Draw the keto and enol forms of 3,3-dimethylbutan-2-one, circle the more stable tautomer, and propose a mechanism for the conversion of one tautomer to the other in acidic conditions.

Convex lenses refract light inward toward a focal point. Light rays progress through the edges of a convex lens and are bent most.

convex lenses are across in the middle. Rays of light that pass through the lens are conducted closer together (they converge). A convex lens is a meet lens. When parallel rays of light pass through a convex lens the refracted rays cross at one point called the principal focus.

. A convex lens is a connecting lens. When parallel rays of light pass through a convex lens the refracted rays merge at one point called the principal focus.

So we can conclude that Convex Lens. Concave Lens; It is known as a converging lens as light rays, when passed through this lens, tends to bend towards each other. It is known as a Convex Lens

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Atom consists of electron, proton and neutron.  The number of proton and electrons is 25. The number of neutron is 30. The given element is manganese, Mn.

Atom is the smallest particle of any matter. Atom combines to form element and element combine to form molecule or compound.

Mass number =Number of neutrons+ number of protons

Mass number =55

Number of neutrons=atomic number+5

Substituting the given values in the above formula

55 =atomic number+5+ number of protons

50=2×atomic number

Atomic number=25

Therefore, the number of proton and electrons is 25. The number of neutron is 30. The given element is manganese, Mn.

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Xeriscaping is a landscaping approach that focuses on conserving water by using drought-tolerant plants and efficient irrigation techniques. The goal is to create a visually appealing and sustainable garden while minimizing water usage.

Successful xeriscaping projects involve several key elements. Firstly, careful plant selection is crucial, opting for species that can thrive in arid conditions without excessive watering. Mulching is used to reduce evaporation and retain soil moisture.

Proper soil preparation, such as improving drainage and adding organic matter, promotes healthier plant growth. Efficient irrigation systems, like drip irrigation or soaker hoses, deliver water directly to plant roots, minimizing wastage.

Additionally, controlling erosion through the use of retaining walls or terracing is important. Lastly, regular maintenance, including appropriate pruning and weed control, ensures the longevity and vitality of the xeriscape garden. Overall, a successful xeriscaping project harmonizes sustainable practices with a beautiful outdoor environment.

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Answer:

During the early Paleozoic Era, the Earth's surface was very different from what it is today. The continents were arranged differently, forming one large supercontinent called Pangea. This landmass was surrounded by a single large ocean called Panthalassa. The climate was much warmer and wetter than it is today, with no ice caps at the poles.

The first land plants, known as bryophytes, appeared during the early Silurian Period, around 430 million years ago. These plants were small and simple, lacking roots and vascular tissue. They grew in damp environments, such as along the edges of lakes and streams. They were important in the development of soils and in the colonization of land by other organisms, such as insects and other arthropods.

Answer:

1. K₂C₂O₄

2. CuC₂O₄.

3. Bi₂(C₂O₄)₃

4. Pb(C₂O₄)₂.

5. (NH₄)₂C₂O₄

6. H₂C₂O₄

Explanation:

Hello,

In this case, since the oxidation states are exchanged according to the naming rules, the resulting species are shown below:

1. K₂C₂O₄

2. CuC₂O₄ (here, the +2 of copper and -2 of C₂O₄, are simplified).

3. Bi₂(C₂O₄)₃

4. Pb(C₂O₄)₂ (here, the +4 of lead and -2 of C₂O₄, are simplified).

5. (NH₄)₂C₂O₄

6. H₂C₂O₄

Regards.

Answer:

Insulator

Explanation:

By using the ideal gas law and Avogadro's number, we find that there are approximately 6.72 × 10^22 molecules of CO2 in 2.5 L at STP.

To determine the number of molecules of CO2 in 2.5 L at STP (Standard Temperature and Pressure), we can use the ideal gas law and Avogadro's number.

Avogadro's number (N_A) is a fundamental constant representing the number of particles (atoms, molecules, ions) in one mole of substance. Its value is approximately 6.022 × 10^23 particles/mol.

STP conditions are defined as a temperature of 273.15 K (0 °C) and a pressure of 1 atmosphere (1 atm).

