How many hydrogen bonds is his 107 forming within ca ii? study the amino acid structure of his and explain why these h bonds are possible at ph 7.4.

The saturated vapor pressure of methanol in the vessel is 125.9 mmHg (rounded off to 3 significant figures). The volume percent concentration of methanol in the vapor is 73.2% (rounded off to 3 significant figures).

a. Saturated vapor pressure of methanol in the vessel in mmHg: The vapor pressure of methanol (MeOH) in the vessel can be determined using Dalton's law of partial pressures. MetOH psat = MetOH Pabs

Here, MetOH Pabs is the pressure exerted by the vapor of methanol in the vessel. It is equal to the total pressure in the vessel minus the partial pressure of the inert gas (nitrogen). Pabs = 2 inch of water gauge = 2 x 25.4 mm of water gauge / 1 inch of water gauge = 50.8 mm of water gauge

Total pressure in mmHg = Pabs x 760 mmHg / 406.8 inch of water gauge = 94.8 mmHg

Partial pressure of nitrogen = 2 inch of water gauge = 50.8 mm of water gauge

Partial pressure of methanol = MetOH Pabs MetOH psat = 126 mmHg (at 25°C)

Therefore, the saturated vapor pressure of methanol in the vessel is 125.9 mmHg (rounded off to 3 significant figures).

b. Volume percent concentration of methanol in the vapor: The volume percent concentration of methanol in the vapor can be determined using the ideal gas law and the Dalton's law of partial pressures. V = nRT / PV = volume of the vapor (L)n = number of moles of methanol gas R = gas constant T = temperature (K)P = pressure (atm)

Let's assume that the total volume of the vapor in the vessel is 1 L and the temperature is 25°C (298 K).

The number of moles of nitrogen gas (N2) in the vapor can be determined using the Dalton's law of partial pressures. N2 Pabs = Pabs - MetOH PabsN2 Pabs = 50.8 mm of water gaugeN2 Pabs = 50.8 mm of water gauge x 1 atm / 760 mmHg = 0.067 atmN2 V = nRT / PN2N2 V = (0.067 atm x 1 L) / (0.082 L atm/K mol x 298 K)N2 V = 0.0022 mol

The number of moles of methanol gas (MeOH) in the vapor can be determined using the ideal gas law. MetOH Pabs = MetOH psatMetOH Pabs = 125.9 mm Hg MetOH Pabs = 125.9 mmHg x 1 atm / 760 mmHg = 0.165 atmMetOH V = nRT / PMetOH V = (0.165 atm x 1 L) / (0.082 L atm/K mol x 298 K)MetOH V = 0.006 mol

The volume percent concentration of methanol in the vapor can be determined using the following equation.

Volume percent concentration of MeOH = MetOH V / (N2 V + MetOH V) x 100%

Volume percent concentration of MeOH = 0.006 mol / (0.0022 mol + 0.006 mol) x 100%

Therefore, the volume percent concentration of methanol in the vapor is 73.2% (rounded off to 3 significant figures).

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Answer:

It is noticed that after five days, there is elongation of the root cells in the roots of onion in the jam bottle

Explanation:

It is noticed that after five days, there is elongation of the root cells in the roots of onion in the jam bottle. The growth of a plant is due to the availability of the growing and  dividing cells in the meristematic regions of the apical meristems. (stem cells). This site are active site for elongation and differentiation  of the stem tissues. In the apical meristem of the root cells, the cells have the potential to carry out mitosis and cell division repeatedly in order to generate newer cells for extension and development of stems and leaves. With each cell division, one cell remains in the meristem which the other increases in size and differentiate due to different genomic expression to form the meristem regions which eventually becomes the stem structure.

The pKa of the α-amino group in the hydrophobic interior of the protein  with a nearby Asp residue, would be lower compared to the pKa of the α-amino group in the hydrophobic interior without a nearby Asp residue.

