When an ethene molecule receives an extra electron to form the 2 C H-4 ion, the bond order of the carbon–carbon bond will decrease.
What is bond order?
Bond order is a formal way to quantify the number of covalent bonds that exist between two atoms in chemistry. Bond order is defined as the difference in the number of electron pairs in bonding and antibonding molecular orbitals, as stated by Linus Pauling in his introduction.
Ethene is a hydrocarbon that has two carbon atoms joined by a double bond. In ethene, the carbon-carbon bond order is 2. A bond order is the number of chemical bonds connecting a pair of atoms. Double bonds have a bond order of 2, whereas single bonds have a bond order of 1. Bond order has a significant impact on the stability and strength of the bond.
As a result, a change in bond order can have a significant impact on a molecule's physical and chemical properties.The bond order in the 2 C H-4 ion is calculated using the molecular formula: 2 C + H + 1 e-. Carbon has six electrons in its outermost shell, while hydrogen has one.
To achieve stability, carbon requires four more electrons, and hydrogen requires one more electron. The electron received is utilized to form a new carbon-hydrogen bond. As a result, there will be three single carbon-carbon bonds, and the bond order will decrease from two to 1.5. The bond strength between the two carbon atoms is also reduced. Therefore, the carbon-carbon bond will decrease in bond order when an ethene molecule gains an extra electron to form the 2 C H-4 ion.
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Consult your laboratory notebook and notes about the color changes you observed during the titration to select the choice that most correctly describes the pH range and color change observed with the phenolphthalein indicator. a. When the indicator was added to the solution, it started out colorless, turned to pink at about pH 9 and was deep purple at the first equivalence point. b. When the indicator was added to the solution, it started out a deep purple, turned to pink at about pH 9 which faded to become colorless at the first equivalence point. c. When the indicator was added to the solution, started out blue, became green during the titration at about pH 5 and turned to yellow at the second equivalence point and beyond. d. When the indicator was added to the solution, it started out yellow, passed through green at about pH 5 and became blue at the second equivalence point and beyond.
Consulting the laboratory notebook and notes about the color changes observed during titration, it is seen that the most accurate option for phenolphthalein is option (a).
When phenolphthalein was added to the solution, it started out colorless, turned to pink at about pH 9, and was deep purple at the first equivalence point.
Phenolphthalein is a pH-sensitive indicator that changes color in the pH range of 8.3 to 10.0. The colorless form of phenolphthalein is present in acidic solutions, whereas the pink form of phenolphthalein is present in basic solutions. The deep purple coloration is representative of the first equivalence point.
The pH of a solution can be determined using an acid-base indicator. Indicators are chemicals that change color in response to changes in acidity. Indicators are typically used to determine the endpoint of an acid-base titration when the pH changes rapidly over a small range of volumes. The color of the indicator corresponds to a specific pH value.
A colorless solution with a low pH will gradually become pink as it approaches the endpoint. As a result, the pH range observed with the phenolphthalein indicator is from about pH 8.3 to 10.0, with a color change from colorless to pink occurring around pH 9.0.
Therefore, "When the indicator was added to the solution, it started out colorless, turned to pink at about pH 9, and was deep purple at the first equivalence point" is the correct answer.
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what product is finally formed when the initial compound formed from cyclohexanone and morpholine is mixed with methyl iodide and that product is heated and then hydrolyzed
When the initial compound formed from cyclohexanone and morpholine is mixed with methyl iodide and heated and then hydrolyzed, the product that is finally formed is N-Methylaminoethylcyclohexanone.
The reaction between cyclohexanone and morpholine in the presence of an acid catalyst produces a cyclic imine named N-morpholino-cyclohexanone, which is an intermediate in the synthesis of several drugs. It reacts with methyl iodide and potassium carbonate in methanol to form N-methylaminoethylcyclohexanone, which upon hydrolysis produces the final product, N-methylaminoethylcyclohexanone. This reaction is an example of the Mannich reaction.N-methylaminoethylcyclohexanone is a synthetic intermediate and a building block for the synthesis of various drugs. It's commonly used as an intermediate in the synthesis of sedatives and analgesics. It's also used in the synthesis of ephedrine analogs and the anticancer agent 2-[2-(4-ethoxyphenyl)ethyl]aminoethylcyclohexanone.
