Manganese sulfate forms a pale pink hydrate with the formula MnSO 4 ⋅ n H 2 O ( s ) . If this hydrate is heated to a high enough temperature, H 2 O ( g ) can be driven off, leaving the grey‑white anhydrous salt MnSO 4 ( s ) . A 16.260 g sample of the hydrate was heated to 300 ∘ C . The resulting MnSO 4 ( s ) had a mass of 14.527 g . Calculate the value of n in MnSO 4 ⋅ n H 2 O ( s ) .

Answer:

The aerobic cellular respiration of the glucose where glucose is converted to energy via four steps as follows

1. Glycolysis (glucose break down to pyruvic acid)

2. Link reaction

3. Krebs cycle

4. Electron transport chain, or ETC

The four pyruvic acid produces Four ATP, twenty NADH, and four [tex]FADH_{2}[/tex] molecules

Explanation:

When four pyruvic acid enters step two of the aerobic cellular respiration, they are converted by Oxidative decarboxylation into acetyl-CoA, four molecules of NADH and four molecule of CO2 are formed. This process is otherwise called the link reaction or transition  step, because it connects or links the Krebs cycle and glycolysis.

From the chemical reactions involved in cellular respiration of one glucose molecule, from two pyruvic acid molecules we have 2 ATP molecules, 10 NADH molecules, and 2 FADH2 molecules

Hence from four pyruvic acid molecules we have that the acetyl-CoA produced from the four pyruvic acid enters the the Krebs cycle and forms four ATP molecules, twenty NADH molecules, and four [tex]FADH_{2}[/tex] molecules.

In aerobic cellular respiration, four molecules of pyruvic acid will generate a total of four molecules of ATP, sixteen molecules of NADH (four from the conversion to Acetyl CoA, and twelve from the Krebs cycle) and four molecules of FADH2

In aerobic cellular respiration, pyruvic acid is converted into acetyl CoA. One molecule of pyruvic acid generates one molecule of NADH during this conversion so four molecules of pyruvic acid will yield four molecules of NADH. Acetyl CoA then enters the Krebs cycle, for each molecule of Acetyl CoA that goes through the Krebs cycle, three molecules of NADH, one molecule of FADH2, and one molecule of ATP is formed. Therefore, the four molecules of pyruvic acid would end up generating four molecules of ATP, twelve molecules of NADH and four molecules of FADH2 during the Krebs cycle (not including the NADH generated during the conversion to Acetyl CoA).

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Answer:

The answer to your question is  36.9 pounds

Explanation:

Data

density = 1 g/ml

mass = ?

volume = 16.743 L

- To solve this problem use the formula of density.

density = mass / volume

- Solve for mass

Mass = density x volume

- Convert volume to ml

                    1000 ml --------------- 1 L

                      x          ---------------- 16.743 L

                      x = (16.743 x 1000) / 1

                      x = 16743 ml

- Substitution

Mass = (1 g/ml)(16743 ml)

Simplification and result

Mass = 16743 g

- Convert mass to pounds

               1 pound ------------------ 453.58 g

                x            ------------------ 16743 g

                x = (16743 x 1) / 453.58

                x = 36.9 pounds

The mass of 16.743 L of water in pounds can be calculated as approximately 36.904 pounds.

To calculate the mass of water in pounds, we first determine the mass in grams using the given water volume and density.

Since the density of water is 1.00000 g/mL, and we know that 1 L is equal to 1000 mL, we multiply the volume in liters by the density in g/mL and by 1000 to get mass in grams.

This gives us 16.743 L * 1.00000 g/mL * 1000 = 16743 grams.

Converting grams to pounds, we know 1 pound is approximately 453.592 grams

So, the mass of water is then 16743 g / 453.592 g/lbs = 36.904 lbs. Therefore, the mass of 16.743 L of water is approximately 36.904 pounds.

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Answer:

The density of air at 22 °C and 760 torr is 1.195 KG/m³

Explanation:

The solution to the above  question is arrived t by considering the given variables and calculating the number of moles in a 1 m³ sample of air by plugging values into the universal gas equation  from which the number of moles of the constituent gases can be calculated by Dalton's law of partial pressure, then their masses and lastly the density of air is calculated using the formula, Density = mass/volume

The given variables are

Percentage Nitrogen = 78.1% by volume

Percentage oxygen = 20.9% by volume

Percentage argon = 0.934% by volume

The molar mass of nitrogen = 14.006g/mol

The molar mass of oxygen = 15.999g/mol

The molar mass of argon = 39.948 g/mol hence Considering a unit volume of air of one cubic meter (1m^3) we have

0.781 m³ of nitrogen, 0.209 m³ of oxygen and 0.00934 m³ of argon

The number of moles in 1 m³ of gas at 22 °C and 760 torr is given by

PV = nRT or n = [tex]\frac{PV}{RT}[/tex] = where 760 torr = 101325Pa we have n = [tex]\frac{(101325)(0.001)}{(8.314)(295.15)}[/tex] = 0.00413 mols per liter or 41.29 moles/m³

thus we have number of moles of nitrogen = 42.129 × 78.1% = 32.25 moles and the mass of nitrogen = 32.25×28.02 = 903.6 g

number of moles of oxygen= 42.129 × 20.1% = 8.63 moles and the mass of nitrogen = 8.63×32 = 276.16 g

number of moles of argon= 42.129 × 0.934% = 0.386 moles and the mass of nitrogen = 0.386×40 = 15.43 g

Therefore, mass of one cubic meter of air (1 m³), has a mass of

903.6 g + 276.16 g + 15.43 g = 1195.2 g or 1.195 KG Hence the density of air

is given by Density = [tex]\frac{mass}{volume}[/tex] =[tex]\frac{1.195 KG}{1 m^{3} }[/tex]  = 1.195 KG/m³

The density of air at 22 °C and 760 Torr is 1.19 g/L.