In the hydrophobic interior of a protein, the α-amino group can form an ionic bond with a carboxylate group of a charged aspartic acid (Asp) residue. This interaction can significantly affect the pKa of the α-amino group compared to its pKa in the absence of the nearby Asp residue.
When the α-amino group forms an ionic bond with the Asp residue, it becomes positively charged by donating a proton to the carboxylate group. This ionic bond stabilizes the charged form of the α-amino group and facilitates the release of a proton, resulting in a lower pKa value. The presence of the ionic bond enhances the stability of the charged form of the α-amino group, making it more acidic and lowering its pKa value even further.
Therefore, the pKa of the α-amino group in the hydrophobic interior of the protein, when it forms an ionic bond with a nearby Asp residue, would be lower compared to the pKa of the α-amino group in the hydrophobic interior without a nearby Asp residue. The ionic bond formation enhances the acidity of the α-amino group, making it more likely to release a proton and lowering its pKa value.

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I admit that I'm unsure, but I believe the answer is D: Sun heating car sitting in a parking lot. It makes sense to me, but I'm blonde. I hope it is correct. I also have that question in my quiz.

Answer:

File down there

Explanation:

Answer:

1.93 molar, or M

Explanation:

By definition, molarity is defined as moles/liter.  A 0.5 molar (or M) solution will have 0.5 moles of solute per 1.0 liter.  To answer this question, first calculate how many moles of NaOH there are in 11.6 grams of the compound.  We'll need the miolar mass of NaOH, which is 40 grams/mole.  This is found by adding the atomic masses of each atom in the compound:

Na:     23

O:      16

H:         1

Total: 40 g/mole

The moles of NaOH is found by dividing the mass by the molar mass:

        (11.6 grams NaOH)/(40 g/mole NaOH) = 0.29 moles NaOH

0.29 moles NaOH is contained in 15cm^3 of solution.  Convert 15cm^3 into liters (L):

 (15 cm^3)*(1 ml/cm^3)*(1 liter/1000 ml) = 0.015 liters

cm^3 and ml both cancel with these conversion factors, leaving only liters, L.

The solution consists of 0.29 moles of NaOH in 0.15L.  It's concentration is therefore:

  (0.29 moles)/(0.015L) = 1.93 molar, or M

Given, Manganese dioxide (MnO2) = 28gHydrochloric acid (HCl) = 42g

To find, Theoretical yield of Cl2, We will find the limiting reagent and use its mole to find the theoretical yield of Cl2.MnO2 + 4HCl → MnCl2 + 2H2O + Cl2First we will find the limiting reagent.

The number of moles of MnO2 is, Moles of MnO2 = Mass/Molar mass = 28/86.94 = 0.322 mol The number of moles of HCl is, Moles of HCl = Mass/Molar mass = 42/36.46 = 1.151 mole will use the mole ratio of MnO2 and HCl to find the mole of Cl2. The mole of HCl required to react with 0.322 mol of MnO2 = 0.322 x 4 = 1.288 mol The number of moles of Cl2 will be, Moles of Cl2 = 0.322 x 1 / 4 = 0.0805 mol. The theoretical yield of Cl2 from the given reactants is,

Theoretical yield of Cl2 = Moles x Molar mass = 0.0805 x 70.9 = 5.71 g Chlorine gas will be produced with a theoretical yield of 5.71 g when 28 g of MnO2 and 42 g of HCl are mixed.

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Answer:

191.11 grams of oxygen gas should be produced.

Explanation:

The balanced reaction is:

2 Al₂O₃ → 4 Al + 3 O₂

By stoichiometry of the reaction (that is, the relationship between the amount of reagents and products in a chemical reaction), the following amounts of moles of each compound participate in the reaction:

Being the molar mass of each compound:

By reaction stoichiometry, the following mass quantities of each compound participate in the reaction:

Then you can apply the following rule of three: if by stoichiometry 108 grams of aluminum are produced along with 96 grams of oxygen, 215 grams of aluminum are produced along with how much mass of oxygen?