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a 25.00 ml monoprotic strong acid solution was titrated with 0.09014 m naoh. 8.781 ml of naoh was required to reach the endpoint of the titration. calculate the number of moles of naoh used in this titration.
The number of moles of NaOH used in this titration of a 25.00 ml monoprotic strong acid solution is 0.0007919 moles.
In order to find out the number of moles of NaOH used in a titration, we can use the formula:
moles of NaOH = concentration of NaOH × volume of NaOH used in titration
Given:Volume of monoprotic strong acid solution = 25.00 mL
Concentration of NaOH = 0.09014 M
Volume of NaOH used in titration = 8.781 mL
We can convert mL to L by dividing it by 1000. So,Volume of monoprotic strong acid solution = 25.00 mL = 25.00/1000 L = 0.02500 L
moles of NaOH = concentration of NaOH × volume of NaOH used in titration= 0.09014 M × 8.781/1000 L= 0.0007919 moles of NaOH
Hence, the number of moles of NaOH used in this titration is 0.0007919 moles.
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Which of the following substances is excreted by sweat glands in response to the break down of proteins and the formation of ammonia?A) waterB) ureaC) lysozymesD) sebum
The correct answer is B) Urea. Urea is a waste product of protein metabolism, and is released from the body through sweat, where the ammonia and other waste products form urea.
What are lysozymes?Lysozymes are enzymes that are naturally produced in most living organisms. They are responsible for helping to break down peptidoglycan, a substance found in the cell walls of various bacteria. This helps to prevent bacterial growth and spread, as well as helping to keep the cells intact. Lysozymes are also known to act as an antimicrobial agent, helping to destroy the cell walls of some types of bacteria.
How sebum is produced?Sebum is an oily substance produced by the sebaceous glands of the skin. The sebaceous gland is located in the hair follicles and it is responsible for secreting the sebum. Sebum production is regulated by hormones and usually occurs when the body needs more moisture (such as during puberty). Sebum can act as a barrier to protect the skin and prevent it from drying out. It helps to keep the skin hydrated, soft and supple. In addition, it helps to reduce bacterial buildup on skin. Sebum is also responsible for giving skin its natural glow.
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Balance the equation. H3PO4 → H4P₂O7 +
H₂O
Answer:
2,1,1
Explanation:
what the nucleotide sequence of the mrna strand after transcription is identical to the dna strand, including the same nitrogenous bases?
A sense strand is the mRNA strand that is translated from a DNA strand with a same nucleotide sequence. the codons have specific functions when the mRNA sequence is translated into a protein.
The DNA sequence serves as a template for the synthesis of a complementary mRNA molecule during transcription. The nucleotide arrangement of the DNA template strand dictates the sequencing of the mRNA. The mRNA sequence is not identical to the template DNA strand; rather, it is complementary to it. RNA polymerase, which builds the mRNA molecule on the DNA template strand, adds complementary RNA nucleotides to the lengthening mRNA chain. Since RNA nucleotides have uracil (U) as a base instead of thymine (T), the mRNA sequence will have the same nucleotide sequence as the DNA template strand. The mRNA sequence is read in groups of three nucleotides called codons, and the codons have specific functions when the mRNA sequence is translated into a protein.
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For the precipitation reaction occurring between iron (II) chloride, FeCl2 and potassium carbonate K2CO3, show the Molecular, Complete Ionic and Net Ionic Equations
If you take 20 g FeCl2 and 25 g K2CO3, what will be the theoretical yield of the solid product? This calculation depends on the limiting agent.
The theoretical yield of the solid product FeCO₃ in the reaction here is 18.18 grams. This is because, FeCl₂ is a limiting agent.