Air is 78.1% nitrogen, 20.9% oxygen, and 0.934% argon by moles. We will calculate the average molar mass of the air (M) as a weighted average of the molar masses of its constituents.

[tex]M = 78.1\% \times M(N_2) + 20.9\% \times M(O_2) + 0.934\% \times M(Ar)\\\\M = 78.1\% \times 28.00g/mol + 20.9\% \times 32.00g/mol + 0.934\% \times 39.95 g/mol = 28.93 g/mol[/tex]

Then, we will convert 22 °C to Kelvin using the following expression.

[tex]K = \° C + 273.15 = 22\° C + 273.15 = 295 K[/tex]

Assuming ideal behavior, we can calculate the density (ρ) of the air at 295 K (T) and 760 Torr (P) using the following expression.

[tex]\rho = \frac{P \times M }{R \times T} = \frac{760 Torr \times 28.93g/mol }{(62.4mmHg.L/mol.K) \times 295K} = 1.19 g/L[/tex]

where,

The density of air at 22 °C and 760 Torr is 1.19 g/L.

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Answer:

N2 element, pure substance

O2 element, pure substance

N2O Compound, pure substance

Air  (mostly N2 and O2 ) homogeneous mixture

Explanation:

N2, Nitrogen is known as the chemical element that is characterized by having atomic number 7 and that is symbolized by the letter N, in its molecular version, it is recognized as N2.

O2, Oxygen is the chemical element of atomic number 8, this molecular form is composed of two atoms of this element.

A chemical element is a type of matter, consisting of atoms of the same class.

N2O, Nitrous oxide is formed by the union of two molecules of nitrogen and one of oxygen, which is considered a chemical compound since it is a substance formed by the chemical combination of two different elements of the periodic table.

A pure substance is one that cannot change state or divide into other substances, except for a chemical reaction.

Air (mostly N2 and O2 ),  it is a homogeneous mixture of gases that constitutes the earth's atmosphere. A homogeneous mixture is a type of mixture in which its components are not distinguished and in which the composition is uniform and each part of the solution has the same properties.

Substances can be classified based on their composition and uniformity: N2 and O2 are elements and pure substances, N2O is a compound and a pure substance, and air is a homogeneous mixture or solution.

When classifying substances, we take into account their composition and uniformity. Here are the classifications for the mentioned substances:

N2 (Nitrogen) is an element and a pure substance since it is composed of only one type of atom.

O2 (Oxygen) is also an element and a pure substance, with two oxygen atoms bonded together.

N2O (Nitrous Oxide) is a compound as it is made up of two different elements, nitrogen and oxygen, in a fixed ratio and a pure substance due to its uniform composition.

Air, which is mostly made up of Nitrogen (N2) and Oxygen (O2), is a homogeneous mixture or solution because the composition is uniform throughout, and it is a mixture of multiple gases.

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Explanation:

Conservation of momentum :

[tex]m_1u_1+m_2u_2=m_1v_1+m_1v_2[/tex]

Where :

[tex]m_1, m_2[/tex] = masses of object collided

[tex]u_1,u_2[/tex] = initial velocity before collision

[tex]v_1,v_2[/tex] = final velocity after collision

We have :

Mass of an engine = [tex]m_1=4M[/tex]

Mass of an car= [tex]m_2=M[/tex]

Initial velocity of railroad engine [tex]m_1=u_1=5 km/h[/tex]

Initial velocity of car [tex]m_2=u_2=0 km/h[/tex] (rest)

Final velocity of  railroad engine [tex]m_1=v_1=v[/tex] (same direction )

Final velocity of car [tex]m_2=v_2=v[/tex] (same direction)

[tex]4M\times 5km/h+M(0 km/h)=4Mv+Mv[/tex]

[tex]4\times 5 km/h=5M[/tex]

v = 4 km/h

The speed of the engine and car after they coupled together is 4 km/h.

Using the principle of conservation of momentum, it's shown that the diesel engine and freight car will coast together at 4 km/h after coupling. This is calculated by equating the initial and final momentums of the system.

The conservation of momentum can be used to solve this problem. The principle states that the total linear momentum of a closed system remains constant, regardless of any internal changes. Here, the system consists of the diesel engine and the freight car.

Let's denote the weight of the freight car as 'm'. Given that the diesel engine weighs four times as much as a freight car, the weight of the engine would be '4m'.

If the diesel engine is coasting at 5 km/hr, the initial momentum of the system is the momentum of the engine, because the freight car is at rest. Therefore, the initial momentum (Pi) is the weight of the engine times its velocity, which is 4m*5 km/hr = 20m km/hr.

