A sample of nitrogen gas occupies 1.55 L at 27.0°C and 1.00 atm. What will the volume be at −100.0°C and the same pressure?

Answer:

2H2S + 3O2 → 2SO2 + 2H2O

Explanation:

Step 1: Data given

Hydrogen sulfide = H2S

Oxygen = O2

sulfur dioxide = SO2

water = H2O

Step 2: The unbalanced equation

H2S + O2 → SO2 + H2O

Step 3: Balancing the equation

H2S + O2 → SO2 + H2O

On the left side we have 2x O (in O2) and on the right side we have 3x O (2x in SO2 and 1x in H2O). To balance the amount of O, we have to multiply O2 (on the left side) by 3 and SO2 and H2O on the right side by 3.

H2S + 3O2 → 2SO2 + 2H2O

On the right side we have 4x H and on the left side we have 2x H. To balance the amount of H, we have to multiply H2S by 2.

Now the equation is balanced.

2H2S + 3O2 → 2SO2 + 2H2O

Explanation:

It is given that the total units of polymixn B silfate are present as follows.

              (10 + 250) ml = 260 ml

Therefore, total volume necessary to inject the whole volume is as follows.

              [tex]260 \times 15[/tex] drops

               = 3900

It is given that total time required to inject the solution is 2 hours. Converting it into minutes as follows.

                     [tex]2 hrs \times \frac{60 min}{1 hr}[/tex]

                       = 1200 minutes

Hence, rate of injection is calculated as follows.

      Rate of injection (r) = [tex]\frac{\text{total drops}}{\text{total time}}[/tex]

                   r = [tex]\frac{3900}{120}[/tex]

                     = 32.5 drops/min

Thus, we can conclude that the rate of injection is 32.5 drops/min.

To infuse 260 mL of solution over 2 hours with a drop factor of 15 drops/mL, the flow rate should be adjusted to 32.5 drops per minute.

We need to calculate the rate at which the solution should be administered in order to complete the infusion over 2 hours. First, we look at the total volume of the solution to be infused which is 250 mL of 5% dextrose injection plus the initial 10 mL, making it 260 mL in total. To calculate the rate, we'll use the formula:

Rate (drops/min) = Total volume (mL) × Administration set drop factor (drops/mL) ÷ Time (min)

Since we are told that the administration set delivers 15 drops/mL and that the infusion needs to be administered over 2 hours (which is 120 minutes), we can substitute the values into the formula:

Rate (drops/min) = (260 mL) × (15 drops/mL) ÷ (120 min)

After calculating, we find:

Rate (drops/min) = 32.5 drops/min

Note: This calculation is based on the student's given values and intended to demonstrate the process of determining the flow rate for an IV infusion.

Answer:

The energy required to ionize the ground-state hydrogen atom is 2.18 x 10^-18 J or 13.6 eV.

Explanation:

To find the energy required to ionize ground-state hydrogen atom first we calculate the wavelength of photon required for this operation.

It is given by Bohr's Theory as:

1/λ = Rh (1/n1² - 1/n2²)

where,

λ = wavelength of photon

n1 = initial state = 1 (ground-state of hydrogen)

n2 = final state = ∞ (since, electron goes far away from atom after ionization)

Rh = Rhydberg's Constant = 1.097 x 10^7 /m

Therefore,

1/λ = (1.097 x 10^7 /m)(1/1² - 1/∞²)

λ = 9.115 x 10^-8 m = 91.15 nm

Now, for energy (E) we know that:

E = hc/λ

where,

h = Plank's Constant = 6.625 x 10^-34 J.s

c = speed of light = 3 x 10^8 m/s

Therefore,

E = (6.625 x 10^-34 J.s)(3 x 10^8 m/s)/(9.115 x 10^-8 m)

E = 2.18 x 10^-18 J

E = (2.18 x 10^-18 J)(1 eV/1.6 x 10^-19 J)

E = 13.6 eV

Answer: The atomic mass of E is 16

Explanation:Please see attachment for explanation

Answer:

The molar mass of E = 15.96 g/mol

Explanation:

Step 1: Data given

The compound Cr2E3 contains 68.47 % Cr and 31.53 % E

The molar mass of Cr = 52 g/mol

⇒ 2*52 = 104 g/mol

The molar mass of Cr2E3 = 2*52 g/mol + 3X

Step 2: Calculate mass of Cr2E3

Cr is 68.47 %

Mass of Cr = 104 grams

Cr2E3 is 100%

Mass of Cr2E3 = 151.89 grams

Molar mass of Cr2E3 = 151.89 g/mol

Step 3: Calculate mass of E

Mass of E3 = mass of Cr2E3 - mass of Cr

Mass of E3 = 151.89 grams - 104 grams

Mass of E3= 47.89 grams

Mass of E = 15.96 grams

The molar mass of E = 15.96 g/mol

Answer:

The answer is Inductive effect

Explanation:

To determine the acidity or alkalinity of an organic compound. We have to keep in mind that the whole analysis is based on the comparison between the compounds, and we must work with the conjugated base of the molecule. Keeping in mind, the more unstable the base, the less acidic the molecule is. Thus, to determine instability, the Inductive Effect of the molecule can be used.

This type of effect occurs when atoms of different electronegativities are linked or very close in the compound. The most electronegative atom has a tendency to bring electrons close to it, thus creating a dipole. This dipole can have a stabilizing effect on the molecule, as it “relieves” the excessive charge on some occasions, better accommodating the charges.

