what is the prupose of the buffer solution that will beadded to each reaction in the experiemnet you will preform

Answer:

i think its B but im not so sure

Large organic molecules are usually assembled by polymerization of a few kinds of simple subunits belonging to the same class of chemicals.  The following is an exception to this statement is:

c) Steroids

Large organic molecules, such as proteins, nucleic acids, and carbohydrates, are typically formed through the process of polymerization. Polymerization involves the repetitive bonding of smaller subunits, known as monomers, to form a long chain or polymer. These monomers usually belong to the same class of chemicals, meaning they have similar structures and functional groups.

In the case of DNA, the monomers are nucleotides, which consist of a sugar molecule, a phosphate group, and a nitrogenous base. The repetitive bonding of nucleotides creates a long chain of DNA.

Similarly, cellulose, a major component of plant cell walls, is composed of repeating units of glucose monomers. The polymerization of glucose molecules forms long cellulose chains.

Contractile proteins, such as actin and myosin found in muscle fibers, are also assembled through the polymerization of monomers. These monomers, called amino acids, are linked together by peptide bonds to form polypeptide chains, which then fold into the functional protein structure.

However, steroids, including molecules like cholesterol, estrogen, and testosterone, are an exception to this general pattern of polymerization. Steroids have a distinct structure consisting of four fused carbon rings. They are not formed through repetitive bonding of identical subunits like proteins or nucleic acids. Instead, steroids are synthesized through specific biosynthetic pathways in living organisms.

While steroids play crucial roles in various physiological processes, they do not follow the typical pattern of polymerization seen in other organic polymers.

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The complete question is:

Large organic molecules are usually assembled by polymerization of a few kinds of simple subunits belonging to the same class of chemicals. which of the following is an exception to this statement?

a) DNA

b) cellulose

c) steroids

d) a contractile protein

Based on the given isotopic composition, the relative contribution of C3 plants is higher compared to C4 plants in the soil organic matter of Kenya.

To determine the relative contribution of C3 and C4 plants to the soil organic matter in Kenya based on their stable carbon isotopic composition, we can use the concept of isotopic discrimination.

C3 and C4 plants have different photosynthetic pathways, and they exhibit distinct carbon isotope signatures. C3 plants typically have a more negative δ13C value (around -30 permil to -22 permil), while C4 plants have a less negative δ13C value (around -16 permil to -9 permil).

In this case, the soil organic matter in Kenya has a δ13C value of -18 permil, while the air δ13C value is -7 permil. The difference between these values (-18 permil - (-7 permil)) gives us the isotopic discrimination between the atmosphere and the soil organic matter.

δ13C discrimination = δ13C organic matter - δ13C atmosphere

δ13C discrimination = -18 permil - (-7 permil)

δ13C discrimination = -11 permil

Since the δ13C discrimination is negative, it suggests that C3 plants have a dominant contribution to the soil organic matter. C4 plants, with their less negative δ13C values, are less likely to contribute significantly to the organic matter in this case.

Therefore, based on the given isotopic composition, the relative contribution of C3 plants is higher compared to C4 plants in the soil organic matter of Kenya.

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Mass of hydrate + crucible = 47.29 g

Mass of anhydrous salt = 2.7 g

Molar mass of anhydrous salt CuSO4 = 159.5 g

Given,

mass of empty crucible = 42.45 g

mass of hydrate salt= 4.84 g

mass of crucible after first heating = 46.1 g

mass of crucible after second heating= 45.153 g

mass of crucible after third heating= 45.15 g

so, as per the question we need to find...

Mass of hydrate + crucible = ? g

Mass of anhydrous salt = ? g

Molar mass of anhydrous salt CuSO4 = ? g

∴Mass of hydrate + crucible = 42.45 + 4.48 = 47.29 g

The given salt is in hydrate form, to remove water from this molecule we need to perform heating .

So we are taking the substance into the crucible as it is in less quantity.

Here, we performed heating 3 times and note the weight after every heating.

