What mass of phosphoric acid (h3po4, 98.0 g/mol) is produced from the reaction of 10.0 g of p4o10 (284 g/mol) with excess water?

Ephedrine, a central nervous system stimulant, is used in nasal sprays as a decongestant. This compound is a weak organic base:

C10H15ON (aq) + H2O (l) -> C10H15ONH+ (aq) + OH- (aq)

A 0.035 M solution of ephedrine has a pH of 11.33.

a) What are the equilibrium concentrations of C10H15ON, C10H15ONH+, and OH-?

b) Calculate Kb for ephedrine.

c(C₁₀H₁₅NO) = 0,035 M.
pH = 11,33.
pOH = 14 - 11,33 = 2,67.
[OH⁻] = 10∧(-2,67) = 0,00213 M.
[OH⁻] = [C₁₀H₁₅NOH⁺] = 0,00213 M.
[C₁₀H₁₅NO] = 0,035 M - 0,00213 M = 0,03287 M.
Kb = [OH⁻] · [C₁₀H₁₅NOH⁺] / [C₁₀H₁₅NO].
Kb = (0,00213 M)² / 0,03287 M = 1,38·10⁻⁴.

Answer:

5.32*10⁻³M

Explanation:

Given:

Rate constant of the First order reaction, k = 1.80*10-2 s-1

Initial concentration of PH3, [A]₀ = 3.16*10-2 M

Reaction time, t = 99 s

Formula:

For a first order reaction:

[tex][A] = [A]_{0} e^{-kt}[/tex]

where [A] and [A]₀ are concentration of reactant at time t and t = 0

k = rate constant

For the given reaction"

[tex][A] = 3.16*10^{-2}  e^{-1.80*10^{-2} *99} = 5.32*10^{-3} M[/tex]

The most likely species of beetles to become extinct when existing conditions change is the Green-colored beetles, pesticide-resistant gene absent; option 3.

Favorable traits are those traits possessed by organisms which enables them to survive or gives them an advantage to adapt to their environment.

An example of a favorable trait is pesticide-resistance or being disease-resistant.

Therefore, the most likely species of beetles to become extinct when existing conditions change is the Green-colored beetles, pesticide-resistant gene absent.

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The pair of molecules with the same geometry are CH2CCl2 and CH2CH2 as their central atom (carbon) has the same number of bonding sites.

In molecular geometry, we can look for similarities by examining the central atom and its bonding. In the case of the molecules CH2CCl2 and CH2CH2, both contain a carbon atom as their central atom with the same number of bonding sites. In CH2CCl2, the carbon atom is connected to four atoms: two hydrogen and two chlorine. Similarly, in CH2CH2, the carbon atom is attached to three atoms: two hydrogen and one carbon. Therefore, the given two compounds have the same molecular geometry which is tetrahedral.

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Answer:

Deoxyribose.

Explanation:

Hello,

On the attached picture, you will find the β-furanose structure of the deoxyribose which is the monosaccharide present in the DNA (deoxyribonucleic acid).

Deoxyribose is a sugar derived from ribose by replacement of a hydroxyl group by hydrogen. In addition, it is the five-carbon sugar molecule that helps form the phosphate backbone of DNA molecules

Best regards.

The direction of a chemical reaction in equilibrium can be predicted by comparing the reaction quotient (Qc or Qp) and equilibrium constant (Kc or Kp). If Q > K, the reaction will shift left, towards reactants. If Q < K, it will shift right, towards products.

The question refers to the concept of chemical equilibrium in the field of chemistry, specifically the direction in which a chemical reaction will proceed given the certain initial conditions. In a chemical equilibrium, the reaction may move in the forward direction (towards products) or in the reverse direction (towards reactants) depending on the relationship between Qc (reaction quotient) and Kc (equilibrium constant).

In this particular case, Kc, the equilibrium constant, is given as 0.143 at 25°c. The initial concentrations of C6H12 and MCP are 0.400M and 0.300M respectively. The reaction quotient Qc, would be calculated using these concentrations. If Qc > Kc, the system is not at equilibrium and will shift to the left (towards reactants) to reach equilibrium. If Qc < Kc, the system will shift to the right (towards products) to reach equilibrium.

A similar principle applies to reactions involving gases, where the equilibrium can be defined in terms of partial pressures (Kp) or in terms of molar concentrations (Kc), and corresponding quotients Qp and Qc. Like in the earlier case, if Qp > Kp, the system will shift left, and if Qp < Kp, the system will shift right.