However, in some cases, instead of chains with chlorine radicals, we may have chains with methyl radicals. This has a major impact on the inductive effect, keeping in mind that alkyl groups are electron donors.

Answer:

Dipole-Dipole force

Explanation:

Dipole - Dipole force -

These are the force of attraction , that occurs between two dipole , i.e. ,a species with two poles , hence , the attraction between the delta positive charge of first species with the delta negative charge of the second species , arises to a dipole - dipole force of attraction.

Hence, from the question,

SO₂ , is a polar compound , where O is more electronegative in comparison to S , thus , O attains a delta negative charge and S attains a delta positive charge and therefore , generates a dipole , and interacts with the dipole of the second molecule of SO₂ , arising a  dipole - dipole force of attraction .

Answer:

D is the true statement.

Explanation:

Recall that during a phase change the temperature remains constant until all the material has changed its phase. The energy put into the system is utilized to make the phase change, and afterward the temperature of the vliquid will start to increase.

In this case we have a mixture of liquid water and ice at  0ºC  ( assume standard pressure) which is the temperature in which liquid water and ice coexist. The ice will  melt until consumed at constant T = 0ºC, and then the temperature of the liquid water will start to increase at a uniform rate since we are heating at a constant rate.

Now we are in position to answer this question.

A- False the temperature does not increase as the ice melts.

B- False for the reasons given above.

C- False the temperature does not increase slowly as the ice melts, but remains constant.

D-True the temperature of the mixture does not change until all the ice has melted, and then the temperature of liquid water will start to increase uniformly.

Heat applied to a water-ice mixture initially goes into breaking the hydrogen bonds in ice (latent heat of fusion), without causing a temperature increase. Only after all the ice has melted, the added heat increases the water temperature.

The correct option is D: The temperature of the mixture does not change at all until all the ice has melted, at which point it increases at a constant rate. This is because the energy from the heat being applied is first used to break the hydrogen bonds in ice, a process known as latent heat of fusion, which does not involve an increase in temperature. Only after all the ice has melted will the added heat then lead to an increase in the temperature of the water.

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Answer:

a) 1.44 kg

b) 1.47 kg

Explanation:

a) By the reaction given, the stoichiometry, the molar ratio, is:

2 moles of C3H6 : 2 moles of NH3 : 3 moles of O2 : 2 moles of C3H3N : 6 moles of H2O, or 2:2:3:2:6.

By the mixture given, one or two of the reactants may be in excess, so, we must found which of them is the limiting reactant, the reactant that will be totally consumed.

The molar masses of the compounds are:

C3H6 = 42.0 g/mol

NH3 = 17.0 g/mol

O2 = 32.0 g/mol

C3H3N = 53.1 g/mol

H2O = 18.0 g/mol

Thus, the mass ratio between the reactants will be the molar ratio multiplied by the molar mass:

2 moles of C3H6*42.0 g/mol --- 2 moles of NH3*17.0 g/mol -- 3 moles of O2* 32.0 g/mol

84.0 g of C3H6 -- 34.0 g of NH3 -- 96.0 g of O2

Thus, assuming C3H6 as limiting, let's use the rule of three to find out which would be the masses of the other reactants needed:

84.0 g of C3H6 -- 34.0 g of NH3

1.14 kg -- x

84.0x = 38.76 kg

x = 0.46 kg of NH3

Because there's more NH3 than it's required, NH3 is in excess, and C3H6 is limiting. Let's test using O2 as reference:

84.0 g of C3H6 -- 96.0 f of O2

1.14 kg -- x

84.0x = 109.44

x = 1.30 kg of O2

So, O2 is also in excess, and the limiting reactant is C3H6.

Thus, the molar ratio between C3H6 and C3H3N is 2:2, which is simplified by 1:1, and so the mass ratio is 42 g of C3H6 -- 53.1 g of C3H3N, so the mass of acrylonitrile can be found by a rule of three:

42 g of C3H6 -- 53.1 f of C3H3N

1.14 kg -- x

42x = 60.534

x = 1.44 kg

b) The molar ratio between C3H6 and water is 2:6, which can be simplified by 1:3, thus the mass ratio is 1*42g/mol of C3H6 -- 3*18 g/mol of H2O

42 g of C3H6 -- 54 g of H2O

By a rule of three the mass of water is:

42 g of C3H6 -- 54 g of H2O

1.14 kg -- x

42x = 61.56

x = 1.47 kg

Final answer:

The mass of acrylonitrile that can be produced from the given mixture of reactants is approximately 1437.73 g, and the mass of water produced is 1464.68 g, assuming a 100% yield and propylene as the limiting reactant.

Explanation:

Calculating Mass of Acrylonitrile and Water Produced

To answer your questions about the production of acrylonitrile and water, we first need to use the provided stoichiometry and the masses of the reactants. The balanced chemical reaction is:

2C₃H₆(g) + 2NH₃(g) + 3O₂(g) → 2C₃H₃N(g) + 6H₂O(g)

This informs us that two moles of propylene (C₃H₆) react with two moles of ammonia (NH₃) and three moles of oxygen (O₂) to produce two moles of acrylonitrile (C₃H₃N) and six moles of water (H₂O).