First, we need to convert the volume from liters to moles of CO2. To do this, we use the ideal gas law equation:

PV = nRT,

where P is the pressure, V is the volume, n is the number of moles, R is the ideal gas constant, and T is the temperature in Kelvin.

Since we have STP conditions, we can substitute the values:

(1 atm) × (2.5 L) = n × (0.0821 L·atm/(mol·K)) × (273.15 K).

Simplifying the equation:

2.5 = n × 22.4149.

Solving for n (the number of moles):

n = 2.5 / 22.4149 ≈ 0.1116 moles.

Next, we can calculate the number of molecules using Avogadro's number:

Number of molecules = n × N_A.

Number of molecules = 0.1116 moles × (6.022 × 10^23 particles/mol).

Number of molecules ≈ 6.72 × 10^22 molecules.

Therefore, there are approximately 6.72 × 10^22 molecules of CO2 in 2.5 L at STP.

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Answer:

see below

Explanation:

The lake will have more total thermal energy but the particles in the coffee will be moving faster.

A comparison of hot coffee and cold lake is, the heat from the coffee will be absorbed by the cold lake through convection method of heat transfer.

The principle of conservation of energy states that the total energy of an isolated system is always conserved.

Heat lost by the hot coffee = Heat absorbed by the cold lake

Heat from the coffee will be absorbed by the cold lake through convection method of heat transfer.

Thus, a comparison of hot coffee and cold lake is, the heat from the coffee will be absorbed by the cold lake through convection method of heat transfer.

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Answer:In ionic compounds, electrons are transferred between atoms of different elements to form ions. But this is not the only way that compounds can be formed. Atoms can also make chemical bonds by sharing electrons equally between each other. Such bonds are called covalent bonds. Covalent bonds are formed between two atoms when both have similar tendencies to attract electrons to themselves (i.e., when both atoms have identical or fairly similar ionization energies and electron affinities). For example, two hydrogen atoms bond covalently to form an H2 molecule; each hydrogen atom in the H2 molecule has two electrons stabilizing it, giving each atom the same number of valence electrons as the noble gas He.

Compounds that contain covalent bonds exhibit different physical properties than ionic compounds. Because the attraction between molecules, which are electrically neutral, is weaker than that between electrically charged ions, covalent compounds generally have much lower melting and boiling points than ionic compounds. In fact, many covalent compounds are liquids or gases at room temperature, and, in their solid states, they are typically much softer than ionic solids. Furthermore, whereas ionic compounds are good conductors of electricity when dissolved in water, most covalent compounds are insoluble in water; since they are electrically neutral, they are poor conductors of electricity in any state.

C. 51.5%  is the percent yield for the reaction below when 364 g SO2 and 42.0 g O2 produce 408 g SO3

The actual yield is determined by calculating the quantity of the product created. We can estimate the percentage yield by dividing the actual yield by the theoretical value. The percentage yield is the difference between the quantity of product that was actually created and the maximum calculated yield.

2SO₂(g) + O₂(g) → 2SO3(g)

SO₂ mass => 364g

O₂ mass => 42g

SO3 mass => 408g

Theoretical yield of SO3:

mole of  O₂ => 42g/32g => 1.31

according to the chemical equation:

1 mole of oxygen => 2 moles of SO3

1.31 mole O2 => x moles of SO3

x=> 2 x 1.31 => 2.62

Hence, the mass of SO3 => 2.62 x 80.06 => 209.75g

The percentage of yield => Actual yield / theoretical yield x 100

                                  => 209.75/408 x 100 => 51.45 => 51.5%

The theoretical yield is the yield that is derived using a balanced chemical reaction. What you actually acquire from a chemical reaction is the actual yield.

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Answer:

the answer is c

Energy is transferred from organisms in a food chain and chemical compounds are recycled

The balanced chemical equation for the reaction between strontium and oxygen can be written as follows: 2 Sr (s) + \(O_2\)(g) → 2 SrO (s).

In this equation, solid strontium (Sr) reacts with gaseous oxygen (\(O_2\)) to produce solid strontium oxide (SrO).