After this, assume that the water is totally evaporated and the remaining salt is in anhydrous form,

∴ Mass of anhydrous salt = 45.15 - 42.45 = 2.7 g

To find the molar mass of anhydrous salt of CuSO4,

atomic weight of Cu = 63.5 g

atomic weight of S = 32 g

atomic weight of O =16 g

∴ molar mass of anhydrous salt of CuSO4 = 63.5 + 32 + (16 ×3)

                                                                   =159.5 gm

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Answer:

The volume of the base is 0.05819·x L, where x is the number of moles of base that combines with one mole of sulfuric acid

Explanation:

The volume of the sulfuric acid, V = 25.3 mL = 25.3 × 10⁻³ L

The concentration of the sulfuric acid, c = 2.30 M

The concentration of the base, \(c_{base}\) = 1.00 M

Let the mole ratio of the acid to base be 1 : x

The number of moles of sulfuric acid present, n = c × V

∴ n = 2.30 M/L × 25.3 × 10⁻³ L = 0.05819 moles

The number of moles of sulfuric acid present, n = 0.05819 moles

1 mole of sulfuric acid combines with x moles of base

Therefore, 0.05819 moles of sulfuric acid will combine with 0.05819·x moles of base

The number of moles of base, \(n_{base}\) = 0.05819·x moles

Therefore, the volume of base, \(V_{base}\) = \(n_{base}\)/\(c_{base}\)

∴  \(V_{base}\) = 0.05819·x/1 ≈ 0.05819·x L

The volume of base,  \(V_{base}\) ≈ 0.05819·x L.

Answer:

HEAT OF COMBUSTION PER GRAM OF OCTANE IS 1723.08 J OR 1.72 KJ/G OF HEAT

HEAT OFF COMBUSTION PER MOLE OF OCTANE IS 196.4 KJ/ MOL OF HEAT

Explanation:

Mass of water = 100 g

Change in temperature = 28.78 °C - 21.36°C = 7.42 °C

Heat capcacity of water = 4.18 J/g°C

Mass of octane = 1.8 g

Molar mass of octane = C8H18 = (12 * 8 + 1 * 18) g/mol= 96 + 18 = 114 g/mol

First is to calculate the heat evolved when 100 g of water is used:

Heat = mass * specific heat capacity * change in temperature

Heat = 100 * 4.18 * 7.42

Heat = 3101.56 J

In other words, 3101.56 J of heat was evolved from the reaction of 1.8 g octane with water.

Heat of combustion of octane per gram:

1.8 g of octane produces 3101.56 J of heat

1 g of octane will produce ( 3101.56 * 1 / 1.8)

= 1723.08 J of heat

So, heat of combustion of octane per gram is 1723.08 J

Heat of combustion per mole:

1.8 g of octane produces 3101.56 J of heat

1 mole of octane will produce X J of heat

1 mole of octane = 114 g/ mol of octane

So we have:

1.8 g of octane = 3101.56 J

114 g of octane = (3101.56 * 114 / 1.8) J of heat

= 196 432.13 J

= 196. 4 kJ of heat

The heat of combustion of octane per mole is 196.4 kJ /mol.

Mass of water = 100 g

Change in temperature = 28.78 °C - 21.36°C = 7.42 °C

Heat capcacity of water = 4.18 J/g°C

Mass of octane = 1.8 g

Molar mass of octane = C8H18 = (12 * 8 + 1 * 18) g/mol= 96 + 18 = 114 g/mol

First is to calculate the heat evolved when 100 g of water is used:

Heat = mass * specific heat capacity * change in temperature

Heat = 100 * 4.18 * 7.42

Heat = 3101.56 J

The empirical formula for lysine, which demonstrates the chemical formula, is C₁₈H₄₂N₆O₆

Nitrogen, sometimes known as N, is an element that has no taste or color. Nitrogen is present in the air we breathe, the water we drink, and the ground beneath our feet. In reality, nitrogen is the most common element in the atmosphere of the Earth, accounting up 78% of it. Nitrogen is essential for the synthesis of amino acids, proteins, nucleic acids, and other essential elements. Stone fruit trees require an appropriate supply of nitrogen to grow and produce fruit every year. The primary types of nitrogen that are absorbed by fine roots are ammonium or nitrate.

On nitrogen, the chemical industry depends. It is used to make fertilizers, dyes, nylon, nitric acid, and explosives. To make these products, nitrogen must first be coupled with hydrogen to make ammonia.