Firstly, we determine the number of moles of each reactant:

Propylene (C₃H₆): Molar mass = 42.08 g/mol, Moles = 1140 g / 42.08 g/mol = 27.11 moles

Ammonia (NH₃): Molar mass = 17.031 g/mol, Moles = 1650 g / 17.031 g/mol = 96.91 moles

Oxygen (O₂): Molar mass = 32.00 g/mol, Moles = 1850 g / 32.00 g/mol = 57.81 moles

Since the limiting reactant determines the amount of product formed, we can find it by comparing the mole ratios. In this case, propylene is the limiting reactant. Using the stoichiometry, we calculate the mass of acrylonitrile produced:

Moles of acrylonitrile produced = Moles of propylene used = 27.11 moles

Mass of acrylonitrile produced = Moles of acrylonitrile produced x Molar mass of acrylonitrile = 27.11 moles x 53.06 g/mol = 1437.73 g

For the mass of water produced:

Moles of water produced = 3 x Moles of propylene used = 3 x 27.11 mol = 81.33 moles

Mass of water produced = Moles of water produced x Molar mass of water = 81.33 moles x 18.015 g/mol = 1464.68 g

Assuming 100% yield, we can produce approximately 1437.73 g of acrylonitrile and 1464.68 g of water from the given reactants.

Answer:

Explanation:Isobaric is a thermodynamic process in which the pressure of the system is zero. It is a process where there is no work done in the system. Therefore the temperature of the system according to the ideal gas law will vary linearly with pressure and inversely with volume.

Answer:

Hi

A conjugate acid is a chemical compound formed by the reception of a proton by a base; therefore, it is a base with an added hydrogen ion. On the other hand, a conjugate base is what remains after an acid has donated its proton during a chemical reaction, so it can be said that a conjugate base is a species modified by extracting a proton from an acid.

In pyrrole, the electron pair can be relocated to the ring and its availability is very small. In addition, the concept of aromaticity comes into play.

In the attached file are the schemas of the conjugate bases.

Explanation:

Answer: 2,5-dimethyl octane.

Explanation:

The basic rules for naming of organic compounds are :

1. First select the longest possible carbon chain.

2. The longest possible carbon chain should include the carbons of double or triple bonds.

3. The naming of alkane is done by adding the suffix -ane, alkene by adding the suffix -ene, alkyne by adding the suffix -yne and carboxylic acid by adding the suffix -oic acid.

4. The numbering is done in such a way that first carbon of double or triple bond gets the lowest number.

5. The carbon atoms of the double or triple bond get the preference over the other substituents present in the parent chain.

6. If two or more similar alkyl groups are present in a compound, the words di-, tri-, tetra- and so on are used to specify the number of times of the alkyl groups in the chain.

Thus the name for straight chain made of 8 carbons with [tex]-CH_3[/tex] groups on the second and fifth carbons will be 2,5-dimethyl octane.

Answer:

4. Oxygen

Explanation:

The question is incomplete, here is the complete question.

A chemist adds 345.0 mL of a 0.0013 mM (MIllimolar) copper(II) fluoride [tex]CuF_2[/tex] solution to a reaction flask.

Calculate the mass in micrograms of copper(II) fluoride the chemist has added to the flask. Be sure your answer has the correct number of significant digits.

Answer : The mass of copper(II) fluoride is, 0.13 mg

Explanation :  Given,

Millimolarity of copper (II) fluoride = 0.0013 mM

This means that 0.0013 millimoles of copper (II) fluoride is present in 1 L of solution

Converting millimoles into moles, we use the conversion factor:

1 moles = 1000 millimoles

So, [tex]0.0013mmol\times \frac{1mol}{1000mmol}=1.3\times 10^{-6}mol[/tex]

To calculate the number of moles, we use the equation:

[tex]\text{Number of moles}=\frac{\text{Given mass}}{\text{Molar mass}}[/tex]

We are given:

Moles of copper (II) fluoride solution = [tex]1.3\times 10^{-6}mol[/tex]

Molar mass of copper (II) fluoride = 101.5 g/mol

Putting values in above equation, we get:

[tex]1.3\times 10^{-6}mol=\frac{\text{Mass of copper (II) fluoride}}{101.5g/mol}\\\\\text{Mass of copper (II) fluoride}=(1.3\times 10^{-6}mol\times 101.5g/mol)=1.32\times 10^{-4}g[/tex]

Converting this into milligrams, we use the conversion factor:

1 g = 1000 mg

So,

[tex]\Rightarrow 1.32\times 10^{-4}g\times (\frac{1000mg}{1g})=0.13mg[/tex]

Therefore, the mass of copper(II) fluoride is, 0.13 mg

Answer:

the mass of copper(II) fluoride added to the flask is approximately 476,250 micrograms.

Explanation:

To calculate the mass of copper(II) fluoride (CuF2) added to the reaction flask, you need to multiply the volume (in liters) of the solution by its molarity and then convert the result to micrograms.