After they couple together, there's no external force, so the total momentum should remain the same (the conservation of momentum principle). Let's denote the final velocity of the engine + car (now moving together) as 'v'. The final momentum (Pf) = (m + 4m) * v = 20m km/hr.

Therefore, we can establish the equation: Pi = Pf, meaning 20m km/hr = 5m * v. Solving for v, we find that v = 4 km/h. Therefore, the engine + car coast together at 4 km/h after coupling, demonstrating the conservation of momentum.

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Answer:

d. Localized.

Explanation:

You posted this question in the chemistry section. The whole exercise requires you to understand the terms in a chemical context. So let's look at the three terms: homogeneous, localized, and radicalized.

Homogeneous

A chemical substance is said to be homogeneous if the chemical composition is the same throughout. It means the substance must have the same states. For example, a mixture of undiluted orange juice forms a homogeneous solution once it is diluted in water. It means that there is perfect mixing to have a uniform liquid state.

Localized

The terminology is often applied to atoms or chemical structures, particularly metals and acids. For example, atoms have localized electrons in the orbitals. It means that the electron orbits in the region for 95 % of its time. Localized means belonging to one region. This applies to the religion.

Radical

The term applies to highly reactive atomic species, normally called ions. These elements seek electrons and are highly "reactive." The ions are called radicals in this sense.

Answer: The unknown element is potassium.

Explanation:

Average atomic mass of an element is defined as the sum of masses of each isotope each multiplied by their natural fractional abundance.

Formula used to calculate average atomic mass follows:

[tex]\text{Average atomic mass }=\sum_{i=1}^n\text{(Atomic mass of an isotopes)}_i\times \text{(Fractional abundance})_i[/tex] .....(1)

Mass of isotope 1 = 38.9637 amu

Percentage abundance of isotope 1 = 93.08 %

Fractional abundance of isotope 1 = 0.9308

Mass of isotope 2 = 39.9640 amu

Percentage abundance of isotope 2 = 0.012 %

Fractional abundance of isotope 2 = 0.00012

Mass of isotope 3 = 40.9618 amu

Percentage abundance of isotope 3 = 6.91 %

Fractional abundance of isotope 3 = 0.0691

Putting values in equation 1, we get:

[tex]\text{Average atomic mass of Z}=[(38.9637\times 0.9308)+(39.9640\times 0.00012)+(40.9618\times 0.0691)][/tex]

[tex]\text{Average atomic mass of Z}=38.85amu[/tex]

For the given options:

Option a: Average atomic mass of Sulfur = 32.065 amu

Option b: Average atomic mass of Potassium = 39.09 amu

Option c: Average atomic mass of Chlorine = 35.45 amu

Option d: Average atomic mass of Calcium = 40.078 amu

Option e: Average atomic mass of Argon = 39.94 amu

As, the average atomic mass of unknown element is near to the average atomic mass of potassium. So, the unknown element is potassium.

Hence, the unknown element is potassium.

The element with isotopes of masses 38.9637 amu, 39.9640 amu, and 40.9618 amu, and natural abundances of 93.08%, 0.012%, and 6.91% respectively, is Calcium (Ca).

To identify the element based on its isotopes, we need to calculate the average atomic mass of the element. This can be done by multiplying the mass of each isotope by its natural abundance (expressed as a decimal), and then summing up these products. In this case, the element with isotopes of masses 38.9637 amu, 39.9640 amu, and 40.9618 amu, and natural abundances of 93.08%, 0.012%, and 6.91% respectively, is Calcium (Ca). The average atomic mass of calcium can be calculated as follows:

(38.9637 amu * 0.9308) + (39.9640 amu * 0.00012) + (40.9618 amu * 0.0691) = 40.08 amu

Therefore, the element is calcium (Ca).

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E) two or more atoms chemically joined together.

The following information should be considered:

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Answer:

two or more atoms chemically joined together.

Explanation:

Answer:

403.3 kPa is the new pressure

Explanation:

This problem is solved by this formula:

P₁ . V₁ = P₂ . V₂

101kPa . 575 mL = P₂ . 144mL

(101kPa . 575 mL) / 144 mL = P₂

403.3 kPa = P₂

Final answer:

Using Boyle's Law, the new pressure of neon gas compressed from 575 mL at 101 kPa to 144 mL, with temperature held constant, is found to be 403.125 kPa.

Explanation:

The student asked for the new pressure of neon gas that was compressed from 575 mL at 101 kPa to a volume of 144 mL, given that the temperature remained constant. This type of problem involves ideal gas behavior and can be solved using Boyle's Law, which states that for a given mass of gas at constant temperature, the volume of the gas is inversely proportional to its pressure (P₁V₁ = P₂V₂).

Starting with the initial conditions:

Initial volume (V₁) = 575 mL

Initial pressure (P₁) = 101 kPa

And the final condition:

Final volume (V₂) = 144 mL

Since temperature remains constant, we can calculate the final pressure (P₂) using the formula:

P₂ = (P₁* V₁) /V₂

Substitute the given values into the equation:

P₂ = (101 kPa* 575 mL) \/ 144 mL = 403.125 kPa

Hence, the new pressure of the neon gas after compression is 403.125 kPa.