\(mass of oxygen=\frac{215 grams of aluminum*96 grams of oxygen}{108grams of aluminum}\)

mass of oxygen= 191.11 grams

191.11 grams of oxygen gas should be produced.

vapor pressure = 5 mb, saturation vapor pressure = 10 mb, RH = 50%

mixing ratio = 15 g/kg, saturation mixing ratio = 20 g/kg, RH = 75%

mixing ratio = 25 g/kg, saturation mixing ratio = 25 g/kg, RH = 100%

For each condition, we can calculate the relative humidity (RH) using the formula:

RH = (vapor pressure/saturation vapor pressure) × 100%

1. For vapor pressure = 5 mb and saturation vapor pressure = 10 mb:

RH = (5 mb / 10 mb) × 100% = 50%

2. For mixing ratio = 15 g/kg and saturation mixing ratio = 20 g/kg:

RH = (15 g/kg / 20 g/kg) × 100% = 75%

3. For mixing ratio = 25 g/kg and saturation mixing ratio = 25 g/kg:

RH = (25 g/kg / 25 g/kg) × 100% = 100%

In the first case, the vapor pressure is half of the saturation vapor pressure, resulting in an RH of 50%. This indicates that the air is holding 50% of the maximum amount of water vapor it can hold at that temperature.

In the second case, the mixing ratio is 75% of the saturation mixing ratio, resulting in an RH of 75%. This means the air is holding 75% of the maximum amount of water vapor it can hold at that temperature.

In the third case, the mixing ratio is equal to the saturation mixing ratio, resulting in an RH of 100%. This indicates that the air is holding the maximum amount of water vapor it can hold at that temperature, leading to saturated conditions.

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The particle released in the reaction has been an alpha particle. Thus, option A is correct.

The release of protons and neutrons in a reaction has been nuclear reactions. The reactions results in the change in the atomic mass of the atom, with the formation of new atom. The extra mass has been released in the form of energy.

The release from the given reaction has been:

\(\rm ^{222}_{66}Rn\;\rightarrow\;^{218}_{84}Po\;+\;^A_ZX\)

Where, X has been the particles that has been released with the nuclear reaction. The superscript A has been the Mass of particle, and subscript Z has been the atomic number.

According to the law of conservation, the mass has been constant on both the product and the reactant side. Thus, the mass and atomic number of eliminated particles has been given as:

\(\rm 222=218+A\\A=222-218\\A=4\)

The atomic mass of the particle has been 4.

\(\rm 86=84+Z\\Z=86-84\\Z=2\)

The atomic number of the particle has been 2.

The element with atomic mass 4, and atomic number 2 has been the Helium particles, \(\rm ^4_2He\). It has been an alpha particle. Thus, the particle released in the reaction has been an alpha particle. Thus, option A is correct.

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Answer:

A

Explanation:

False. The statement that the species BF3 does not exist because its central boron (B) atom has only 6 valence electrons .

The existence and stability of chemical species are determined by the electron configuration and bonding of the atoms involved. In the case of BF3 (boron trifluoride), boron is the central atom.

Boron, located in Group 13 of the periodic table, has an atomic number of 5. As a result, it has 5 electrons in its neutral state. However, when boron forms chemical compounds, it can utilize vacant orbitals to accommodate additional electrons.

In the case of BF3, boron forms three covalent bonds with three fluorine (F) atoms, resulting in a total of 8 electrons around the boron atom. This satisfies the octet rule, which states that atoms tend to gain, lose, or share electrons in order to achieve a stable electron configuration with 8 valence electrons.

Therefore, BF3 does exist, and its central boron atom accommodates 8 valence electrons, rather than just 6.

