrons during the reaction.

Conversely, the oxidizing agent is the species that undergoes reduction in the course of the reaction, and here, it's the iron(III) ion (Fe3+). As the reaction unfolds, electrons are effectively transferred from iodide ions to iron(III) ions, signifying a fundamental aspect of electron transfer reactions. This intricate interplay between the reducing and oxidizing agents, with the exchange of electrons, characterizes the essence of redox reactions in chemistry.

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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