Answer and Explanation

The final reaction is the production of propane from Carbon and Hydrogen.

3C (s) + 4H2 (g) ---> C3H8 (g)

So, reverse eq. A,

3CO2 (g) + 4H20 (l) ---> C3H8 (g) + 502 (g)

Add 3 × eq. B,

3C (s) + 3O2 (g) ----> 3CO2 (g)

Add 4 × eq. C,

4H2 (g) + 2O2 (g) ---> 4H20 (l)

Writing them together,

3CO2 (g) + 4H20 (l) ---> C3H8 (g) + 502 (g)

+ 3C (s) + 3O2 (g) ----> 3CO2 (g)

+ 4H2 (g) + 2O2 (g) ---> 4H20 (l)

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

3CO2 (g) + 4H20 (l) + 3C (s) + 3O2 (g) + 4H2 (g) + 2O2 (g) ---> C3H8 (g) + 502 (g) + 3CO2 (g) + 4H20 (l)

The compounds that exist on both sides cancel out and we're left with

3C (s) + 4H2 (g) ---> C3H8 (g)

So, mathematically, the final reaction can be written as:

(-eq. A + 3(eq. B) + 4(eq. C))

-A+3B+4C = 3C (s) + 4H2 (g) ---> C3H8 (g)

QED!

[tex]\[{3\text{C(s, graphite)} + 4\text{H2(g)} \rightarrow \text{C3H8(g)} + 5\text{O2(g)}} \][/tex]

This balanced equation shows the production of propane (C3H8) from its elements carbon and hydrogen gas.

To combine the given reactions to form the balanced chemical equation describing the production of propane (C3H8) from its elements carbon (C, as graphite) and hydrogen (H2), let's follow these steps:

Given reactions:

(a) [tex]\( \text{C3H8(g)} + 5\text{O2(g)} \rightarrow 3\text{CO2(g)} + 4\text{H2O(g)} \)[/tex]

(b) [tex]\( \text{C(s, graphite)} + \text{O2(g)} \rightarrow \text{CO2(g)} \)[/tex]

(c) [tex]\( \text{H2(g)} + \frac{1}{2}\text{O2(g)} \rightarrow \text{H2O(g)} \)[/tex]

We need to manipulate these reactions to combine them into one overall balanced equation that shows the formation of propane (C3H8) from carbon and hydrogen.

Step-by-Step Combination:

1. Reverse Reaction (a):

[tex]\( 3\text{CO2(g)} + 4\text{H2O(g)} \rightarrow \text{C3H8(g)} + 5\text{O2(g)} \)[/tex]

This is the reverse of reaction (a), which is necessary to show the formation of propane.

2. Multiply Reaction (b) by 3:

[tex]\( 3\text{C(s, graphite)} + 3\text{O2(g)} \rightarrow 3\text{CO2(g)} \)[/tex]

Multiply reaction (b) by 3 to balance the carbon atoms with the propane formation reaction.

3. Multiply Reaction (c) by 4:

[tex]\( 4\text{H2(g)} + 2\text{O2(g)} \rightarrow 4\text{H2O(g)} \)[/tex]

Multiply reaction (c) by 4 to balance the hydrogen atoms with the propane formation reaction.

4. Combine the Reactions:

Now, add the balanced reactions (reverse of (a), multiplied (b), and multiplied (c)) to get the overall balanced equation for the formation of propane:

[tex]\( 3\text{C(s, graphite)} + 3\text{O2(g)} + 4\text{H2(g)} + 2\text{O2(g)} \rightarrow 3\text{CO2(g)} + 4\text{H2O(g)} + 5\text{O2(g)} \)[/tex]

5. Simplify the Equation:

Combine like terms (oxygen on both sides):

[tex]\( 3\text{C(s, graphite)} + 3\text{O2(g)} + 4\text{H2(g)} \rightarrow 3\text{CO2(g)} + 4\text{H2O(g)} + 5\text{O2(g)} \)[/tex]

Intramolecular Sn2 reaction is a bimolecular, second-order, elementary reaction. It involves a single, concerted step in which a nucleophile attacks the substrate, leading to a transition state, and then to expulsion of a leaving group. The stereochemistry of the molecule is usually inverted at the reaction centre.

To best describe the properties of an intramolecular Sn2 reaction mechanism, we can say that it's a bimolecular reaction, which means it involves two reactant species. In this case, the reaction mechanism involves a single, concerted step where a nucleophile attacks the substrate, leading to a transition state and finally the expulsion of a leaving group. This makes Sn2 a type of elementary reaction.

In these reactions, the rate is dependent on the concentration of both reactants, leading to a second-order rate law. Further, the rate-determining step (the slowest in the mechanism) for an Sn2 reaction is the single concerted step itself. One important aspect to remember about Sn2 mechanisms is the stereochemical alteration that takes place, typically resulting in inversion of configuration at the reaction centre.

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The intramolecular SN2 reaction mechanism is a bimolecular and concerted process characterized by inversion of configuration at the reaction center, second-order kinetics, and sensitivity to steric hindrance.