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Explanation:

Oxidation Numbers. It is often useful to follow chemical reactions by looking at changes in the oxidation numbers of the atoms in each compound during the reaction. Oxidation numbers also play an important role in the systematic nomenclature of chemical compounds.

brainliest please

Answer:

800 cubic cm

Explanation:

Answer:c

Explanation:

As the combined mass of the HCl and NaOH is 15 grams before the reaction. Therefore the mass after the reaction will be 15 grams according to the law of conservation of mass. Therefore, option (C) is correct.

Matter can be transformed form via physical changes and chemical changes from one form to another form, during any of these changes, the total mass is conserved. The same quantity of matter exists before and after the chemical or physical as none of the matter is created or destroyed.

The balanced equation between the reaction of HCl and NaOH:

\(HCl +NaOH \longrightarrow H_2O +NaCl\)

According to the law of conservation of mass, the mass of HCl and NaOH will be equal to the mass of the products water and NaCl.

As mentioned in the question the combined mass of HCl and NaOH measured before the reaction is 15 grams. Therefore, the mass of the products in the closed container will be equal to 15 grams as well.

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Your question was incomplete,  most probably the complete question was,

Fernando places 15 ml of HCl and 50 ml of NaOH in 100 ml of a beaker. He places them on a scale together and measures the combined mass of 15 grams.

After the HCl and NaOH react, Fernando measures the mass again. Using the mass before the reaction, what is the mass after the reaction? Remember, It is in a closed system.

A. 5.00 grams

B. 10.00 grams

C. 15.00 grams

D. 20.00 grams

Answer:

It is the third option.

Explanation:

Codominance is when both traits can be seen.

The volume of the rectangular box is 250 cubic centimeters (cm³) or 0.25 liters (L).

To find the volume of a box with a length of 5cm, a width of 5cm and a height of 10cm, we use the formula for the volume of a rectangular box, which is given as;Volume of rectangular box = Length × Width × HeightGiven that the length of the box is 5cm, the width is also 5cm, and the height is 10cm.

Therefore, we substitute the values into the formula above;Volume of rectangular box = 5cm × 5cm × 10cm= 250cm³.

Therefore, the volume of the rectangular box is 250 cubic centimeters (cm³).

We can also represent this volume in liters (L) by converting from cubic centimeters to liters, since 1L is equal to 1000cm³.

Thus, to convert 250cm³ to liters;Volume in liters = Volume in cm³ / 1000cm³/L= 250cm³ / 1000cm³/L= 0.25L.

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The significant difference between inorganic and organic compounds is that an organic compound contains carbon; few inorganic compounds do. The correct answer is option(B).

Organic compounds are molecules made up of a combination of carbon, hydrogen, and oxygen atoms. Organic compounds are said to be more volatile and also highly combustible. These compounds are often associated with living organisms and are found in foods and medicines. These are insoluble in water.

Examples of organic compounds are carbohydrates, fats (lipids), proteins, and nucleic acids, which are the basis for the molecules of life.

Inorganic compounds are made up of atoms from different elements, such as metals, sulfur, or nitrogen, and are not associated with living things. These compounds are not flammable and are non-volatile in nature. These are soluble in water and also non-soluble in some of the organic solutions.

           Examples of inorganic compounds include table salt (sodium chloride), bleach (sodium hypochlorite), and ammonia (NH3). The compound which despite having carbon are inorganic are graphite, carbon dioxide, carbon monoxide, carbides (e.g, Silicon Carbide), carbonates ( Calcium Carbonate), and some cyanides.

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Answer:

1800 g Zr(NO₃)₄

General Formulas and Concepts:

Chemistry - Atomic Structure

Explanation:

Step 1: Define

5.3 mol Zr(NO₃)₄

Step 2: Identify Conversions

Molar Mass of Zr - 91.22 g/mol

Molar Mass of N - 14.01 g/mol

Molar Mass of O - 16.00 g/mol

Molar Mass of Zr(NO₃)₄ - 91.22 + 4(14.01) + 12(16.00) = 338.26 g/mol

Step 3: Convert

\(5.3 \ mol \ Zr(NO_3)_4(\frac{339.26 \ g\ Zr(NO_3)_4}{1 \ mol \ Zr(NO_3)_4} )\) = 1798.08 g Zr(NO₃)₄

Step 4: Check

We are given 2 sig figs. Follow sig fig rules and round.