To determine the mass of product expected to form in this reaction, we need to consider the molar ratio between strontium and strontium oxide. From the balanced equation, we can see that 2 moles of strontium react to produce 2 moles of strontium oxide.

The molar mass of strontium (Sr) is 87.62 g/mol, and the molar mass of strontium oxide (SrO) is 119.62 g/mol. Since the molar ratio is 1:1 between strontium and strontium oxide, the mass of strontium oxide formed will be equal to the mass of strontium used.

In this case, the student added 4.93 g of strontium to the crucible. Therefore, the expected mass of strontium oxide formed will also be 4.93 g.

It's important to note that this calculation assumes that the reaction proceeds to completion, meaning that all of the strontium reacts with oxygen. In actual laboratory conditions, the yield of the reaction may be less than 100% due to factors such as incomplete reaction, side reactions, or product loss.

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Answer:

c

Explanation:

Answer:

1. Increase in the temperature of the water

2. Increasing the surface area of the lithium

Explanation:

1. Increase in the temperature of the water

The activation energy for the lithium water reaction is +161 kJ/mol while the activation energy for the sodium is +109 kJ/mol, hence for increased reaction rate, the water temperature will be raised to enable more lithium atoms enter into reaction with the water molecules as their energy is increased lowering the activation energy required for the reaction.

2. Increasing the surface area of the lithium

As the lithium floats on the water, due to its low temperature and the heat evolved from the reaction of lithium with the cold water is below the melting point of lithium, the reaction rate can be increased by increasing the surface area of lithium sample by grinding so as to increase the number of lithium water reaction sites.

Answer

The percent composition of Fe = 26.73%

The percent composition of C = 22.99%

The percent composition of H = 4.34%

The percent composition of O = 45.94%

Explanation

Given:

The compound is Fe(C₂H₃O₂)₃

What to find:

The percent composition of each of the elements in Fe(C₂H₃O₂)₃

Step-by-step solution:

Step 1: Calculate the total of Fe(C₂H₃O₂)₃ using the atomic masses of each of the elements in the periodic table.

Atomic mass of: (H = 1.008, C = 12.01, O = 15.999, Fe = 55.845)

Total mass of Fe(C₂H₃O₂)₃ = Fe + 4(C) + 9(H) + 6(O)

Total mass of Fe(C₂H₃O₂)₃ = 55.845 + 4(12.01) + 9(1.008) + 6(15.999)

Total mass of Fe(C₂H₃O₂)₃ = 55.845 + 48.04 + 9.072 + 95.994

Total mass of Fe(C₂H₃O₂)₃ = 208.951

Step 2: Determine the percent composition of Fe in Fe(C₂H₃O₂)₃

Mass of Fe in Fe(C₂H₃O₂)₃ = 55.845

The percent composition of Fe = 26.73%

Step 3: Determine the percent composition of C in Fe(C₂H₃O₂)₃

Mass of C in Fe(C₂H₃O₂)₃ = 4(C) = 48.04

The percent composition of C = 22.99%

Step 4: Determine the percent composition of H in Fe(C₂H₃O₂)₃

Mass of H in Fe(C₂H₃O₂)₃ = 9(H) = 9.072

The percent composition of H = 4.34%

Step 5: Determine the percent composition of O in Fe(C₂H₃O₂)₃

Mass of O in Fe(C₂H₃O₂)₃ = 6(O) = 95.994

The percent composition of O = 45.94%

Answer:

Paper is Not Reuseable like plastic.

Explanation:

Photosynthesis is a chemical reaction in which energy from sunlight is used to convert carbon dioxide and water into glucose and oxygen. This process occurs in the chloroplasts of plant cells and is essential for the survival of plants, as well as for many other organisms that depend on plants for food.

The percent by mass of nitrogen in ammonia (NH3) is approximately 82.15%

Calculating the mass of nitrogen to the total mass of the compound and then expressing the result as a percentage will allow us to determine the percent by mass of nitrogen in NH3 (ammonia).