The intramolecular SN2 reaction mechanism is characterized by several distinct properties. Firstly, it is a bimolecular reaction, meaning the rate of the reaction depends on the concentration of two reactants: the nucleophile and the electrophile. Secondly, the reaction proceeds via a concerted process where bond-forming and bond-breaking occur simultaneously, leading to an inversion of configuration at the carbon center where the substitution takes place.

Lastly, the SN2 mechanism exhibits second-order kinetics, as the reaction rate depends on the concentration of both the nucleophile and the electrophile. It is important to note that SN2 reactions are sensitive to steric hindrance; bulky groups near the reactive site can inhibit the reaction by limiting the nucleophile's access to the electrophile.

Answer:The isotope is Carbon-12 and its atomic mass is 12.

Explanation:

Mass number is the total number of protons and neutrons in a nucleus.

Atomic number is the number of protons in the nucleus of an atom.

An isotope of a chemical element is an atom that has a different mass number but the same atomic number as the element. The difference in mass number is from the number of neutrons (that is, a greater or lesser atomic mass) than the standard for that element.

Carbon-12 is an isotope of carbon it has 6 neutrons and 6 protons, giving it a mass number of 12 and atomic number of 6. Carbon-12 is a stable isotope of carbon, it has the same mass number and atomic number as carbon.

The isotope of carbon that has exactly the same mass and atomic number as used in the definition of the atomic mass unit is;

Isotope Carbon-12.

In chemistry, we know that;

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Answer:

33.78 g

Explanation:

NaClO3 decomposes to NaCl and O2 by this reaction:

2NaClO3 --> 2NaCl + 3O2

Let's determines the mole of chlorate we used (mass / molar mass)

61.5 g / 106.45 g/mol = 0.578 moles.

Ratio is 2:2, so x amount of chlorate will produce x amount of chloride. In conclussion we made 0.578 moles of NaCl from 0.578 moles of chlorate. Let's convert the moles to mass:

0.578 mol . 58.45g/1mol = 33.78 g

That is the theoretical yield of NaCl.

Answer: adhesion

Explanation:

Cohesion is the attraction between similar molecules. Example: Force of attraction between water molecules.

Thus hydrogen bond formed between the molecules of water due to the development of partial negative charge on oxygen and partial positive charge on hydrogen is cohesion.

Adhesion is the attraction between different molecules. Example: Force of attraction between HCl and water.

The hydrogen bond formed between H of HCl and O of water due to developments of partial positive and partial negative charge respectively is adhesion.

Answer:

The mass of water did Doc placed in the bottle is 288 grams.

Explanation:

Mass of water filled in water bottle = M

Volume of the water filled in water bottle = V = 288 mL

Density of the water at 20°C , d= 0.99823 g/ml

[tex]D=\frac{M}{V}[/tex]

[tex]M=D\times V=0.99823 g/ml\times 288 mL=287.49024 g\approx 288 g[/tex]

The mass of water did Doc placed in the bottle is 288 grams.

Final answer:

The mass of water at 20 °C with a density of 0.99823 g/mL that Doc placed in the bottle is 287 g when rounded to three significant digits.

Explanation:

The mass of water Doc placed in the bottle can be calculated using the density of liquid water at 20 °C, which is 0.99823 g/mL. Given he filled it with exactly 288 mL of water, the mass of the water is calculated by multiplying the volume by the density: The mass of water at 20 °C with a density of 0.99823 g/mL that Doc placed in the bottle is 287 g when rounded to three significant digits.

Mass = volume × density

Mass = 288 mL × 0.99823 g/mL

Mass = 287.63064 g

Since we are asked to provide the mass to three significant digits, the mass of water Doc placed in the bottle is 287 g.

In hemoglobin, a single amino acid change at position 6 from Glu to Val has major consequences on hemoglobin structure that makes the molecule defective leading to sickle cell anemia. Predict whether the following hypothetical change would or would not have a major effect at position 6. Briefly explain (1-2 sentences). Glu to Leu Hint: Look at the structures of the R groups and consider their chemical properties

The structure of the haemoglobin, hence the RBC won't be same as normal.

Both the leucine and glutamic acid are alpha amino acids which have an alpha carboxylic acid group and an alpha amino group. The variable in case of glutamic acid is propyl acid while the variable in case of leucine is isobutyl.

The glutamic acid is the normal amino acid of the 6th position of Beta chain of hemoglobin. Its an acid group, so can form bonds with another base inside the haemoglobin, or can form other hydrogen bonds. But the isobutyl group is an alkyl group. So it doesn't have that much effect in the recovering the structure, and sickle cell anemia prevails.

Answer:

option (e), dichloromethane, methene chloride

Explanation:

(a)  [tex]CHCl_3[/tex]

Common name:  chloroform

IUPAC name: one carbon atom, therefore, root word is meth.

Its is saturated compounds, so add ane after root word.

Three chlorine atoms are present.

Therefore, IUPAC name: Trichloromethane

(b)  [tex]CCl_4[/tex]

Common name:  Carbon tetrachloride

IUPAC name: one carbon atom, therefore, root word is meth.

Its is saturated compounds, so add ane after root word.

Four chlorine atoms are present, chlorine atoms are named as prefixes.