1798.08 g Zr(NO₃)₄ ≈ 1800 g Zr(NO₃)₄

Answer:

the answer would be B :)

The correct answer is; Bonds create new elements

Chemical bonds are formed when two or more atoms are joined together to form a compound.

Chemical compounds are more stable than the respective elements from which they are formed.

Chemical bonds do not create new elements.

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Answer:

protons are +1

Explanation:

protons are positive 1 while electrons are -1, they are equal but opposite charge

Dissolving a solute such as KOH in a solvent such as water results in- option A] a decrease in the vapor pressure of the liquid.

We can say that vapor pressure is defined as the pressure exerted by a vapor in thermodynamic equilibrium with its condensed phases (here liquid) at a given temperature in a closed system. Since the equilibrium vapor pressure is more like an indication to a liquid's thermodynamic tendency to evaporate, as it relates to the balance of particles escaping from the liquid in equilibrium with those in a coexisting vapor phase. With this a substance with a high vapor pressure at normal temperatures is often referred to as commonly known as volatile. Therefore, we know that the pressure exhibited by vapor present above a liquid surface is known as vapor pressure. As the temperature of a liquid increases, the attractive interactions between liquid molecules become less significant in comparison to the entropy of those molecules in the gas phase, also increasing the vapor pressure along with it. Thus, we know that liquids with strong intermolecular interactions are likely to have smaller vapor pressures, and vice versa for weaker interactions.

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Answer

Explanation

Given that:

The mass of pure silver needed = 800 g

Mass of zinc = 300 g

Mass of silver nitrate = 600 g

What to find:

Will the mass of zinc and silver nitrate be able to make 800 g of pure silver.

Step-by-step solution:

Step 1: Write the balanced equation for the reaction.

Zn + 2AgNO₃ → 2Ag + Zn(NO₃)₂

Step 2: Determine the moles of the reactants.

Using the mole formula, the moles of the reactants will be:

Step 3: Determine the moles of pure silver produced.

Using the mole ratio of Zn to AgNO₃ in the equation and the moles in step 2, we

The solubility (in grams per liter) of magnesium fluoride in 0. 013 M sodium fluoride, NaF is 6.5 x 10⁻⁹ g/L.

Sodium is a chemical element with the symbol Na and atomic number 11. It is a soft, silvery-white, highly reactive metal that is a member of the alkali group of elements. Sodium is essential to the human body and is a major component of many minerals and salts. It is used in many industrial processes, such as in the production of glass, paper, and textiles, and is also used in the production of certain medical treatments and medicines.

The solubility of MgF₂ in 0.013 M sodium fluoride solution is calculated using the expression:

Ksp = [Mg²⁺][F⁻]2

[Mg²⁺] = [F⁻]2 x Ksp

[Mg²⁺] = (0.013)2 x 7.1 x 10⁻⁹

[Mg²⁺] = 1.05 x 10⁻¹⁰ M

The solubility of MgF₂ in 0.013 M sodium fluoride solution is then calculated as follows:

Solubility (g/L) = [Mg²⁺] x Molar mass of MgF₂

Solubility (g/L) = 1.05 x 10⁻¹⁰ x 62.3

Solubility (g/L) = 6.5 x 10⁻⁹ g/L

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Wavelength (L) is calculated using: L = gT²/2π, here g=9.8 m/s2 and T is wave period in seconds.

Wavelength of a wave describes how long the wave is and the distance from the "crest" (top) of one wave to the crest of next wave is called wavelength. We can also measure from the "trough" (bottom) of one wave to trough of next wave and get the same value for the wavelength.