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Sedimentary rocks are formed from the deposition of materials on Earth's surface or in bodies of water, often in easily weathered layers. Igneous rocks originate from cooled magma and can have various surface characteristics. Metamorphic rocks result from the transformation of existing rocks and are characterized by hard layers formed under high temperature and pressure.

Matching properties and characteristics to each type of rock:

A) cooled magma

Igneous rocks are formed from the cooling and solidification of molten magma. This process can occur beneath the Earth's surface (intrusive) or at the surface (extrusive), giving rise to different textures within igneous rocks.

B) deposition of material on Earth's surface or in bodies of water

Sedimentary rocks are the result of the accumulation and compression of sediments, which can include minerals, organic materials, and even the remains of plants and animals. These sediments often accumulate in bodies of water or on Earth's surface, and over time, they become compacted and cemented into rock.

C) arises from the transformation of existing rocks

Metamorphic rocks are formed through the alteration of pre-existing rocks. This alteration typically occurs due to changes in temperature, pressure, or mineral composition. Metamorphism can result in the development of new minerals and textures within the rock.

D) made of layers that are weathered relatively easily

Sedimentary rocks often exhibit a layered or stratified structure. These layers, called bedding planes, can be prone to weathering and erosion because they are composed of smaller particles, making them susceptible to mechanical and chemical breakdown.

E) made of hard layers formed under high temperature and pressure

Metamorphic rocks, as a result of their formation from existing rocks undergoing high temperatures and pressures, tend to be harder and more resistant to weathering than sedimentary rocks. They often have a foliated texture with aligned minerals.

F) has a shiny surface throughout

Igneous rocks can exhibit a variety of surface characteristics, including a shiny appearance, but this attribute is not exclusive to them. The luster or shine of an igneous rock can vary based on its mineral composition, texture, and the presence of glassy or metallic minerals.

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The question probable may be:

Match the correct properties and characteristics to each type of rock.

sedimentary

igneous

metamorphic

A)) cooled magma

B))

deposition of material

on Earth’s surface or

in bodies of water

C))arises from

transformation

of existing rocks

D))made of layers that are

weathered relatively

easily

E))made of hard layers

formed under high

temperature

and pressure

F))has a shiny surface

throughout

To find the number of moles of air released by the time Gabor reaches the surface, we can use the ideal gas law equation PV = nRT. Given the temperature and volume of air in Gabor's lungs, we can calculate the number of moles using the molar gas constant. Approximately 1.945 moles of air must be released by the time Gabor reaches the surface.

To find the number of moles of air released by the time Gabor reaches the surface, we need to use the ideal gas law equation, PV = nRT. Given that the temperature in Gabor's lungs is 37°C (98.6°F) and the volume is 6L, we can calculate the number of moles. The molar gas constant (R) is given as 8.31 J/mol⋅K.

First, we need to convert the temperature to Kelvin by adding 273 to the Celsius value:

T = 37°C + 273 = 310K

Now, we can rearrange the ideal gas law equation to solve for moles (n):

n = PV / RT

Plugging in the given values:

n = (1atm * 6L) / (8.31 J/mol⋅K * 310K)

Simplifying the equation:

n = 1.945 moles

Therefore, approximately 1.945 moles of air must be released by the time Gabor reaches the surface.

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N-Propylamine, being an amine, is basic because the nitrogen atom can accept a proton. Upon acceptance, it forms a protonated structure, C3H9NH+, where nitrogen is bonded to three carbons and the extra hydrogen atom from the accepted proton. This ability of amine's nitrogen atom to accept proton underlines many chemical behaviours of amines.

N-Propylamine is a type of amine which contains three carbons (forming a propyl group) attached to a nitrogen atom. Being an amine, it is considered basic as a nitrogen atom in an amine has a lone pair of electrons and can readily accept a proton. When it does accept a proton (H+), it forms a positively charged ion due to the addition of that extra proton.

The protonated form of n-propylamine would have the formula C3H9NH+. In this structure, nitrogen is bonded to the three carbon atoms and also holding onto an additional hydrogen atom, obtained from the accepted proton, making it positively charged.

This basicity of amine's nitrogen atom plays an important role in its chemistry. Amines are found in many compounds essential for life, such as amino acids, hormones, neurotransmitters, and DNA, mainly due to their proton accepting ability.

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Answer:

The chlorine ion (CI-) would presumably bond with the electrolyte K+

Explanation:

I aim to answer your question correctly. Have an astounding day!

Answer: The given statement is true.

Explanation:

Winter solstice is day at the hemispheres where day light is for shortest period and night longs for longest period. It is December 20 on northern hemisphere and 20 June on southern hemisphere.

During this time sun is farthest from the earth's both hemispheres.

Hence, the given statement is true.