Ammonia's molecular structure, NH3, indicates that it is made up of one nitrogen atom (N) and three hydrogen atoms (H). We must take both the molar masses of nitrogen and ammonia into account when calculating the percent by mass of nitrogen.

Nitrogen's (N) molar mass is roughly 14.01 g/mol. The molar masses of nitrogen and hydrogen are added to determine the molar mass of ammonia (NH3). Since hydrogen's molar mass is around 1.01 g/mol, ammonia's molar mass is:

(3 mol H 1.01 g/mol) + (1 mol N 14.01 g/mol) = 17.03 g/mol = NH3.

Now, we can use the following formula to get the nitrogen content of ammonia in percent by mass:

(Mass of nitrogen / Mass of ammonia) / 100% is the percentage of nitrogen by mass.

Ammonia weighs 17.03 g/mol and contains 14.01 g/mol of nitrogen by mass. By entering these values, we obtain:

(14.01 g/mol / 17.03 g/mol) 100% 82.15 % of nitrogen by mass

Ammonia (NH3) has a nitrogen content that is roughly 82.15 percent by mass.

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Answer:

a

Explanation:

Answer:

pH change is -0.07

Explanation:

Using H-H equation for acetic acid:

pH = pKa + log [Acetate salt] / [Acetic acid]

Replacing:

pH = 4.74 + log[1.188M] / [1.188M]

pH = 4.74

The HCl reacts with sodium acetate producing acetic acid, thus:

HCl + CH₃COONa → CH₃COOH + NaCl

That means the final moles of sodium acetate are initial moles - moles of HCl and moles of acetic acid are initial moles + moles of HCl.

As the volume of the buffer is 1.0L, initial moles of both substances are 1.188moles. After reaction, the moles are:

sodium acetate: 1.188mol - 0.1mol = 1.088mol

Acetic acid: 1.188mol + 0.1mol = 1.288mol

Using again H-H equation:

pH = 4.74 + log[1.088M] / [1.288M]

pH = 4.67

pH change is: 4.67 - 4.74 = -0.07

Syringe tips and titrant solution in a container,  unused solid reagent in the manufacturer's container and  broken flask in the cardboard box.

Syringe tips are dispose into the disposal containers present in the lab or hospitals.  Unused solid reagent is dispose in the manufacturer's container and then label the container with a hazardous waste tag so that people can't use it and stay away from it.  

Broken flask is dispose in the cardboard box or a container.  Titrant solution can also be dispose by placing in a container which is leak proof and closed the opening tightly so that it can't pollute and damaged the outside world.

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The places to dispose each type of laboratory waste are as follows :

Waste                                           Disposal destination

syringe tips                                      container ( disposal )

unused solid reagent                      Manufacturer's container

Broken flask                                     Cardboard box

Titrant solution                                 leak proof container

These types of wastes are found in a science laboratory or can be found in a research or medical laboratory. To avoid/reduce the level of potential hazardous situations in the Laboratory proper waste disposal should be practiced.

Unused solid reagent should be properly disposed into the manufacturer's container and the container should be labeled properly ( using a hazardous waste tag ).  and

 For proper disposal of titrant solution it should be placed in a leak proof container and sealed very tight to prevent air pollution of the laboratory and  the environment.

Hence we can conclude that  where to dispose each of the type of waste respectively are : container ( disposal ),  Manufacturer's container,  Cardboard box and  leak proof container

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Answer:

I  forgot that but I think boron is first d ff\

Explanation:

Explanation:

if it was meant for [Ar]4s²3d¹ :

1s²2s²2p⁶3s²3p⁶4s²3d¹

number of e‐ : 21

atomic number is 21.. element is scandium

ion: Sc³+

Answer:

0.856

Explanation:

Given data:

Feed molar rate ( Fao ) = 10 mol/min

Feed concentration ( Cao ) = 0.4 mol/dm^3

current conversion rate = 80%  

Temperature = 305 K

catalyst = 100 kg

entering pressure = 20 atm

exit pressure = 2 atm

Determine the conversion if the catalyst particle size were doubled and everything else remained the same

The conversion if the catalyst particle size were doubled and other factors remains the same = 0.856

attached below is a detailed solution

sorry what us this i don't get it

The calculated formula masses to 133.5 amu, we find that the substance with a formula mass closest to 133.5 amu is (d) AlCl3. Therefore, the answer is option D.