We measure wavelength in following ways:

Use photometer to measure the energy of wave.

Convert energy into joules (J).

Divide energy by Planck's constant, 6.626 x 10⁻³⁴, to get the frequency of  wave.

Divide speed of light, ~300,000,000 m/s, by frequency to get wavelength.

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To calculate the pH of the solution, we need to consider the ionization of acetic acid (HC₂H₃O₂) and the dissociation of its conjugate base (C₂H₃O₂⁻) in water. Let's proceed with the calculations:

First, we need to determine the concentration of H3O+ ions resulting from the ionization of HC₂H₃O₂ . The Ka expression for acetic acid is:

Ka = [H₃O+][C₂H₃O₂⁻] / [HC₂H₃O₂]

Given that the initial concentration of HC₂H₃O₂ is 0.10 M and the initial concentration of C₂H₃O₂⁻ (provided by the sodium acetate) is 0.050 M, we can assume that the concentration of [H₃O+] formed is x M.

Using the Ka value of 1.8 x 10^-5, we can set up the equation:

1.8 x 10^-5 = x * 0.050 / 0.10

Simplifying and solving for x, we find:

x = (1.8 x 10^-5) * (0.10 / 0.050)

x = 3.6 x 10^-5 M

Now, we have the concentration of [H₃O+] in the solution resulting from the ionization of HC₂H₃O₂. To calculate the pH, we use the equation:

pH = -log[H₃O+]

Substituting the value of [H₃O+] into the equation, we get:

pH = -log(3.6 x 10^-5)

pH ≈ 4.44

Therefore, the pH of the solution containing 0.10 M HC₂H₃O₂ and 0.050 M NaC₂H₃O₂ is approximately 4.44.

The pH of a solution is a measure of its acidity. In this case, we have a solution of acetic acid (HC₂H₃O₂) and its conjugate base (C₂H₃O₂⁻) formed by the dissociation of sodium acetate (NaC₂H₃O₂). Acetic acid is a weak acid that partially ionizes in water.

To calculate the pH, we first determined the concentration of H3O+ ions resulting from the ionization of HC₂H₃O₂. By using the Ka expression and the initial concentrations of HC₂H₃O₂ and C₂H₃O₂⁻, we found the concentration of [H₃O+] to be 3.6 x 10^-5 M.

Then, we used the pH equation, which is the negative logarithm of [H₃O+], to calculate the pH of the solution. Taking the negative logarithm of 3.6 x 10^-5 M, we obtained a pH value of approximately 4.44.

This pH value indicates that the solution is slightly acidic, as it falls below the neutral pH of 7. The presence of the weak acid (HC₂H₃O₂) and its conjugate base (C₂H₃O₂⁻) contributes to the acidic nature of the solution.

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Answer:

dioxide hexachloride

The number of molecules of HI  that are needed to produce 72.54g of BaL2 is6.365 x 10^23.

To determine the number of molecules of HI needed to produce 72.54g of BaL2, we need to first find the moles of BaL2 and then use the balanced chemical equation to find the moles of HI.

First, we'll find the moles of BaL2:

72.54 g of BaL2 / 137.3 g/mol = 0.529 moles of BaL2

Next, we'll use the balanced chemical equation to find the moles of HI:

BaSO4 + 2HI -> BaL2 + H2SO4

From this equation, we can see that for every 2 moles of HI that react, 1 mole of BaL2 is produced. Therefore, to produce 0.529 moles of BaL2, we need 0.529 x 2 = 1.058 moles of HI.

Finally, we'll convert the moles of HI to the number of molecules:

1.058 moles of HI x 6.022 x 10^23 molecules/mole = 6.365 x 10^23 molecules of HI.

Therefore  the number of molecules of HI needed to produce 72.54g of BaL2 is approximately 6.365 x 10^23.

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