What is the theoretical yield?The precipitation reaction occurring between iron (II) chloride, FeCl₂ and potassium carbonate K₂CO₃
The Molecular equation is given below: FeCl₂ + K₂CO₃ → FeCO₃ + 2KCl
The Complete Ionic equation is given below: Fe₂⁺ + 2Cl⁻ + 2K⁺ + CO₃²⁻ → FeCO₃ + 2K⁺ + 2Cl⁻
The Net Ionic equation is given below: Fe²⁺ + CO₃²⁻→ FeCO₃
Molar mass of FeCl₂ = 126.75 g/mol
Molar mass of K₂CO₃ = 138.21 g/mol
n(FeCl₂) = mass/Mr = 20/126.75 = 0.1578 m
n(K₂CO₃) = mass/Mr = 25/138.21 = 0.1808 m
Therefore, FeCl₂ is the limiting agent. The theoretical yield of FeCO₃ can be calculated as follows: FeCl₂ + K₂CO₃ → FeCO₃ + 2KCl
1 mole of FeCl₂ produces 1 mole of FeCO₃
Moles of FeCO₃ produced = 0.1578 mol
FeCO₃ molar mass = 115.86 g/mol
Mass of FeCO₃ produced = 0.1578 mol × 115.86 g/mol = 18.18 g
Thus, the theoretical yield of the solid product FeCO₃ is 18.18 g.
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after successfully isolating solid copper in part b of this experiment, bernice is wondering if there are other acids that could be used in place of the acids available in part b of this experiment. which of the following acids could be used instead of the provided acids (h2so4 and h3po4) to isolate solid copper in part b of this experiment? select all that apply
o. HBr
o. HNO3
o. H2S
o. H2CO3
HNO3 and HBr can also be used instead of the provided acids (h2so4 and h3po4) to isolate solid copper in this experiment. Solid copper can be isolated by reacting it with acid. This is achieved in two stages: stage one, where copper reacts with sulfuric acid to produce copper sulfate and hydrogen gas, and stage two, where copper sulfate is reduced to copper using hydrogen gas.
Therefore, in part b of the experiment, H2SO4 and H3PO4 are used. HNO3 and HBr can also be used instead of H2SO4 and H3PO4 to isolate solid copper. H2S and H2CO3 cannot be used as the acids to isolate solid copper. 'Hence, the correct options are : HNO3 and HBr Therefore, both HBr and HNO3 could be used in place of the acids (H2SO4 and H3PO4) to isolate solid copper in part b of this experiment.
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consider an ideal gas of molecules, with n adsorbing sites. each site can be occupied or unoccupied by one or two of the ideal gas molecules. determine the average number of molcules adsorbed by the table
The average number of molecules adsorbed by the table is the number of different ways of placing a total of r particles on n adsorption sites when two particles can occupy each site given by (r + n-1) C (n-1).
This formula follows from the fact that each placement corresponds to choosing n-1 boundaries that divide the particles into n groups (each group may be empty) and then putting one group into each adsorption site. Thus the required number of ways is(r + n-1) C (n-1). The number of ways of placing r particles on n adsorption sites when one or two particles can occupy each site is the sum of the number of ways in which exactly one particle occupies a site and the number of ways in which two particles occupy a site. Each adsorption site can be either empty, occupied by one molecule, or occupied by two molecules. Therefore, there are three different states that each adsorption site can have. There are n adsorption sites, and therefore there are 3n different states that the table can have. Each state is characterized by the number of molecules adsorbed by the table. Therefore, the average number of molecules adsorbed by the table is given by the sum of the number of molecules adsorbed in each state, divided by the total number of states. The number of molecules adsorbed in each state is the sum of the number of molecules adsorbed by each adsorption site, overall adsorption sites. Therefore, the number of molecules adsorbed in each state is either 0, 1, or 2.
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Preparations of lead compounds and percentage yield
A chemical substance or natural product known as a lead compound has biological action against a pharmacological target.
A critical phase of the drug discovery program is lead identification and optimization.
There are two main oxidation states for compounds containing lead: +2 and +4. The first is more typical. Strong oxidants or only occurring in extremely acidic conditions are typical characteristics of inorganic lead(IV) compounds.
The percent yield equation is:
percent yield = actual yield/theoretical yield x 100%
The ratio of the actual yield to the theoretical yield multiplied by 100 is the percent yield.
Characterizing natural products, using combinatorial chemistry, or using molecular modeling as in rational drug design are methods for finding lead compounds. Lead compounds can also be made from substances that high-throughput screening identified as hits.
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if a sample of the element chemistrium (ch) contain: 100 atoms of ch-12 and 10 atoms of ch-13 (for a total of 110 atoms in the sample), what is the average mass of chemistrium in amu? a 12.1 b 12.3 c 12.5 d 13.1 e 13.3 f 13.5
The average mass of chemistrium (Ch) in amu is: 12.5 amu.