Given:

Volume of the solution (V) = 0.015 L

Molarity of the solution (M) = 0.500 M

First, calculate the number of moles of copper(II) fluoride using the formula:

Moles (n) = Molarity (M) × Volume (V)

n = 0.500 M × 0.015 L = 0.0075 moles

Now, you need to convert moles to micrograms. One mole of any substance contains Avogadro's number of molecules (6.022 × 10^23). You also need to consider the molar mass of copper(II) fluoride (CuF2), which is the sum of the atomic masses of copper (Cu) and two fluorine (F) atoms:

Molar mass of CuF2 = (1 * Cu) + (2 * F) = 63.5 g/mol (approximately)

Now, calculate the mass in grams:

Mass (g) = Moles (n) × Molar mass (Molar mass of CuF2)

Mass (g) = 0.0075 moles × 63.5 g/mol ≈ 0.47625 g

Now, convert grams to micrograms. There are 1,000,000 micrograms in a gram:

Mass (micrograms) = Mass (g) × 1,000,000

Mass (micrograms) = 0.47625 g × 1,000,000 µg/g = 476,250 µg

So, the mass of copper(II) fluoride added to the flask is approximately 476,250 micrograms.

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Answer: The vapor pressure of water at [tex]50^0C[/tex] is 93.8 torr

Explanation:

The vapor pressure is determined by Clausius Clapeyron equation:

[tex]ln(\frac{P_2}{P_1})=\frac{\Delta H_{vap}}{R}(\frac{1}{T_1}-\frac{1}{T_2})[/tex]

where,

[tex]P_1[/tex] = initial pressure at [tex]25^oC[/tex] = 23.8 torr

[tex]P_2[/tex] = final pressure at [tex]50^oC[/tex] = ?

[tex]\Delta H_{vap}[/tex] = enthalpy of vaporisation = 43.9 kJ/mol = 43900 J/mol

R = gas constant = 8.314 J/mole.K

[tex]T_1[/tex] = initial temperature = [tex]25^oC=273+25=298K[/tex]

[tex]T_2[/tex] = final temperature = [tex]50^oC=273+50=323K[/tex]

Now put all the given values in this formula, we get

[tex]\log (\frac{P_2}{23.8}=\frac{43900}{2.303\times 8.314J/mole.K}[\frac{1}{298K}-\frac{1}{323K}][/tex]

[tex]\frac{P_2}{23.8}=antilog(0.5955)[/tex]

[tex]P_2=93.8torr[/tex]

Therefore, the vapor pressure of water at [tex]50^0C[/tex] is 93.8 torr

The correct formula unit equation for the reaction of potassium sulfate and barium chloride forming potassium chloride and barium sulfate in water is: K2SO4 (aq) + BaCl2 (aq) → 2 KCl (aq) + BaSO4 (s). This is because both potassium sulfate and barium chloride are soluble in water, but barium sulfate is not.

The correct formula unit equation for the reaction of potassium sulfate and barium chloride forming potassium chloride and barium sulfate in water is: K2SO4 (aq) + BaCl2 (aq) → 2 KCl (aq) + BaSO4 (s).

This can be justified as when potassium sulfate (K2SO4) and barium chloride (BaCl2) are dissolved in water they dissociate into their respective ions, making them aqueous. The formation of potassium chloride (KCl) acknowledges that potassium ions and chloride ions can be attracted to each other in the water solution, so this is also aqueous. However, barium sulfate (BaSO4) is known to be insoluble in water, and hence, precipitates as a solid (s). Hence, option 2 is correct.

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Answer:

1) chemical change

2) filtration

Explanation:

A chemical change involves the formation of a new substance. In this case, the pure white solid formed is an entirely new substance, with a different chemical identity from those of the two solutions mixed to form it. The new solid is a precipitate. Precipitates are easily separated by filtration of the reaction mixture. Another name for chemical change is chemical reaction.

1) This is an example of chemical change.

2) The most convenient way to separate the newly formed white solid from the liquid mixture is filtration

A chemical change involves the formation of a new substance. In this case, the pure white solid formed is an entirely new substance, with a different chemical identity from those of the two solutions mixed to form it. The new solid is a precipitate. Precipitates are easily separated by filtration of the reaction mixture. Another name for chemical change is chemical reaction.

The most convenient way  to separate the newly formed white solid from the liquid mixture is filtration since precipitates are formed.

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Q: What is the change of entropy for 3.0 kg of water when the 3.0 kg of water is changed to ice at 0 °C? (Lf = 3.34 x 105 J/kg)

Answer:

-3670.33 J/K

Explanation:

Entropy: This can be defined as the degree of randomness or disorderliness of a substance. The S.I unit of Entropy is J/K.

Mathematically,  change of Entropy can be expressed as,

ΔS = ΔH/T ....................................... Equation 1

Where ΔS = Change of entropy, ΔH = heat change, T = temperature.

ΔH = -(Lf×m).................................... Equation 2

Note: ΔH is negative because heat is lost.

Where Lf = latent heat of ice = 3.34×10⁵ J/kg, m = 3.0 kg, m = mass of water = 3.0 kg

Substitute into equation

ΔH = -(3.34×10⁵×3.0)

ΔH = - 1002000 J.

But T = 0 °C = (0+273) K = 273 K.

Substitute into equation 1

ΔS = -1002000/273

ΔS = -3670.33 J/K

Note: The negative value of ΔS shows that the entropy of water decreases when it is changed to ice at 0 °C

The change in entropy for 3.0 kg of water changing into ice at 0 degrees Celsius is approximately 3658 J/K. This is calculated using the formula for entropy change and the known values for the heat of fusion for water and the temperature in Kelvin.