Therefore, IUPAC name: tetrachloromethane

(c)  [tex]C_6H_5I[/tex]

Common name: Phenyl iodide

IUPAC name:

The given compound is an aryl halides. Aryl haildes are named as haloarenes. The prefix halo is placed before aromatic hydrocarbon. Here, prefix is iodo and aromatic hydrocarbon is benzene.

Therefore, IUPAC name of the compound is iodobenzene.

(d)  [tex]CH_3Cl[/tex]

Common name:  Methyl chloride

IUPAC name: one carbon atom, therefore, root word is meth.

Its is saturated compounds, so add ane after root word.

One chlorine atom is present.  

Therefore, IUPAC name: chloromethane

(e)  [tex]CH_2Cl_2[/tex]

Common name: Methylene chloride

IUPAC name: one carbon atom, therefore, root word is meth.

Its is saturated compounds, so add ane after root word.

Two chlorine atoms are present, chlorine atoms are named as prefixes.

Therefore, IUPAC name: dichloromethane.

Therefore, the correct option is option (e), dichloromethane, methene chloride

Answer: False

Explanation:

Molecular formula is the chemical formula which depicts the actual number of atoms of each element present in the compound.  

Empirical formula is the simplest chemical formula which depicts the whole number of atoms of each element present in the compound.

Example: [tex]CH_4[/tex] has similar molecular formula and empirical formula as the elements are already present in simplest of the ratios.

[tex]C_2H_2[/tex] has molecular formula of [tex]C_2H_2[/tex] but [tex]CH[/tex] as the empirical formula.

Answer:

Calcite

Explanation:

Calcite is a mineral formed by calcium carbonate (CaCO3), class 05 of the Strunz classification, the so-called carbonate and nitrate minerals.

All members of this group crystallize in the trigonal system, have a perfect rhombohedral cleavage and exhibit a strong double refraction in transparent rhombohedra.

It presents a variety of shapes and colors. It is characterized by its low hardness, 3 on the Mohs scale, and by its high reactivity even with weak acids.

The best property to identify calcite is the acid test, since this mineral produces effervescence with acids. The reason for this is the following reaction:

CaCO3 + 2H + → Ca2+ + H2O + CO2 (gas)

Final answer:

Using the 68-95-99.7 rule, the percentage of eggs weighing between 46.5 grams and 65.7 grams, given a mean weight of 51.3 grams and a standard deviation of 4.8 grams, is estimated to be between 95% and 99%.

Explanation:

The question asks us to calculate the percentage of eggs that weigh between 46.5 grams and 65.7 grams, given that the weights are normally distributed with a mean of 51.3 grams and a standard deviation of 4.8 grams. Utilizing the 68-95-99.7 rule (also known as the Empirical Rule), we can determine percentages for different ranges from the mean in a normal distribution.

Firstly, to find the specific range that includes 46.5g to 65.7g from our mean of 51.3g, we calculate the number of standard deviations each value is from the mean. However, without doing the math, we see that 46.5g is less than one standard deviation away (since one standard deviation is 4.8g), and 65.7g is significantly more than two but less than three standard deviations away.

According to the 68-95-99.7 rule, 68% of data falls within one standard deviation, 95% within two, and 99.7% within three. Thus, intuitively, without precise calculation, we can say that the percentage of eggs weighing between 46.5g and 65.7g would be a bit less than 99.7%, as the upper limit is not yet reaching three standard deviations from the mean but is beyond the two-standard deviation mark that covers 95% of the distribution. Thus, it's reasonable to conclude that approximately 95-99% of eggs will fall within this weight range.

Answer : The maximum mass of sucralose is, 7952.8 grams.

Explanation :  Given,

Molal-freezing-point-depression constant [tex](K_f)[/tex] for water = [tex]1.86^oC/m[/tex]

Mass of water (solvent) = 2.00 kg

Molar mass of sucralose = 397.64 g/mole

Formula used :  

[tex]\Delta T_f=i\times K_f\times m\\\\T^o-T_s=i\times K_f\times\frac{\text{Mass of sucralose}}{\text{Molar mass of sucralose}\times \text{Mass of water in Kg}}[/tex]

where,

[tex]\Delta T_f[/tex] = change in freezing point

[tex]\Delta T_s[/tex] = freezing point of solution = [tex]-18.6^oC[/tex]

[tex]\Delta T^o[/tex] = freezing point of water = [tex]0^oC[/tex]

i = Van't Hoff factor = 1 (for sucralose non-electrolyte)

[tex]K_f[/tex] = freezing point constant for water = [tex]1.86^oC/m[/tex]

m = molality

Now put all the given values in this formula, we get

[tex](0-(-18.6)^oC)=1\times (1.86^oC/m)\times \frac{\text{Mass of sucralose}}{397.64g/mol\times 2.00kg}[/tex]

[tex]\text{Mass of sucralose}=7952.8g[/tex]

Therefore, the maximum mass of sucralose is, 7952.8 grams.

Answer:

The given chemical compound has 2 atoms of hydrogen and one atom of oxygen for each atom of carbon. The mass of CH2O is 12 + 2*1 + 16 = 30. The molecular weight of the compound is 180.18 which is approximately 180. This gives the molecular formula of the chemical compound as C6H12O6.