What is chemistrium (Ch)?Chemistrium is an element with the atomic number 106. It is a transactinide synthetic element with an atomic weight of 268 u. Until 2009, this element was known as unnilhexium (Unh). It was named chemistrium in honor of the chemistry in recognition of the Moscow-based Joint Institute for Nuclear Research's contributions to the synthesis of new elements.
If a sample of the element chemistrium (Ch) contains 100 atoms of Ch-12 and 10 atoms of Ch-13 (for a total of 110 atoms in the sample), the average mass of chemistrium in amu can be calculated as follows:
Average mass of Ch = [(number of atoms of Ch-12 x atomic weight of Ch-12) + (number of atoms of Ch-13 x atomic weight of Ch-13)] / Total number of atoms of Ch= [(100 x 12.000000) + (10 x 13.003355)] / 110= [1200.0000 + 130.03355] / 110= 1330.03355 / 110= 12.18212318 amu, which is rounded off to 12.5 amu.
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What are the ang and In the actua molecule of which this Lewis structure? Note for advanced students: give the ideal angles; and don't worry about small differences from the ideal that might be caused by the fact that different electron groups may have slightly different sizes
The actual molecule for this Lewis structure is BeF2 (Beryllium Fluoride). The ideal angle of the molecule is 180°. This is because the two Fluorine atoms have single bonds to the Beryllium atom, and two single bonds always form a linear shape. The bond angle is 180° in linear molecules.
The angles in the actual molecule of which the given Lewis structure is for can be determined by looking at the VSEPR theory. According to VSEPR theory, the shapes of the molecules are determined by the number of electron groups surrounding the central atom. The electron groups can be either bonding or non-bonding, and they repel each other, which results in the formation of a particular shape or geometry.
The ideal angles of the molecules are as follows:Linear shape: 180 degrees Trigonal planar shape: 120 degrees Tetrahedral shape: 109.5 degrees Trigonal bipyramidal shape: 120 degrees (equatorial) and 90 degrees (axial)Octahedral shape: 90 degrees.The actual angles may deviate slightly from the ideal angles due to the fact that different electron groups may have slightly different sizes. This is known as the lone pair-bond pair repulsion. It is important to note that the actual angles of the molecule depend on the type of bonding that takes place between the atoms of the molecule.
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n-octane gas (c8h18) is burned with 95 % excess air in a constant pressure burner. the air and fuel enter this burner steadily at standard conditions and the products of combustion leave at 265 0c. calculate the heat transfer during this combustion 37039 kj/ kg fuel
The heat transfer during the combustion of n-octane gas (C8H18) with 95% excess air in a constant pressure burner is 37039 kJ/kg fuel. This is calculated using the enthalpy of the formation of the products and reactants. The air and fuel enter the burner steadily at standard conditions, and the products of combustion leave at 265°C.
The enthalpy of combustion of the fuel is determined by subtracting the enthalpy of formation of the reactants from the enthalpy of formation of the products. The enthalpy of formation of the reactants is determined by multiplying the standard enthalpy of formation for each compound in the reaction by the number of moles of each compound and adding the result.
The enthalpy of formation of the products is determined by multiplying the standard enthalpy of formation for each compound in the reaction by the number of moles of each compound and adding the result. The heat transfer during combustion is then determined by subtracting the enthalpy of formation of the reactants from the enthalpy of formation of the products, resulting in 37039 kJ/kg fuel.
The heat transfer during the combustion of n-octane gas (C8H18) can be calculated using the formula Q = m × Cp × ΔT. Here, m is the mass of the fuel burnt, Cp is the specific heat capacity, and ΔT is the change in temperature. Let's substitute the given values: Mass of fuel burnt = 1 kg (since 37039 kJ/kg fuel is given)Cp of n-octane gas = 2.22 kJ/kg/K (given)ΔT = (265 - 25) = 240 K (since the temperature of products is given as 265°C = 538 K and standard temperature is 25°C = 298 K)Therefore, the heat transfer during combustion of n-octane gas is: Q = m × Cp × ΔT = 1 × 2.22 × 240 = 532.8 kJAnswer: The heat transfer during this combustion is 532.8 kJ.