To identify the substance with a formula mass of 133.5 amu, we need to calculate the formula mass of each given substance and compare it to 133.5 amu.

(a) MgCl2:

The formula mass of MgCl2 can be calculated by adding the atomic masses of magnesium (Mg) and chlorine (Cl).

Mg: atomic mass = 24.31 amu

Cl: atomic mass = 35.45 amu

Formula mass of MgCl2 = (24.31 amu) + 2(35.45 amu) = 95.21 amu

(b) SCl:

The formula mass of SCl can be calculated by adding the atomic masses of sulfur (S) and chlorine (Cl).

S: atomic mass = 32.07 amu

Cl: atomic mass = 35.45 amu

Formula mass of SCl = 32.07 amu + 35.45 amu = 67.52 amu

(c) BCl:

The formula mass of BCl can be calculated by adding the atomic mass of boron (B) and chlorine (Cl).

B: atomic mass = 10.81 amu

Cl: atomic mass = 35.45 amu

Formula mass of BCl = 10.81 amu + 35.45 amu = 46.26 amu

(d) AlCl3:

The formula mass of AlCl3 can be calculated by adding the atomic mass of aluminum (Al) and 3 times the atomic mass of chlorine (Cl).

Al: atomic mass = 26.98 amu

Cl: atomic mass = 35.45 amu

Formula mass of AlCl3 = 26.98 amu + 3(35.45 amu) = 133.78 amu. Option D

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Answer:

77.53 g of Na₂O is the answer to your question

The SO2 Lewis structure would consist of two oxygen (O) atoms and one sulfur atom.

Both the sulfur and oxygen atoms have six reactivity electrons. The molecular geometry of sulfur dioxide is a bent form. The sulfur-to-oxygen ratio in sulfur dioxide. Sulfur dioxide and SiCl4 have 1 lone pair and 2 bond pairs.SO2 molecular addition is considered to V-shaped or bent. otherwise, the electron geometry of sulfur dioxide is in the shape of a trigonal planar. The three pairs of bonding electrons lie at an angle of 119o.

The bond sequence and 1.2 in SO2 and the sulfonyl group, separately. Thus, electrostatic forces (and not multiple covalent bonding) give more to the SÀO bond shortening. In the SO2 lewis structure, there are two double bonds linking the sulfur atoms and oxygen atoms. The sulfur dioxide lewis structure is drawn step by step using VESPR

So we can conclude that The electron dot form of the SO2 molecule is also known as the SO2 Lewis structure.

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Answer: The changes are the mL of methanol ingestion (30mL) and the proof of the whiskey (40% ethanol; 80 proof), and the percentage of reduction (1%). I applied the formulas from the last post in reference to this question but am completely lost. The formulas used last time were: 2. Formulas Applied alpha=1 + ([etOH]/KetOH) (V[meOH]/V[etOH])= (Vmax*[meOH]/KmeOH+[meOH])/(Vmax*[meOH]/alpha*KmeOH+[meOH]) which reduces to (V[meOH]/V[etOH])=(alpha*KmeOH+[meOH])/(KmeOH+[meOH]) 3. My attempt Molarity of methanol: 30mL; which equates to 23.7g of methanol; in 40L that is equal to 0.5925 g/L Dividing the molecular weight by 32.04g/mol I get 0.0184925 which is approximately 0.02M; Km is 0.01M Since the molar mass of methanol and ethanol are two fold, I can multiply the g/l by 4. However, unlike the previous problem, I cannot multiply by 2 because I do not have 50% EtOH, so because 40 is less than 50 I assume to multiply by 2.5 yielding: (30mL)(4)(2.5)=300mL 300mL of EtOH to effectively reduce the Methanol to 1%.

Answer:

Primary alcohols can be oxidized to form aldehydes and carboxylic acids; secondary alcohols can be oxidized to give ketones. Tertiary alcohols, in contrast, cannot be oxidized without breaking the molecule's C–C bonds.