Explanation:

To calculate the percent composition of CH2O, determine the molar mass of each element and the total molar mass of the compound. The percent composition is approximately 40.0% Carbon, 6.7% Hydrogen, and 53.3% Oxygen.

To calculate the percent composition of CH2O, we will need to determine the molar mass of each element in the compound and the total molar mass of the compound. The molar masses from the periodic table are approximately 12.01 g/mol for Carbon (C), 1.01 g/mol for Hydrogen (H), and 16.00 g/mol for Oxygen (O).

First, let's calculate the total molar mass of CH2O: (1 × 12.01) + (2 × 1.01) + (1 × 16.00) = 12.01 + 2.02 + 16.00 = 30.03 g/mol.

Now, let's find the percent composition for each element:

The percent composition of CH2O is therefore approximately 40.0% Carbon, 6.7% Hydrogen, and 53.3% Oxygen.

Answer:

The answer to your question is the fourth one.

Explanation:

From the description, we know that the reactants are two liquids and the products are a solid and a gas.

The first option is incorrect because it mentions that the reactants are one liquid and solid.

The second option is also incorrect for the same reason as the first one and one of the products must be a solid.

The third option is incorrect besides there are two liquids in the reactants and  a solid and gas in the liquids there is a minus sign that is not possible.

The fourth option is the correct one.

A sigma bond b. overlap of two d orbitals end-to-end overlap of p orbital.

A sigma bond is a strong bond that is made up of overlapping orbitals. In fact, these bonds are the strongest known bonds in chemical reactions. The overlapping orbitals are in the form of covalent bonding and this gives us the idea that sigma bonds are strong in nature.

In addition, sigma bonds are mostly common with diatomic elements and compounds. There are three orbitals where the sigma bond can be found and they are the p-p, s-p, and the s-s orbitals. These are known for forming symmetry groups.

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Answer and Explanation:

The rate constant (K) is related to activation energy (Ea), frequency factor (A) and temperature (T) in Kelvin by the equation

R = molar gas constant

K = A(e^(-Ea/RT))

Taking natural log of both sides

In K = In A - (Ea/RT)

In K = (-Ea/R)(1/T) + In A

Comparing this to the equation of a straight line; y = mx + c

y = In K, slope, m = (-Ea/R), x = (1/T) and intercept, c = In A

a) From the question, m = (-Ea/R) = -1.10 × (10^4) K

(-Ea/R) = -1.10 × (10^4) = -11000

R = 8.314 J/K.mol

Ea = -11000 × 8.314 = 91454 J/mol = 91.454 KJ/mol

b) c = In A = 33.5

A = e^33.5 = (3.54 × (10^14))/s

c) K = A(e^(-Ea/RT))

A = (3.54 × (10^14))/s, Ea = 91454 J/mol, T = 25°C = 298.15 K, R = 8.314 J/K.mol

K = (3.54 × (10^14))(e^(-91454/(8.314×298.15))) = 0.0336/s

QED!

This question pertains to the rate of a chemical reaction, which is  first order with respect to (CH3)3CBr and zero order with respect to OH-. An Arrhenius plot indicating temperature and rate constant is used to find activation energy and frequency factor.

The reaction you're describing is a typical chemical reaction. This type of reaction is first order with respect to (CH3)3CBr, which means the rate of the reaction depends on the concentration of this compound. OH-, on the other hand, is zero order, meaning its concentration doesn't affect the reaction's rate.

The plot you've mentioned is an Arrhenius plot, and it is used to determine the activation energy and frequency factor of a reaction from the slop and y-intercept respectively. Given the slope value of –1.10 x 10^4 K you mentioned, you can find the activation energy (Ea) from the formula Ea = -slope * R , where the slope is –1.10 x 10^4 K and R is the universal gas constant (8.3145 Joule/(mole*K)). Similarly, from the y-intercept value, you can find the frequency factor by the formula A=e^(y-intercept), where A is frequency factor and e is natural base.

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Answer:

The percentage yield of O2 is 66.7%

Explanation:

Reaction for decomposition of potassium chlorate is:

2KClO₃ →  2KCl  +  3O₂

The products are potassium chloride and oxygen.

Let's find out the moles of chlorate.

Mass / Molar mass = Moles

12.3 g / 123 g/mol = 0.1 mol

So ratio is 2:3, 2 moles of chlorate produce 3 mol of oxygen.

Then, 0.1 mol of chlorate may produce (0.1  .3)/ 2 = 0.15 moles

Let's convert the moles of produced oxygen, as to find out the theoretical yield.

0.15 mol . 32 g/ 1mol = 4.8 g

To calculate the percentage yield, the formula is

(Produced Yield / Theoretical yield) . 100 =

(3.2g / 4.8g) . 100 = 66.7 %

The branch of science which deals with the study of chemicals and their bond is called chemistry.

The correct percentage yield of O2 is 66.7%

The Reaction for decomposition of potassium chlorate is as follows:-

[tex]2KClO_3 <---> 2KCl + 3O_2[/tex]

The formula is as follows:- [tex]\frac{Mass }{Molar\ mass} = Moles[/tex]

After putting the value in the question is:-

[tex]\frac{12.3 g}{123} = 0.1 mol[/tex]

So the ratio present in the reaction is 2:3.