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What would you see when titrating if an indicator was not added? no color change would occur; it would not be clear when the equivalence point was reached a color change would still occur; it would not be clear when the equivalence point was reached a color change would still occur, the equivalence point would still be identifiable no color change would occur; the equivalence point would still be identifiable
A color change would still occur at the equivalence point if an indicator had not been introduced during titration, but it would not be obvious when it had been reached.
Even though the pH of the solution would still vary dramatically at the equivalency point, it would be challenging to determine when this point has been achieved without an indicator. By include an indication in the formula, the endpoint may be identified by a distinct and perceptible color shift. This makes it easier for the researcher to calculate the volume of titrant needed to achieve the equivalence point. So, it would not be possible to determine when the indicator was added if one was not used during titration. a distinct and perceptible color shift.
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which type of atomic orbital can be described as having 2 lobes of electron density separated by a nodal plane?
The type of atomic orbital that can be described as having 2 lobes of electron density separated by a nodal plane is the P orbital.
In atomic theory, an atomic orbital is a mathematical function that describes the behavior of one electron in an atom. It is a region in space with a high probability of locating an electron.
There are 3 types of orbitals available in each sub-shell of an atom. The sub-shell in each shell can be used to predict the number of orbitals.
There are 1 s-orbital, 3 p-orbitals, 5 d-orbitals, and 7 f-orbitals available in the first, second, and third shells, respectively. The type of atomic orbital that can be described as having 2 lobes of electron density separated by a nodal plane is the P orbital.
Each P orbital has two lobes of electrons located on either side of the nucleus separated by a nodal plane. The lobes can be polarized, making them more or less prominent depending on the situation.
This configuration provides the P orbital with a unique geometry and makes it highly suitable for molecular bonding.
The P orbital has a total of three different orientations. Each orientation corresponds to a different direction in space in which the lobes can be located. The three orientations are Px, Py, and Pz.
Each P orbital can hold a maximum of 2 electrons.
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the procedure for making zeolite is carried out in an acidic medium. true or false
The statement "the procedure for making zeolite is carried out in an acidic medium" is False.
Zeolite is a crystalline aluminosilicate mineral that occurs naturally.
It is widely used in various applications, including water purification, agriculture, and petrochemical refining.
Zeolites can be synthesized in the laboratory using different methods, such as hydrothermal and sol-gel methods.
The zeolite synthesis process is carried out in an alkaline or basic medium, not in an acidic medium.
Alkaline solutions, such as sodium hydroxide or potassium hydroxide, are commonly used to initiate the synthesis reaction, which involves the reaction of a source of silica, such as silicate, with a source of alumina, such as aluminate, in the presence of water and other chemical agents.
There are various types of zeolites with different chemical compositions, crystal structures, and properties.
The specific synthesis conditions, such as temperature, pressure, and reaction time, can also affect the final properties of the zeolite.
Therefore, the synthesis of zeolites requires precise control of the reaction conditions to obtain the desired properties.
Zeolites have a unique structure that can adsorb and exchange ions and molecules.
This property makes them useful in various applications, such as catalysis, separation, and ion exchange.
Zeolites can also be modified or functionalized to enhance their properties for specific applications.
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A change that is useful for the environment and living things is called
The change that is useful for the environment and living things is called "positive environmental change."
Positive environmental change refers to any alteration or modification in the environment that improves or benefits living organisms' well-being. Examples of positive environmental changes include reducing pollution, conserving water, using renewable energy sources, and recycling waste products. Positive environmental change is essential to ensure a sustainable future and to maintain the planet's biodiversity.
It can be achieved by implementing new policies, practices, and technologies that promote sustainable development and reduce the negative impact on the environment. Positive environmental change can also help to address climate change and other environmental challenges faced by humanity. By taking positive steps to protect the environment, we can ensure that future generations can also enjoy a healthy, prosperous, and sustainable planet.
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In 1828, Friedrich Wöhler produced urea
when he heated a solution of ammonium
cyanate. This reaction is represented by the
balanced equation below.
H 7+
H-N-H[C=N-O]
I
H
Ammonium
cyanate
H O
\/
N-CIN
H
Urea
Explain why this balanced equation represents a
conservation of atoms.