Molecular Formula is calculated using following formula,

Molecular Formula  =  n (Empirical Formula)     ----- (1)

Calculating n;

                     n  =  Molecular Weight / Empirical Formula Weight  --- (2)

Empirical formula weight = (1.50×3) + (2.72×7) + (1.29×1) + (0.287×1)

Empirical formula weight = (4.5) + (19.04) + (1.29) + (0.287)

Empirical formula weight = 25.117

Putting values in Eq. 2,

                               n  =  146.19 / 25.117

                               n  =  6

Putting Value of n and Empirical Formula in Eq. 1,

Molecular Formula  =  6 × C₃H₇NO

Molecular Formula  =  C₁₈H₄₂N₆O₆

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The genotype and phenotype ratio for the first cross will be 100% Rr and 100% round, while the second cross will be 1RR:1Rr, 50/50 round and wrinkled respectively.

Assuming that the seed shape allele is R(r).

For the first cross:             RR    x    rr

                                        Rr   Rr   Rr   Rr

Genotype ratio = 100% Rr

Phenotype ratio = 100% round

For the second cross:           RR    x    Rr

                                             RR   Rr   RR   Rr

Genotype ratio = 1RR : 1Rr

Phenotype ratio = 50% round and 20% wrinkled.

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Answer:

The histosols of peat pogs are soils rich in organic material but poor in mineral nutrients, and are not fertile. Conversely, aridosols (desert soils) have an abundance of mineral nutrients but a dearth of organic matter and are also not fertile.

hope i help✨!

The concentration is measured by molarity, the formula of molarity is:

Based on the given data, we've already had the volume in liters and we have to convert 6.25 grams of HCl to moles. We have to use the molar mass of HCl which is 36.4 g/mol (you can find the molar mass using the periodic table). The conversion would be:

And the final step is to replace the values we have in the formula of molarity:

The molarity of 6.25 g of HCl in 0.300 L of solution is 0.573 M.

The mass of magnesium that would produce the maximum amount of heat is 0.547 grams.

The chemical reaction between magnesium and HCl is

Mg + 2 HCl → MgCl₂ + H₂

In stoichiometry, the Avogadro's law said that the coefficient for every subtances in the reaction is the ratio for the number of moles for every subtances.

For HCl

For Mg

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Ba(N03)2 is the unknown compound. correct option is (C).

Na2So4 -----> precipitate

Nacl ---------> No reaction

Na2Co3 --------> precipitate

Ba (No3)2 + Na2So4 -----------> BaSo4 (s) + 2 NaNo3 (aq)

Here white precipitate of barium sulphate will be produced.

Ba (No3)2 + Na2Co3 ------------> BaCo3 (s) + 2NaNO3 (aq)

Here white precipitate of barium carbonate will be produced.

Titration is a process of chemical analysis in which the quantity of some constituent of a sample is determined by adding to measured sample to an exactly known quantity of another substance with which the desired constituent reacts in a definite, known proportion.

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Answer:

D.

The concentration of reactants and the concentration of products are constant.

Explanation:

pls mark as brainliest

The correct answer is:

B) Na⁺; Ca²⁺ for influx of ions.

An influx action potential that enters a presynaptic terminal activates Ca2+ channels and momentarily raises the local Ca2+ concentration in the presynaptic active zone. After activating synaptotagmins Ca2+, neurotransmitter release occurs within a few hundred microseconds. Through the interaction of their C2-domains with phospholipids and SNARE proteins, synaptotagmins' two C2-domains bind Ca2+ and translate the Ca2+ signal into a nanomechanical activation of the membrane fusion machinery. Synaptotagmins cannot initiate exocytosis on their own; instead, they need a necessary cofactor known as complexin, a tiny protein that binds to SNARE complexes and simultaneously activates and clamps the SNARE complexes, setting them up for later synaptotagmin action.

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Answer:

1.89g of HNO3

Explanation:

Data obtained from the question include:

Mass of water = 31.5 g

Molality of HNO3 = 0.950 m

Mass of HNO3 =.?

Next, we shall determine the number of mole of HNO3 in the solution.

Mass of water = 31.5 g = 31.5/1000 = 0.0315 Kg

Molality of HNO3 = 0.950 m

Number of mole of HNO3 =..?

Molality is simply defined as the mole of solute per unit kilogram of solvent (water) i.e

Molality = mole /kg of water

With the above formula, the mole of HNO3 can be obtained as follow:

Molality = mole /kg of water

0.950 = mole of HNO3 /0.0315

Cross multiply

Mole of HNO3 = 0.950 x 0.0315

Mole of HNO3 = 0.03 mole

Finally, we shall convert 0.03 mole of HNO3 to grams. This is illustrated below:

Molar mass of HNO3 = 1 + 14 + (16x3) = 63g/mol

Mole of HNO3 = 0.03 mole

Mass of HNO3 =..?