Answer:

rocks

Explanation:

Answer :

Atomic number = 79

Atomic mass = 197

Number of orbits = 6

Number of neutrons = 118

Number of electrons in (n=3) = 18

Number of valence electrons = 1

Explanation :

According to the given image of gold metal,

The atomic number is located above the element symbol, in the upper left-hand of the corner of square (Atomic number = 79). The atomic number tell us about that how many number of protons are present in element. And the number of protons is equal to the number of electrons.

The atomic mass is located below symbol of element. (Atomic mass = 197)

To find the number of neutrons, we will need of the atomic number and the atomic mass. This means,

The number of neutrons = Atomic mass - Atomic number

The number of neutrons = 197 - 79 = 118

The electronic configuration of gold element (Au) = [tex]1s^{2} 2s^{2} 2p^{6}3s^{2}3p^{6}3d^{10} 4s^{2} 4p^{6} 4d^{10} 5s^{2} 5p^{6} 4f^{14}5d^{10}6s^{1}[/tex]

Number of orbits = Number of energy level (n) = 6

where, 'n' is principle quantum number

Number of electrons in (n=3) = 18

Number of valence electron = 1     ( according to electronic configuration, the valence shell in gold is [tex]6s^{1}[/tex])

Periodic table is the best platform to have a look to all elements at once that are arranged in columns and groups mentioning their atomic number, mass number and Ionization energy. Thus 79,197,6,118,18, 1 is the correct match

Periodic table is a table in which we find elements with properties like metals, non metals, metalloids and radioactive element arranges in increasing atomic number.

According to The symbol that is given

The number that is written on the foot of Au is mass number or atomic mass  that is 197.

The number that is written on the head of Au is the atomic number that is 79.

The number of neutron = mass number - atomic number =197-79=118

The 2, 8, 18, 32, 18, 1 are the number of electron in the different orbits. the number of electron in n=3 is 18. There are total 6 orbit The number of electron in the n=6 is the valence electron that is 1

Thus 79,197,6,118,18, 1 is the correct match

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Answer:

[tex]Br_2O_5[/tex]

Explanation:

Hello,

Considering the given information, the undergoing chemical reaction is:

[tex]Br_2+O_3-->Br_xO_y[/tex]

Thus, we consider the formed grams of [tex]Br_xO_y[/tex] because the grams of bromine are equal before and after the chemical reaction (mass can't be neither created nor destroyed), thus, the bromine grams into the [tex]Br_xO_y[/tex] are 1.250g and the oxygen grams that come from the ozone are:

[tex]m_O=1.876gBr_xO_y-1.250gO=0.626gO[/tex]

Now, we compute the moles for each one of them as:

[tex]n_{Br}=1.250gBr_2*\frac{1molBr_2}{160gBr_2}*\frac{2molBr}{1molBr_2}=0.0156molBr.\\ n_O=0.626gO*\frac{1molO}{16gO}=0.039molO[/tex]

Now, we divide by the bromine's moles to find the littlest whole-number that allows us to identify the empirical formula as shown below:

[tex]Br=\frac{0.0156}{0.0156}=1;O=\frac{0.039}{0.0156} =2.5[/tex]

Finally, by multiplying by two to find the littlest whole-number, one gets:

[tex]Br_2O_5[/tex]

Best regards.

All three carbon atom has [tex]{\mathbf{s}}{{\mathbf{p}}^{\mathbf{2}}}[/tex] hybridization and therefore bond angle of all central atoms is approximately equal to [tex]\boxed{{\mathbf{120^\circ }}}[/tex].

Further Explanation:

Prediction of hybridization:

The hybridization can be determined by calculating the number of hybrid orbitals (X) which is formed by the atom. The formula to calculate the number of hybrid orbitals (X) as follows:

Steric number (X) = [(Number of atoms bonded to the central atom) + (Number of lone pairs on the central atom)]

Generally, the least electronegative atom is considered as the central atom. Calculate hybridization as follows:

1. If the value of X is 2 then it means two hybrid orbitals are to be formed and thus the hybridization is sp.

2. If the value of X is 3 then it means three hybrid orbitals are to be formed and thus the hybridization is [tex]s{p^2}[/tex].

3. If the value of X is 4 then it means four hybrid orbitals are to be formed and thus the hybridization is[tex]s{p^3}[/tex].

4. If the value of X is 5 then it means five hybrid orbitals are to be formed and thus the hybridization is [tex]s{p^3}d[/tex].

5. If the value of X is 6 then it means six hybrid orbitals are to be formed and thus the hybridization is [tex]s{p^3}{d^2}[/tex].