Therefore, the 0.1 mole of chlorate produce [tex]\frac{(0.1 *3)}{2} = 0.15 moles[/tex]

Convert them into molar mass is:-

[tex]\frac{0.15*32 g}{1mol} = 4.8 g[/tex]

The percentage of the compound is [tex]\frac{(3.2g}{4.8g}* 100 = 66.7 %[/tex].

Hence, the correct answer is 66.7%

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Answer:

Empirical CHO

molecular C4H4O4

Explanation:

From the question, we know that it contains 41.39% C , 3.47% H and the rest oxygen. To get the % composition of the oxygen, we simply add the carbon and hydrogen together and subtract from 100%.

This means : O = 100 - 41.39 - 3.47 = 55.14%

Next is to divide the percentage compositions by their atomic masses.

C = 41.39/12 = 3.45

O = 55.14/16 = 3.45

H = 3.47/1 = 3.47

Now we divide by the smallest value which is 3.45. We can deduce that this will definitely give us an answer of 1 all through as the values are very similar.

Hence the empirical formula of Maleic acid is CHO

Now we go on to deduce the molecular formula.

To do this we need the molar mass. I.e the amount in grammes per one mole of the compound.

Now we can see that 0.378mole = 43.8g

Then 1 mole = xg

x = (43.8*1)/0.378 = 115.87 = apprx 116

[CHO]n = 116

(12 + 1 + 16]n = 116

29n = 116

n = 116/29 = 4

The molecular formula is thus C4H4O4

Blue-green algae are prokaryotes that carry out photosynthesis in chloroplasts, converting light energy into carbohydrates by using chlorophyll as a pigment for energy absorption.

Some prokaryotes, such as blue-green algae, are photosynthetic and contain chloroplasts where chlorophyll and other pigments absorb energy from the sun to produce carbohydrates via photosynthesis. The algae, specifically, are among the groups of prokaryotes that are capable of photosynthesis, much like plants. The chloroplasts in these cells function as the site where light energy captured by chlorophyll is converted into chemical energy, which is then used to create carbohydrates from carbon dioxide and water.

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Answer:

The answer is 671 litres of carbon dioxide is produced from 0.410 kg of hexane

Explanation:

We first write a balanced reaction for the complete combustion of hexane thus

The stoichiometry of the cumbustion of hexane in air is

2C6H14(g)+18O2(g)→12CO2(g)+14H2O(l) or

C6H14(g)+9O2(g)→6CO2(g)+7H2O(l)

From the above reaction it is observed that one mole of hexane burns completely in the presence of oxygen to produce 6 moles of carbon dioxide

Therefore we calculate the nuber of moles of hexane present in the sample thus

Mass hexane of sample = 0.41 kg

Molar nass of hexane = 86.18 g/mol

number of moles of hexane = (mass of hexane)/(molar mass of hexane) = (0.41×1000)/86.16 = 410/86.16 = 4.76 moles

As we have seen from the chemical reaction, 1 mole of H6H14 produces 6 moles of CO2 hence 4.76 moles of Hexane produces

4.76×6 moles of CO2 which is 28.55 moles of CO2

From the question we have the temperature and the pressure of the production of CO2 as

Temperature of reaction = 13° C converting to kelving gives= 13+273.15 = 286.15 K

and pressure = 1 atmosphere or 101325 Pa

13.0∘C=13.0∘C+273.15=286.15 K

The volume of the produced CO2 can be calculated using the combined ideal gas equation given by

P×V=n×R×T where

Here

P = Gas pressure (of CO2 )

V = Volume (of the CO2)

n = number of moles of gas (CO2) present

R = universal gas constant, equal to 0.0821 atm× L/(mol× K )

T = absolute temperature in Kelvin

Thus we have

1×V = 28.55×0.0821×286.15  or V  = 670.76L

Rounding up the answer to 3 significant digits we have

670.76L ≅ 671L

671 litres of carbon dioxide is produced from 0.410 kg of hexane

Answer:

                    The elements that will be present after the burning of Ethanol are;

                  (i) Carbon

                   (ii)  Oxygen

                   (iii) Hydrogen

Explanation:

                    The balanced chemical equation for the burning of ethanol is as follow;

                            C₂H₅OH + 3 O₂ → 2 CO₂ + 3 H₂O

It can be observed in given balanced chemical equation that there three elements involved in this entire reaction. The elements that present in the reactant side are also found on the product side. It means that the elements have just rearranged going from reactant to product.

This means that this reaction is obeying Law of Conservation of Mass which states that mass can neither be created nor destroyed but it can be changed from one form to another hence, to keep the mass on both sides of the reaction balanced the same elements should be present on the product side too.

Answer: carbon, hydrogen, and oxygen only

Explanation: When a chemical reaction occurs, the atoms in the original set of substances are rearranged to form a new set of substances. The number of atoms of each element does not change.

The elements carbon, hydrogen, and oxygen are the only elements present in the original substances, so the only elements in the final substances after the reaction will be carbon, hydrogen, and oxygen, as well. The number of atoms of each element must be the same before and after the reaction.