The change in entropy during a phase transition, such as the one from water to ice, can be calculated using the formula ΔS = Q/T, where ΔS is the change in entropy, Q is the heat involved during the phase transition, and T is the temperature in Kelvin. For the given scenario, the phase change is from liquid to solid (freezing), and the heat involved (Q) is equal to the mass multiplied by the heat of fusion for water. The heat of fusion for water is 333.55 J/g and the temperature (T) at which this change occurs is 0 degrees Celsius or 273.15 Kelvin. The mass is 3.0 kg or 3000g. Therefore, Q = 3000g x 333.55J/g = 999150 J. Substituting these values in our formula, ΔS = 999150 J / 273.15 K ≈ 3658 J/K.

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A

Individual lipid molecules are free to diffuse laterally in the surface of the bilayer is a general feature of the lipid bilayer in all biological membranes

Explanation:

The inside of the bilipid layer of the cell membrane of cells is made of fatty acid chains and cholesterol which are nonpolar molecules. they are sandwiched between the polar (glycerol) ends of the  chains because they are hydrophobic and cannot interact with the ‘watery’ extracellular fluids.  Non-polar molecules like lipids can easily diffuse laterally, within this lipid layers of the membrane because non-polar molecules interact well with other nonpolar molecules.

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Answer: the empirical formula is C3H4O3

Explanation:Please see attachment for explanation

Answer:

The answer to your question is  C₃H₄O₃

Explanation:

Data

CxHyOz

mass of sample = 0.150 g

mass of CO₂ = 0.225 g

mass of H₂O = 0.0614 g

Reaction

                     CxHyOz + O₂   ⇒   CO₂   + H₂O

Process

1.- Calculate the moles of C

                            44 g of CO₂ ----------------- 12 g of C

                            0.225 g    ---------------- x

                            x = (0.225 x 12) / 44

                            x = 0.0614 g of C

                            12 g of C -------------------- 1 mol

                           0.0614 g of C --------------- x

                                 x = 0.0051 moles of Carbon

2.- Calculate the moles of hydrogen

                            18 g of H₂O ------------------ 2 g of H

                             0.0614 g --------------- x

                             x = 0.0068 g of H

                             1 g of H ----------------------- 1 mol of H

                            0.0068 g --------------------- x

                             x = 0.0068 moles of H

3.- Calculate the mass of Oxygen

    Mass of oxygen = 0.150 - 0.0614 - 0.0068

                               = 0.0818 g

                       16 g of O -------------------  1 mol

                      0.0818 g -------------------- x

                         x = (0.0818 x 1) / 16

                         x = 0.0051 moles of O

4.- Divide by the lowest number of moles

Carbon       0.0051 / 0.0051  = 1

Hydrogen   0.0068 / 0.0051 = 1.33

Oxygen       0.0051 / 0.0051 = 1

Multiply these numbers by 3

Carbon 3

Hydrogen = 4

Oxygen = 3

5.- Write the empirical formula

                                      C₃H₄O₃

Answer:

In the arrangement of particles within any atom, the outermost sort of particle is always the electron.

Explanation:

An atom consist of electron, protons and neutrons. Protons and neutrons are present with in nucleus while the electrons are present out side the nucleus.

All these three subatomic particles construct an atom. A neutral atom have equal number of proton and electron. In other words we can say that negative and positive charges are equal in magnitude and cancel the each other.

The electron is subatomic particle that revolve around outside the nucleus and has negligible mass. It has a negative charge.

Symbol= e-

Mass= 9.10938356×10-31 Kg

It was discovered by j. j. Thomson in 1897 during the study of cathode ray properties.

While neutron and proton are present inside the nucleus. Proton has positive charge while neutron is electrically neutral. Proton is discovered by Rutherford while neutron is discovered by James Chadwick in 1932.

Symbol of proton= P+  

Symbol of neutron= n0  

Mass of proton=1.672623×10-27 Kg

Mass of neutron=1.674929×10-27 Kg

In any atom, the outermost type of particle is the electron, specifically the valence electrons in the outermost shell. These electrons play a significant role in chemical reactions and bonding properties of the element. The organization of the Periodic Table reflects patterns in valence electron configurations that correspond to chemical behaviors.

In the arrangement of particles within any atom, the outermost sort of particle is always the electron. Electrons are the smallest of the three types of sub-atomic particles, carrying a negative charge and occupying the space outside the atomic nucleus. Inside the nucleus, much larger particles—protons and neutrons—are found, with protons having a positive charge and neutrons being electrically neutral.

The outermost electrons are of particular importance because they are the valence electrons. These are the electrons that reside in the outermost shell, or valence shell, of an atom in its uncombined state, and they play critical roles in determining the chemical properties of an element as well as its ability to form bonds with other atoms. The electron configuration of an atom is notably important because atoms with the same outer electron configurations tend to show similar chemical behavior, as demonstrated in the organization of the Periodic Table.

The number of valence electrons in the outermost shell can define an element's chemical reactivity and the types of bonds it can form. The arrangement of electrons in atoms means that the electrons with the highest energy levels, which are the valence electrons, are more likely to interact in chemical reactions than the core electrons which are closer to the nucleus and have lower energy levels.