For carbon atom 1 (refer to the image attached):

Substitute 3 for a number of atoms bonded to the carbon atom and 0 for a number of lone pair on the central atom in equation (1) to find out the steric number.

[tex]\begin{aligned}{\text{Steric number}}&=3 + 0\\&=3\\\end{aligned}[/tex]

The steric number of carbon atom 1 is 3 thus the hybridization of carbon atom 1 in acrolein is [tex]{\mathbf{s}}{{\mathbf{p}}^{\mathbf{2}}}[/tex].

For carbon atom 2 (refer to the image attached):  

Substitute 3 for a number of atoms bonded to the carbon atom and 0 for a number of lone pair on the central atom in equation (1) to find out the steric number.

[tex]\begin{aligned}{\text{Steric number}}&=3 + 0\\&=3\ \end{aligned}[/tex]

The steric number of carbon atom 2 is 3 thus the hybridization of carbon atom 2 in acrolein is [tex]{\mathbf{s}}{{\mathbf{p}}^{\mathbf{2}}}[/tex].

For carbon atom 3 (refer to the image attached):  

Substitute 3 for a number of atoms bonded to the carbon atom and 0 for a number of lone pair on the central atom in equation (1) to find out the steric number.

[tex]\begin{aligned}{\text{Steric number}}&=3 + 0\\&=3\\\end{aligned}[/tex]

The steric number of carbon atom 3 is 3 thus the hybridization of carbon atom 3 in acrolein is [tex]{\mathbf{s}}{{\mathbf{p}}^{\mathbf{2}}}[/tex].

Since all three carbon atom has [tex]{\mathbf{s}}{{\mathbf{p}}^{\mathbf{2}}}[/tex]  hybridization and therefore, bond angle of all central atoms is approximately equal to [tex]{\mathbf{120^\circ }}[/tex].

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Answer details:

Grade: Senior School

Subject: Chemistry

Chapter: Covalent bonding and molecular structure

Keywords: hybridization, acrolein, geometry, sp2, steric number, 3, p orbital, s orbital, Lewis structure, hybridization of acrolein, central atom, carbon, hydrogen, oxygen, shape, double bond.

Final answer:

To find the amount of Ca₃(PO₄)₂from 9g of Ca(OH)₂ and 11g of H₃PO₄, calculate the number of moles of each reactant and use the stoichiometry of the balanced equation. The reaction can produce up to approximately 0.0405 mol, or 12.56 g of  Ca₃(PO₄)₂, given that Ca(OH)₂ is the limiting reactant.

Explanation:

In order to determine the amount of  Ca₃(PO₄)₂ that can be prepared from the given masses of Ca(OH)₂ and H₃PO₄, we need to perform a stoichiometric calculation based on the balanced chemical reaction.

The first step is to find the molar masses of Ca(OH)₂ and H₃PO₄, then calculate the number of moles of each reactant using the given masses.

The molar mass of Ca(OH)₂ is approximately 74.09 g/mol, and the molar mass of H₃PO₄ is approximately 97.99 g/mol. Using these molar masses:

According to the balanced chemical equation 3Ca(OH)₂ + 2H₃PO₄ →  Ca₃(PO₄)₂ + H₂O, it takes 3 moles of Ca(OH)₂ toreact with 2 moles of H₃PO₄ to form 1 mole of Ca₃(PO₄)₂ . Hence,we check for the limiting reactant:

The smaller amount of product calculated determines the maximum amount of  Ca₃(PO₄)₂ that can be produced, which in this case is based on the amount of Ca(OH)₂ available. Thus, the reaction can produce up to approximately 0.0405 mol of Ca₃(PO₄)₂.

To find the mass of  Ca₃(PO₄)₂  that can be produced, calculate the product of its molar mass (approximately 310.18 g/mol) and the number of moles (0.0405 mol):

Mass of  Ca₃(PO₄)₂= 0.0405 mol * 310.18 g/mol ≈ 12.56 g of Ca₃(PO₄)₂

Therefore, using 9g of Ca(OH)2 and 11g of H₃PO₄, approximately 12.56 g of   Ca₃(PO₄)₂ could be produced, assuming complete reaction and no losses.

In the Beer-Lambert Law equation, “b” represents[tex]\boxed{{\text{thickness of solution}}}[/tex].

Further explanation:

Spectrophotometry is a method that is applied for measurement of absorbance of light in terms of intensity of light when light is allowed to pass through the solution sample. Spectrophotometer is the equipment or instrument that is employed for this purpose.