Mole = mass /molar mass

0.03 = mass of HNO3 /63

Cross multiply

Mass of HNO3 = 0.03 x 63

Mass of HNO3 = 1.89g

Therefore, 1.89g of HNO3 is needed to prepare the solution.

Answer: WAIT WHAT- YOU SHOULD BE MY FRIEND BTW SNAP IS kpurdham5

Explanation: ‍♂️

The atmospheric concentration of a gas depends on its emission into the atmosphere and its rates of physical, chemical and biological removal. The time to lower the concentration of the gas to 37% of its original amount is its atmospheric lifetime.

The measurements of CO2 equivalents in parts per million CO2 is termed as atmospheric concentration. The pressure exerted by the atmosphere on the gas is called atmospheric pressure.

The atmospheric lifetime of a species mainly measures the time which is require to restore equilibrium in the atmosphere that follows a sudden decrease or increase in the concentration of the species in the atmosphere.

Emission is something which can be released, emitted or discharge.

Types of emission

Thus, we concluded that the atmospheric concentration of a gas depends on its emission into the atmosphere and its rates of physical, chemical and biological removal. The time to lower the concentration of the gas to 37% of its original amount is its atmospheric lifetime.

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In order to properly mix these two components together, they must be poured into a special holder called a dappen dish.

When it comes to creating artificial nails using acrylics, the process involves mixing a monomer liquid and polymer powder together to form a pliable mixture that can be shaped and molded. However, in order to properly mix these two components together, they must be poured into a special holder called a dappen dish. This small glass or plastic container is designed to hold the monomer and polymer powder until they are ready to be mixed together with a brush.

The dappen dish allows for precise measurements and easy access to the mixture during the application process. Overall, the dappen dish is a crucial tool in the acrylic nail application process, as it provides a convenient and efficient way to mix the monomer liquid and polymer powder together.

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Answer:

Change in energy.

Explanation:

An energy change is required to shift one state to another or do any of the other listed.

Polyatomic ion is OOH

Polyatomic ion are ion which consist of more than one atom and atom in a polyatomic ion are usually covalently bonded to one another and therefore stay together as a single or charged unit and here OOH there are two oxygen atom and one hydrogen atom that's it is polyatomic ion

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Answer:

240 N/m

Explanation:

Using the formula as follows:

U = ½kx²

Where;

U = elastic potential energy (J)

K = spring constant (N/m)

x = stretched displacement (m)

According to the information provided in this question,

U = 35J

x = 0.54m

k = ?

U = ½kx²

35 = ½ × k × 0.54²

35 = ½ × k × 0.2916

35 = 0.2916k/2

70 = 0.2916k

k = 70 ÷ 0.2916

k = 240.05

To 2s.f, the spring constant (k) = 240 N/m.

Answer: 240

Explanation:

Answer:

27 °C

Explanation:

Applying,

V/T = V'/T'................. Equation 2

Where V = Initial volume of oxygen, T = Initial temperature, V' = Final volume of oxygen, T' = Final temperature.

Make T' the subject of the equation

T' = V'T/V................ Equation 2

Form the question,

Given: V' = 600 mL, V = 460 mL, T = -43°C = (-43+273) = 230K

Substitute these values into equation 2

T' = (600×230)/460

T' = 300 K

T' = (300-273) °C

T' = 27 °C

Here's a step-by-step guide:

Gather all the metal strips that you have set out in front of the test tubes.

Make sure you have a weighing scale or balance that is capable of measuring the masses of the metal strips accurately. If you don't have one, you might need to borrow one or use a nearby laboratory's equipment.

Place the weighing scale on a stable and level surface.

Calibrate the weighing scale if necessary, following the manufacturer's instructions.

Place a clean and dry weighing boat or a piece of weighing paper on the weighing scale. Make sure it is properly tared (the scale reads zero) before placing any metal strip on it.

Carefully pick up one metal strip at a time, taking care not to touch the weighing boat or paper with your fingers as it can affect the measurement.

Place the metal strip onto the weighing boat or paper on the scale. Be gentle and avoid any sudden movements that could cause the metal strip to fall off.

Allow the weighing scale to stabilize and display the mass of the metal strip. Once the reading has settled, record the mass in a notebook or any other recording medium you prefer.

Repeat steps 6 to 8 for each metal strip you have.

After measuring and recording the masses of all the metal strips, ensure that you clean the weighing scale and return it to its proper place.