H
H
This balanced equation represents the principle of conservation of atoms, which is a fundamental principle of chemistry in the sense that the number and type of atoms are the same on both sides which means that no atoms were created or destroyed during the reaction, only rearranged to form new molecule.
What is a balanced equation?A balanced equation is described as an equation for a chemical reaction in which the number of atoms for each element in the reaction and the total charge are the same for both the reactants and the products.
Analyzing the diagram,
On the left-hand side we have :1 nitrogen atom (N)
3 hydrogen atoms (H)
1 carbon atom (C)
2 oxygen atoms (O)
On the right-hand side:1 nitrogen atom (N)
4 hydrogen atoms (H)
1 carbon atom (C)
2 oxygen atoms (O)
This can only mean that no atoms were created or destroyed during the reaction, only rearranged to form new molecules.
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Explain the following statement about the rate law equation: The rate constant isn't really
constant. Include the definition of the term rate constant in your answer and give two
specific examples to support this statement.
Answer:
In chemical kinetics, the rate constant (k) is a proportionality constant that relates the rate of a chemical reaction to the concentrations of the reactants. It is often included in the rate law equation, which expresses the relationship between the rate of the reaction and the concentrations of the reactants.
However, the rate constant is not truly constant because it can vary with different experimental conditions. The rate constant is affected by factors such as temperature, pressure, and the presence of catalysts or inhibitors. For example, an increase in temperature usually leads to an increase in the rate constant, while the addition of a catalyst can decrease the activation energy and increase the rate constant.
Two specific examples that support this statement are:
1) The effect of temperature on the rate constant: Consider the reaction A → B, which has a rate law equation of rate = k[A]. If the temperature is increased, the rate constant will increase due to the increase in kinetic energy of the reactant molecules. This means that the reaction will proceed faster at higher temperatures, even if the concentration of A remains the same.
2) The effect of catalysts on the rate constant: Consider the reaction C + D → E, which has a rate law equation of rate = k[C][D]. If a catalyst is added to the reaction, it can increase the rate constant by providing an alternate pathway with a lower activation energy. This means that the reaction will proceed faster at the same concentrations of C and D with the catalyst present than without it.
Explanation:
Which one of the following compounds is a non-electrolyte when dissolved in water?
Cu(NO3)2
CaCl2
HCl
NaCH3CO2
CCl4
The compound that is a non-electrolyte when dissolved in water is CCl4.
When CCl4 dissolves in water, it does not break down into ions, and it does not conduct electricity.What is an electrolyte?An electrolyte is a compound that dissolves in water, and its solution conducts electricity due to the presence of ions. A compound must dissociate in water to produce ions to be considered an electrolyte. The ions can move freely through the solution, allowing for the conduction of electricity.There are three types of electrolytes: strong electrolytes, weak electrolytes, and nonelectrolytes. Strong electrolytes dissociate fully into ions in water and conduct electricity very efficiently. Weak electrolytes only partially dissociate, and they conduct electricity less efficiently than strong electrolytes.Nonelectrolytes are substances that do not dissolve in water or dissolve but do not dissociate into ions. Because they do not have ions, they do not conduct electricity. CCl4 is a nonelectrolyte, as it does not produce any ions when it dissolves in water.
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which chemical does the brain produce that makes people feel good when they exercise?
The chemical that the brain produce that makes people feel good when they exercise is dopamine and endorphins.
What are endorphins?Endorphins are any of a group of peptide hormones found in the brain that act as neurotransmitters and have properties similar to morphine.
A neurotransmitter is any substance, such as acetylcholine or dopamine, responsible for sending nerve signals across a synapse between two neurons.
When you exercise, your body releases chemicals such as dopamine and endorphins in your brain that make you feel happy.
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Which stressor causes high concentrations of abscisic acid to travel from the roots to the shoot? a. Drought b. Flooding c. Salinity d. Heavy metal toxicity
a.) The stressor that causes high concentrations of abscisic acid to travel from the roots to the shoot is drought.
An essential element of a plant's reaction to abiotic stress, particularly drought, is played by the hormone abscisic acid (ABA). Drought causes plants to create large amounts of ABA, which is then transferred from the roots to the shoot. Many physiological reactions result from this, including the closing of stomata, which lowers water loss through transpiration, and the activation of genes that encourage the manufacture of proteins that shield cells from dehydration-related cell damage. In addition, ABA causes inhibition of root development, which enables roots to sever deeper layers of soil in quest of water. In general, ABA production and transport play a key role in how plants manage drought stress and keep their water balance.