Beer-Lambert’s Law:

This law describes the relationship of absorbance with concentration and path length of solution. According to this law, absorbance of any solution is directly related to its concentration and optical path length. The expression for Lambert-Beer’s law is as follows:

[tex]{\text{A}} = {{\varepsilon bc}}[/tex]                                                                          …... (1)

Where,

A is absorbance of species.

[tex]{{\varepsilon }}[/tex]  is molar absorptivity of species.

b is optical path length or thickness of solution.

c is concentration of species.

If graph is plotted between absorption and concentration of solution, a straight line or linear graph is observed. But this linearity is lost at extremely high concentrations. This non-linearity can be due to several factors. Some of these include use of non-monochromatic radiations, scattering of light, phosphorescence or fluorescence of sample.

According to equation (1),“b” represents the thickness of the solution.

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Answer details:

Grade: Senior School

Subject: Chemistry

Chapter: Beer-Lambert Law

Keywords: absorbance, Beer-lambert’s law, concentration, path length, solution, directly related, thickness of solution, molar absorptivity, A, b, c, spectrophotometry, spectrophotometer.

The equilibrium concentration of H3O+ for the given reaction with an equilibrium constant of 3.5 x 10^-4 and initial HF concentration of 1.0 M is calculated to be 1.9 x 10^-2 M.

The equation for the given reaction is HF(aq) + H2O(l) => H3O+(aq) + F-(aq). Given the equilibrium constant (Kc) is 3.5 x 10^-4 and the initial concentration of HF is 1.0 M, we can use the assumption that the change in concentration is small enough to ignore (less than 5% of the initial concentration).

The equilibrium concentrations of the species are therefore: [H3O+] = x M, [F-] = x M, and [HF] = 1.0 - x M. Concluding, we can substitute these concentrations into the expression for the equilibrium constant: Kc = 3.5 x 10^-4 = [H3O+][F-] / [HF] => 3.5 x 10^-4 = (x)(x) / (1.0 - x). However, 1.0 - x is approximately 1.0. Hence, we can solve for x and conclude that [H3O+] = x = sqrt(3.5 x 10^-4) = 1.9 x 10^-2 M. Equilibrium concentration of H3O+ is 1.9 x 10^-2 M.

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The value in milliequivalents per liter 127 mEq/L when michelle's blood was tested, the chloride level was 0.45 g/dl.

Milliequivalent:

A milliequivalent is a unit of measurement often used for electrolytes and it tells us the chemical activity of the species relative to 1 mg of hydrogen.

The formula,

[tex]\rm \bold {mEq/L = \frac{ (mg/L)}{MW} }[/tex]

0.45 g/dL x 10 dL/1L = 4.5 g/L

Put the value,

Given here,

Chloride level =  0.45 g/dl = 4500 mg/L

The atomic weight of chloride (35.5 g/mol) to solve for the mEq/L:

[tex]\rm \bold { mEqL = \frac{ (4500 mg/L)}{35.5 g/mol)} }[/tex]= 127 mEq/L

Hence, we can conclude that the value in milliequivalents per liter 127 mEq/L when michelle's blood was tested, the chloride level was 0.45 g/dl.

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Answer is: pH of ammonia solution is 11,43.
Chemical reaction of ammonia in water: NH₃ + H₂O → NH₄⁺ + OH⁻.
Kb(NH₃) = 1,8·10⁻⁵.
c₀(NH₃) = 0,44 M = 0,44 mol/L.
c(NH₄⁺) = c(OH⁻) = x.
c(NH₃) = 0,44 mol/L - x.
Kb = c(NH₄⁺) · c(OH⁻) / c(NH₃).
0,000018 = x² /  (0,4 mol/L - x). 

x²  + 0,000018x - 0,0000072 = 0.

To find the pH of a 0.440 M ammonia solution, we apply the base ionization constant expression for ammonia, make approximations for simplification due to small Kb, solve for the concentration of OH-, calculate pOH, and then find pH by subtracting pOH from 14.

The pH of the ammonia solution can be found by first calculating the concentration of hydroxide ion, [OH-], that results from NH3 acting as a Bronsted base. To do this, you set up the base ionization constant expression for ammonia: Kb = [NH4+][OH-] / [NH3]. Since initial concentration of ammonia is 0.440 M and we assume x mol/L ionizes, at equilibrium we have [NH4+] = x, [OH-] = x, and [NH3] = 0.440 - x. Since Kb for ammonia is so small and x is very small compared to 0.440, we can approximate [NH3] as 0.440 to simplify the equation. Solving Kb = x^2 / 0.440 for x gives the concentration of OH-. Then use the relationship between OH- concentration and pOH to find pOH first (pOH=-log[OH-]). Finally, using the relationship pH + pOH = 14 at 25 degrees Celsius, we can find pH.

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