Answer:

mm protein = 15365.8183 g/mol

Explanation:

∴ molar mass (mm) ≡ g/mol

∴ π = 0.118 atm

∴ T = 25°C ≅ 298 K

∴ concentration (C ) [=] mol/L

∴ mass protein = 371 mg = 0.371 g

∴ volume sln = 5.00 mL = 5 E-3 L

⇒ C = π / RT = (0.118 atm)/((0.082 atm.L/K.mol)(298 K))

⇒ C = 4.8289 E-3 mol/L

⇒ mol protein = (4.8289 E-3 mol/L)×(5 E-3 L) = 2.4145 E-5 mol

⇒ mm protein = (0.371 g)/(2.4145 E-5 mol) = 15365.8183 g/mol

Final answer:

To find the molar mass of the protein, we begin by converting mass to grams and volume to liters, and then apply the van 't Hoff equation for osmotic pressure. After rearranging the equation, we calculate the molar mass using the given osmotic pressure, volume, temperature, and ideal gas constant.

Explanation:

To calculate the molar mass of the protein using the osmotic pressure, we can use the van 't Hoff equation for osmotic pressure, Π = MRT, where Π is the osmotic pressure, M is the molarity of the solution, R is the ideal gas constant, and T is the temperature in Kelvin. In this scenario, the student provided the mass of the protein (371 mg), the volume of the solution (5.00 mL), the osmotic pressure (0.118 atm), and the temperature at which the osmotic pressure was measured (25 °C which is 298.15 K).

First, convert the mass of the protein to grams:

371 mg = 0.371 g

Next, convert the volume from mL to liters:

5.00 mL = 0.005 L

Now, calculate the molarity (M) of the protein:

Number of moles = mass (g) / molar mass (g/mol)

As we do not know the molar mass yet, let's call it 'Mm'. The molarity (concentration) of the solution is:

Molarity (M) = Number of moles / Volume (L) = (mass (g) / Mm (g/mol)) / Volume (L)

Convert the temperature to Kelvin:

25 °C = 298.15 K

Using the van 't Hoff equation, we solve for the molar mass (Mm):

Π = (mass (g) / Mm (g/mol)) / Volume (L) × R × T

0.118 atm = (0.371 g / Mm (g/mol)) / 0.005 L × 0.0821 (L·atm/K·mol) × 298.15 K

Rearranging the equation and solving for Mm gives us:

Mm = (0.371 g / 0.005 L) / (0.118 atm / (0.0821 L·atm/K·mol × 298.15 K))

After calculating, we find the molar mass of the protein.

Answer: (1). pH = 1.70

(2). pH = 2.3

(3). pH = 3.3

(4). pH = 4.3

(5). pH = 8.41

(6). pH = 10.22

Explanation:

we assume that the formula representation of acid is H₂A

the titration curve has reasonably sharp breaks at both equivalence points, corresponding to the reactions;

H₂A + OH⁻ → HA⁻ + H₂O

HA⁻ + OH⁻ → A²⁻ + H₂O

the volume of NaOH (V₀) at the first equivalent point is,

V₀ = (20.0 mL)(0.100M) / 0.100M = 20.0mL

where volume of NaOH at 1/2 equivalent point is,

V₀/2 = 10.0mL

also Volume of NaOH at the second equivalence (2V₀) point is 40.0mL

the volume of NaOH at 1/2 second equivalent point is,

V₀ + V₀/2 = 30.0mL

Volume of NaOH after second equivalence exceeds 40mL

therefore, at 0 mL NaOH addition;

where the extent of ionization is assumed to be x, we have

                        H₂A   ⇆     HA⁻   +   H⁺

where initial:   0.1 M       -            -

          change:   -x         +x           +x

          Equili:      0.1-x      x             x

Kаl = [HA⁻][H⁺] / [H₂A]

10⁻²³ = (x)(x) / (0.1-x)

x = 0.020

[H⁺] = 0.020 M

pH = -log [H⁺]

pH = -log(0.020)

pH = 1.70

(2). at 10 mL NaOH addition

[H₂A]ini = 0.10 M * 20.0 mL = 2 mmol

[OH⁻] = 0.1 M * 10 mL = 1 mmol

after reaction:

[H₂A] = 1 mmol

[H⁻] = 1 mmol

pH = pKa₁ + log [HA⁻] / [[HA⁻]

pH = 2.3 + log 1mmol / 1mmol

pH = 2.3

(3). pH at the first equivalence point is,

pH = 1/2 (pKa₁ + pKa₂)

pH = 1/2(2.3 + 4.3) = 3.3

pH = 3.3

(4). pH at the second 1/2 equivalence point is

pH = pKa₂ = 4.3

pH = 4.3

(5). pH at the second equivalence point;

all H₂A is converted into A²⁻

[A²⁻] = initial moles of H₂A / total volume = (20.0 mL)(0.10 M) / (20.0 + 40.0) mL = 0.033 M

at equilibrium:

                   A²⁻ + H²O    ⇆   HA⁺ OH⁻

          0.033 - x

from the Kb₁ expression,

Kb₁ = [OH⁻][HA⁻] / [A²]

Kw/Ka₂ = x²/(0.0333 - x)

10⁻¹⁴/10⁻⁴³ = x²/(0.0333 - x)

x = 2.57 * 10⁻⁶

[OH⁻] = 2.57 * 10⁻⁶M

pH = -log Kw/[OH⁻] = 8.41

pH = 8.41

(6). pH after second equivalence point;

assuming the volume of NaOH is 40.10 mL

after second equivalence point OH⁻ in excess

[OH⁻] = 0.10 M * 0.10 mL / (20 + 40.10) mL = 1.66 * 10⁻⁴ M

pH = 0=-log Kw/[OH⁻] = 10.22

pH = 10.22

The volume of a potassium iodide solution with a concentration of 0.0380 M and containing 150 g of potassium iodide is 3.95 L, rounded to three significant digits.