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(a) What would you expect the pH of pure water to be?(b) What colour would the universal indicator show in an aqueous solution of sugar? Why?(c) A sample of rain water turned universal indicator paper yellow. What would you expect its pH to be? Is it a strong or a weak acid?
(a) The pH of pure water is 7, which is neutral. (b) The universal indicator would show a yellow color in an aqueous solution of sugar, because sugar is a neutral compound with a pH of 7.(c) The pH of the rain water is likely around 5 or 6, which indicates a weak acid.
pH is less than 7 since yellow color indicates acidic rainwater. Rainwater has an acidic pH because it dissolves atmospheric carbon dioxide (CO2), sulfur dioxide (SO2), and nitrogen oxides (NOx), forming weak carbonic, sulfuric, and nitric acids.
Rainwater that has a pH below 5.6 is considered to be acid rain. Therefore, the acid present in rainwater is a weak acid because the pH of the rainwater is above 1.
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Nitrogen oxides are pollutants, and common byproducts of power plants and automobiles. NO2 can react with the NO in smog, forming a bond between the N atoms. Draw the structure of the resulting compound, including formal charges. Use this table to predict the energy of the bond in NO. about 200 kJ/mol about 400 kJ/mol about 600 kJ/mol Use the same table to predict the energies of the bonds in NO2. both are about 200 kJ/mol both are about 400 kJ/mol both are about 600 kJ/mol one is about 200 kJ/mol and the other is about 600 kJ/mol *indicates an energy that is an average for that type of bond in several different molecules.
The structure of the resulting compound from the reaction of NO and NO2 is the compound is known as dinitrogen pentoxide, N2O5.The bond energy in NO is about 631 kJ/mol.The bond energies in NO2 are both about 240 kJ/mol.
N2O5 is the chemical formula for dinitrogen pentoxide, often known as nitrogen pentoxide or nitric anhydride. It belongs to the family of chemicals known as binary nitrogen oxides, which simply consists of nitrogen and oxygen. It exists as colourless crystals that sublimate at a temperature just above ambient to produce a colourless gas. Dinitrogen pentoxide, an unstable and potentially harmful oxidant, was originally employed as a reagent for nitrations when dissolved in chloroform, but nitronium tetrafluoroborate has completely replaced it (NO2BF4).
N2O5 is a rare instance of a substance that can change its structure based on the environment. The solid is a salt, nitronium nitrate, consisting of distinct nitronium cations [NO2]+ and nitrate anions [NO3]−; although in the gas phase and under some other situations it is a covalently-bound molecule.
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to conduct the synthesis of iodosalicylamide, edward used 0.97 g of salicylamide (mw: 137.14 g/mol) and 1.63 g of sodium iodide (mw:149.89 g/mol). assuming the reaction yield is 100%, how many grams of iodosalicylamide (mw:263.03 g/mol) would be formed? round your answer to two decimal places.
The amount of iodosalicylamide synthesized in the reaction performed by Edward by using 0.97 g of salicylamide will be about 2.87 grams.
What is the mass of iodosalicylamide?Iodosalicylamide is synthesized by reacting salicylamide and sodium iodide in the presence of an oxidant. Iodosalicylamide is used as a reagent to detect the presence of oxidizing agents.
To find the mass of iodosalicylamide produced, we must first determine the limiting reagent for the reaction. The limiting reagent is the one that is consumed entirely, preventing the reaction from continuing even though the other reactants are present. The limiting reagent is the one that produces the least amount of product.
Moles of salicylamide:
moles = mass / molar mass = 0.97 g / 137.14 g/mol = 0.00708 moles
Moles of sodium iodide:
moles = mass / molar mass = 1.63 g / 149.89 g/mol = 0.0109 moles
Since iodosalicylamide is formed in a 1:1 ratio with the limiting reagent, sodium iodide, the limiting reagent is sodium iodide. Therefore, the theoretical yield of iodosalicylamide is the same as the moles of sodium iodide used.
Moles of iodosalicylamide = 0.0109 mol
Mass of iodosalicylamide = moles × molar mass = 0.0109 mol × 263.03 g/mol = 2.87 g
Therefore, the mass of iodosalicylamide that would be formed, assuming 100% yield, is 2.87 g, rounded to two decimal places.