Remember to handle the metal strips with care and follow any safety protocols that may be applicable in your laboratory or workspace.

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Increasing the temperature in the exothermic production of ammonia from its elements, known as the Haber-Bosch process, affects both the reaction rate and yield. Reaction rate typically increases with an increase in temperature. This is due to the fact that higher temperatures provide more thermal energy to the reactant molecules, enabling them to overcome the activation energy barrier more easily. As a result, more productive collisions occur between the reacting species, leading to a faster reaction rate.

However, the yield of the ammonia production is influenced by temperature in a slightly different manner. While the reaction rate may increase, a higher temperature can also cause a decrease in the equilibrium yield of ammonia. This is because the Haber-Bosch process is an equilibrium reaction, and according to Le Chatelier's principle, increasing the temperature will favor the endothermic direction of the reaction. In this case, it means that the equilibrium will shift towards the reactant side, resulting in a lower yield of ammonia. Therefore, there is a trade-off between reaction rate and yield when adjusting the temperature in the Haber-Bosch process. The optimal temperature is usually chosen to maximize the yield while maintaining a reasonable reaction rate for efficient production.

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First of all, Z is unknown. I hope it is a mistake.

        Now, it is given that the element X has nineteen electrons. This proves that X is actually Potassium.

        As per the periodic table, both Potassium and Lithium belongs to group 1 as their valency is 1 because of the presence of only one electron in the outermost shell of electrons i.e., they lose an electron during a chemical reaction to form a stable compound. Furthermore, both are metallic.

        Magnesium belongs to group 2 and hence its valency is two, which is different from potassium though it is metallic. Similiarly, bromine belongs to group 17 and gains one electron during a reaction in contrast to potassium.

( No internal links available for reference. For clarification, check the periodic table).

1) The reaction of a primary amide with LiAlH₄ would result in the reduction of the amide functional group to a primary amine.

2) The reaction of a primary amide with H₃O⁺ would result in the hydrolysis of the amide functional group to form a carboxylic acid and ammonia.

1) LiAlH₄ is a strong reducing agent commonly used for the reduction of carbonyl compounds. In the presence of LiAlH₄, the primary amide undergoes reduction, where the carbonyl group (-C=O) is transformed into a primary amine (-NH₂), resulting in the removal of the oxygen atom.

2) H₃O⁺ represents an acidic environment and can initiate the hydrolysis of amides. In the presence of H₃O⁺, the amide functional group undergoes hydrolysis through a reaction called acid hydrolysis. This process cleaves the amide bond, breaking it into a carboxylic acid and an amine. The amine formed in this case would be ammonia (NH₃).

Overall, the reaction of a primary amide with LiAlH₄ results in the reduction to a primary amine, while the reaction with H₃O⁺ leads to the hydrolysis of the amide, forming a carboxylic acid and ammonia.

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Answer:

Crystals Grow and Grow

A crystal is an inorganic (not alive, not from something alive) homogeneous solid (meaning a solid with the same properties at all points) with a three-dimensional, repeated ordering of atoms or molecules

When water is lost, but electrolytes are retained, the first thing that happens is that moves water from the ICF to the ECF.

The fluid inside of cells, also called the cytoplasm or cytosol, makes up about 60% of the water in the human body, totaling about 7 gallons. Organelles like the nucleus, endoplasmic reticulum, mitochondria, lysosomes, and Golgi apparatus are suspended in and supported by the ICF. Also found in the ICF are cellular building blocks like sugars, proteins, carbohydrates, and lipids.

ECFs are any body fluids that are not inside cells. The two main components of ECF are plasma and interstitial fluid (IF). The balance consists of cerebrospinal fluid, lymph, the synovial fluid in the joints, pleural fluid in the pleural cavities (lungs), pericardial fluid around the heart, peritoneal fluid in the peritoneal cavity (abdomen), and the aqueous humor of the eye. In mammals, milk is also considered an extracellular fluid.

The ICF has higher amounts of potassium, phosphate, magnesium, and protein compared to the ECF. The plasma has high concentrations of sodium, chloride, and bicarbonate, but lower levels of protein as compared to the ICF. While water moves passively via osmosis, sodium and potassium ions move in and out of cells using active transport ion pumps. The pumps are powered by adenosine triphosphate (ATP) to provide the energy to move the ions against their concentration gradients (i.e. sodium moves out of the cell and potassium is pumped in) and maintain the gradients inside and outside the cell.

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