To calculate the volume V of a solution, you can use the formula:

[tex]\[ V = \frac{\text{mass of solute}}{\text{molarity}} \][/tex]

Given that the molarity M is [tex]\(0.0380 \, \text{M}\)[/tex] and the mass of potassium iodide is [tex]\(150 \, \text{g}\)[/tex], substitute these values into the formula:

[tex]\[ V = \frac{150 \, \text{g}}{0.0380 \, \text{M}} \]\[ V = 3947 \, \text{mL} \][/tex]

Since the answer should have the correct number of significant digits, the volume is [tex]\(3.95 \, \text{L}\)[/tex] (rounded to three significant digits).

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The complete question is:

Calculate the volume in liters of a 0.0380M potassium iodide solution that contains 150 g of potassium iodide. Be sure your answer has the correct number of significant digits.

Answer:

the transition state will resemble the products more than the reactants

Explanation:

Since the free-radical bromination of methane involves the following reactions

Br₂ → 2 Br•  , ΔH⁰ (per mole) = +192 kJ , Ea (per mole) = + 192 kJ

CH₄ +Br•   → CH₃• + HBr  , ΔH⁰ (per mole) = +67 kJ , Ea (per mole) = + 75 kJ

CH₃• +Br₂  → CH₃Br  + Br•  , ΔH⁰ (per mole) = -101 kJ , Ea (per mole) = -4 kJ

without considering the Br dissociation (initiation reaction) the rate-determining step is the second equation ( highest Ea) .

Then since the reaction is endothermic (ΔH⁰ (per mole) = +67 kJ) , the transition state will resemble the products more than the reactants ( Hammond postulate)

The free-radical bromination of methane's transition state resembles both the products and reactants equally. This similarity arises due to the formation and subsequent decay of the high-energy transition state that exists between reactants and products. The reaction's activation energy is always positive and its exothermic nature results in a decrease in system enthalpy.

In the free-radical bromination of methane, the nature of the transition state of the rate-determining step is closest to option b. The transition state resembles both the products and reactants equally. This is because during the process, reactant molecules with enough energy collide to form a high-energy activated complex or transition state. This transition state then decays to yield stable products, maintaining similarities to both the initial reactants and final products.

As per chemical kinetics, the reaction diagram indicating this process highlights the activation energy, Eå, as the energy difference between the reactants and the transition state. The enthalpy change of the reaction, ΔH, is evaluated as the energy difference between the reactants and products. Here, the reaction is exothermic (ΔH < 0) as it results in a decrease in system enthalpy.

Understanding that the activation energy is always positive in these reactions, regardless of whether the reaction is exergonic (releases energy) or endergonic (absorbs energy), further supports the resemblance between the transition state and both reactants and products.

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Answer:

C21H30O5

Explanation:

Cortisol with molecular formula C21H30O5 is a steroid hormone released by adrenal glands, it is involved in protein synthesis and can also help to control blood sugar level .

Answer: 2.02mL

Explanation:

Density = 2.82 g/mL

Mass = 5.71 g

Volume =?

Density = Mass /volume

2.82 = 5.71 / Volume

Volume = 5.71 / 2.82

Volume = 2.02mL

Complete Question

The complete question is show on the first uploaded image

Answer:

This is shown on the second,third , fourth and fifth image

Explanation:

This is shown on the second,third , fourth and fifth image

The desired compound using a stepwise approach from dicyclopentadiene and incorporating a Diels-Alder reaction .

Step 1: Diels-Alder Reaction

In the Diels-Alder reaction, dicyclopentadiene (DCPD) will react with a suitable dienophile to form the desired cycloadduct. The dienophile you choose should have a functional group that can be further manipulated to achieve the final compound. Let's select maleic anhydride (C4H2O3) as the dienophile, which can react with DCPD to yield a cycloadduct.

The reaction would look like this:

Dicyclopentadiene + Maleic Anhydride → Cycloadduct

Step 2: Functionalization of the Cycloadduct

The cycloadduct formed in the Diels-Alder reaction will contain the necessary functional groups for further manipulation. In this case, you want to introduce additional substituents or functional groups.

For example, you can introduce an alcohol group (-OH) through a nucleophilic addition reaction. You can do this by treating the cycloadduct with a strong base (e.g., sodium hydroxide, NaOH) and water (H2O) to open the anhydride ring and form a carboxylic acid intermediate. Then, you can reduce the carboxylic acid using a reducing agent (e.g., lithium aluminum hydride, LiAlH4) to obtain the alcohol group.

The reaction sequence would look like this:

Cycloadduct + NaOH + H2O → Carboxylic Acid Intermediate

Carboxylic Acid Intermediate + LiAlH4 → Alcohol Group

Step 3: Further Functionalization

Depending on your specific requirements, you can further modify the alcohol group by various organic reactions such as esterification, acylation, or oxidation, among others, to achieve the desired final compound.