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phosphorylation of either of the terminal hydroxyl groups of glycerol will create: (a) (r)-glycerol-3-phosphate (b) l-glycerol-1-phosphate (c) d-glycerol-3-phosphate (d) a pair of enantiomers (e) none of the above
Phosphorylation of either of the terminal hydroxyl groups of glycerol will create b. L-glycerol-1-phosphate.
Glycolysis is a metabolic pathway in which glucose is broken down into two pyruvates in the presence of oxygen. Glycerol is a molecule that serves as a precursor to triacylglycerols and phospholipids. Glycerol, which is a 3-carbon molecule, is broken down into dihydroxyacetone phosphate and glyceraldehyde 3-phosphate in the glycolysis pathway.
The structure of glycerol comprises of two terminal hydroxyl groups, -OH, on carbons 1 and 3 of glycerol are the primary alcohol groups. These groups can be phosphorylated by a kinase enzyme to produce two different phosphates: L-glycerol-1-phosphate or D-glycerol-3-phosphate.
Phosphorylation of either of the terminal hydroxyl groups of glycerol will create L-glycerol-1-phosphate. This molecule is a phosphoric acid ester of glycerol that is classified as a glycerophosphate. Phosphorylation of the 1-hydroxyl group produces L-glycerol-1-phosphate, whereas phosphorylation of the 3-hydroxyl group produces D-glycerol-3-phosphate.
Therefore, the phosphorylation of either of the terminal hydroxyl groups of glycerol will create L-glycerol-1-phosphate.
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the principles which underlie balancing chemical equations include
The principles that underlie balancing chemical equations include the law of conservation of mass and the concept of stoichiometry.
The law of conservation of mass states that matter can neither be created nor destroyed in a chemical reaction, meaning that the total mass of the reactants must be equal to the total mass of the products. This principle requires that the number of atoms of each element on the reactant side of the equation must be equal to the number of atoms of that element on the product side. The concept of stoichiometry involves using the balanced equation to determine the quantitative relationships between the reactants and products, including the amounts of each substance involved in the reaction.
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--The complete question is, The principles that underlie balancing chemical equations include the______________ and the concept of stoichiometry. ---
Why is a physical property not a characteristic of matter that you can’t observe or measure without changing the identity of matter
Answer:
Characteristics of matter change due to chemical changes, not physical ones. Physical properties can be determined without changing the substance's chemical identity
Explanation:
WHAT IS THE MASS OF O2 GIVEN THE EQUATION: 4FE + 3O2 --> 2FE2O3
Answer: I think its 111.6
Explanation:
A certain half-reaction has a standard reduction potential EPod=-0.75 V. An engineer proposes using this half-reaction at the cathode of a galvanic cell that must provide at least 0.90 V of electrical power. The cell will operate under standard conditions. Note for advanced students: assume the engineer requires this half-reaction to happen at the cathode of the cell. Is there a minimum standard reduction potential that the half-reaction used at the anode of this cell can have? Is there a maximum standard reduction potential that the half-reaction used at the anode of this cell can have? By using the information in the ALEKS Data tab, write a balanced equation describing a half reaction that could be used at the anode of this cell. Note: write the half reaction as it would actually occur at the anode.
Using the following formula, the total cell potential, Ecell, may be calculated: Ecathode + anode equals Ecell. where Ecathode is the cathode half-reduction reaction's potential and Eanode.
We can determine the minimal Eanode needed to create a cell potential of 0.90 V since the engineer suggests employing a half-reaction with EPod = -0.75 V at the cathode:
Ecathode + anode equals Ecell.
Eanode: 0.90 V = -0.75 V
Eanode = 0.75 0.90 volts
Eanode equals 1.65 V.
The half-reaction employed at the anode must thus have a standard reduction potential of -1.65 V or less.
The typical reduction potential of the half-reaction utilised at the anode, on the other hand, has no upper limit. Yet, a higher Ecell and a more effective galvanic cell would be produced by a larger reduction potential at the anode.
We can utilise the half-reaction to create a balanced equation for the anode half-reaction:
Cu(s) becomes Cu2+(aq) plus 2e-
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