This stepwise synthesis should allow you to obtain the desired compound from dicyclopentadiene, using a Diels-Alder reaction as one of the key steps. Keep in mind that the reaction conditions and reagents may need to be optimized based on the specific compound you are targeting. Always follow safety guidelines and consult relevant chemical literature for detailed reaction conditions.

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To synthesize the given compound from dicyclopentadiene, a stepwise approach involving a Diels-Alder reaction can be followed. First, dicyclopentadiene is converted to cyclopentadiene. Then, cyclopentadiene reacts with maleic anhydride to form the Diels-Alder adduct.

A stepwise synthesis of the given compound from dicyclopentadiene using a Diels-Alder reaction as one step can be achieved as follows:

Note that the specific reagents and conditions for each step may vary depending on the desired compound and the available starting materials. Finally, the adduct can be further modified to obtain the desired compound.

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Answer:

a, b and d

Explanation:

Proper Solvent choice is very important for effective recrystallization.

Therefore, features of a good solvent are.

a. The solvent should dissolve a moderate quantity of the target substance near its boiling point but only.

b. The solvent should not react with the target substance.

d. The solvent should be easily removed from the purified product.

options c and d are not a property of good solvent.

This question is incomplete and the full question can be seen below:  

b) Calculate ΔG for ATP hydrolysis in liver at 18° C. Use the liver concentration From part a.

The equation for the ATP hydrolysis is:  

           H₂0  

ATP ------------> ADP + P₁                      ΔG = -30.5 [tex]\frac{KJ}{mol}[/tex]  

a) calculate ΔG for ATP hydrolysis to rank the following conditions from most favorable to least favorable. Assume a temperature of 37.0° C, R = 8.315 [tex]\frac{J}{mol.k}[/tex]  

muscle: [ATP]= 8.1mM; [ADP]= 0.9mM; [P₁]= 8.1mM  

liver: [ATP]= 3.4mM; [ADP]= 1.3mM; [P₁]= 4.8mM  

brain: [ATP}= 2.6mM; [ADP}= 0.7mM; [P₁]= 2.7mM  

b) Calculate ΔG for ATP hydrolysis in liver at 18° C. Use the liver concentration From part a.

Answer:  

−45.8 KJ/mol

Explanation:  

Equilibrium constant (k) is defined as a measure of the ratio of the equilibrium concentration of the products of a reaction, to the Equilibrium concentration of the reactants with each concentration raised to the power corresponding to the coefficient in the balanced equation of the reaction.

In the reaction in the question given above;  

[tex]K=\frac{[ADP][P_1]}{ATP}[/tex]

For muscle;

ADP= 0.9 × 10⁻³

P₁= 8.1 × 10⁻³

ATP= 8.1 × 10⁻³

∴ K = [tex]\frac{(0.9*10^-3)(8.1*10^-3)}{(8.1*10^-3)}[/tex]

  K = 0.9 × 10⁻³

For liver;

ADP= 1.3 × 10⁻³

P₁= 4.8 × 10⁻³

ATP= 3.4 × 10⁻³

∴ K = [tex]\frac{(1.3*10^-3)(4.8*10^-3)}{(3.4*10^-3)}[/tex]

  K = 1.8 × 10⁻³

For brain;

ADP= 0.7 × 10⁻³

P₁= 2.7 × 10⁻³

ATP= 2.6 × 10⁻³

∴ K = [tex]\frac{(0.7*10^-3)(2.7*10^-3)}{(2.6*10^-3)}[/tex]

  K = 7.3 × 10⁻⁴

b) Since we are concerned about calculating ΔG for ATP hydrolysis in liver at 18° C and we've already obtained the liver concentration from part a; we can therefore calculate ΔG as:

ΔG = ΔG° + RTInK

ΔG° = -30.5

R= 8.315 [tex]\frac{J}{mol.k}[/tex]  

T= 18° C = 18 + 273.15k = 291.15k

K= 1.8 × 10⁻³

InK = In(1.8 × 10⁻³ )

      ≅ -6.32

∴ ΔG = -30.5 +  8.315 [tex]\frac{J}{mol.k}[/tex] × 291.15k × (-6.32)

   ΔG = -30.5 +  (−15.30016542)

   ΔG = −45.80016542

   ΔG ≅ −45.8 KJ/mol

Change in free energy for ATP hydrolysis at 18 degree Celsius   is -45.8 kJ.  

The equation for the ATP hydrolysis is,    

[tex]\bold {ATP \rightarrow ADP + Pi }[/tex]                     ΔG = -30.5 kJ/mol at STP  

Change in free energy for ATP hydrolysis in liver at 18° C can be calculated by the formula,  

[tex]\bold {\Delta G = \Delta G^o + RT \times lnK}[/tex]  

Where,

ΔG° - Free energy at STP = -30.5  

R - gas constant = 8.315    

T - temperature in Kelvin = 18° C = 18 + 273.15k = 291.15k  

K - Equilibrium constant = 1.8 × 10⁻³  

In K = In(1.8 × 10⁻³ ) =      -6.32  

Put the values in the formula,

[tex]\bold {\Delta G = -30.5 + 8.315 \times 291.15 \times (-6.32)}\\\\\bold {\Delta G = -30.5 + (- 15.3)}\\\\\bold {\Delta G = - 45.8 KJ}\\\\[/tex]

Therefore, change in free energy for ATP hydrolysis at 18 degree Celsius   is -45.8 kJ.  

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