Impure crystalline (solid) substances can be purified by recrystallization from a suitable solvent. Arrange the steps of the recrystallization procedure, from start to finish.
Weigh the crystals into a tared flask.
Dissolve the crude substance in a minimum amount of hot solvent.
Filter the hot solution to remove the solid impurities (if present)
Allow the solution to cool slowly so crystals form.
Filter the crystals.
Dry the crystals

The molecule C7H10NBr has an Index of Hydrogen Deficiency (IHD) of 4, indicating 4 degrees of unsaturation. These could be a combination of double bonds, triple bonds, or rings.

The Index of Hydrogen Deficiency (IHD) or Degree of Unsaturation for a molecule can be calculated using the formula IHD = 1/2(2C + 2 + N - X - H), where C is the number of Carbons, N the number of Nitrogens, X the number of Halogens, and H the total Hydrogen count. For the molecule C7H10NBr, applying the formula would yield:

IHD = 1/2(2*7 + 2 + 1 - 1 - 10) = 4

This means there are 4 degrees of unsaturation present in the molecule. The types of unsaturation could be double bonds, triple bonds, or rings. In this case, with a value of 4, it could be 4 double bonds, or 2 double bonds and 2 rings, or 2 rings and 1 triple bond, or 1 ring and 3 double bonds, or 2 triple bonds, and many more combinations.

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The Index of Hydrogen Deficiency for the compound C₇H₁₀N Br is 4, which indicates that the molecule can have a combination of 4 double bonds or 4 rings or a mix of both.

The Index of Hydrogen Deficiency (IHD) is a count of how many molecules of H2 are missing from an alkane having the same number of carbon atoms. It's used to identify the number of rings and/or double bonds in organic compounds. For the compoundC₇H₁₀N Br, consider that Br adds 1 to the hydrogen count, and N subtracts 1. Thus, the IHD calculation would be: IHD= (2*C + 2 + N - X - H)/2 = [2*7 + 2 +1 -0 -10]/2 = 4

Since it has an IHD of 4, it indicates that the molecule can have a combination of 4 double bonds or 4 rings or a mix of both.

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Answer: The pH of the solution is 4.57

Explanation:

To calculate the number of moles for given molarity, we use the equation:

[tex]\text{Molarity of the solution}=\frac{\text{Moles of solute}\times 1000}{\text{Volume of solution (in mL)}}[/tex]

Molarity of benzoic acid = 50 mM = 0.05 M      (Conversion factor:  1 M = 1000 mM)

Volume of solution = 100 mL

Putting values in above equation, we get:

[tex]0.05M=\frac{\text{Moles of benzoic acid}\times 1000}{100mL}\\\\\text{Moles of benozic acid}=\frac{(0.05\times 100)}{1000}=0.005mol[/tex]

Molarity of sodium hydroxide = 50 mM  = 0.05 M

Volume of solution = 70 mL

Putting values in above equation, we get:

[tex]0.05M=\frac{\text{Moles of sodium hydroxide}\times 1000}{50mL}\\\\\text{Moles of sodium hydroxide}=\frac{0.05\times 70}{1000}=0.0035mol[/tex]

The chemical reaction for sodium hydroxide and benzoic acid follows the equation:

                [tex]C_6H_5COOH+NaOH\rightarrow C_6H_5COONa+H_2O[/tex]  

Initial:        0.005            0.0035    

Final:         0.0015        -                   0.0035

Volume of solution = 100 + 70 = 170 mL = 0.170 L    (Conversion factor:  1 L = 1000 mL)

To calculate the pH of acidic buffer, we use the equation given by Henderson Hasselbalch:

[tex]pH=pK_a+\log(\frac{[salt]}{[acid]})[/tex]

[tex]pH=pK_a+\log(\frac{[C_6H_5COONa]}{[C_6H_5COOH]})[/tex]

We are given:

[tex]pK_a[/tex] = negative logarithm of acid dissociation constant of benzoic acid = 4.2

[tex][C_6H_5COONa]=\frac{0.0035}{0.170}[/tex]

[tex][C_6H_5COOH]=\frac{0.0015}{0.170}[/tex]

pH = ?

Putting values in above equation, we get:

[tex]pH=4.2+\log(\frac{0.0035/0.170}{0.0015/0.170})\\\\pH=4.57[/tex]

Hence, the pH of the solution is 4.57

Answer:

I and II.

Explanation:

The interactions between the molecules in a substance are associated with the type of the substance. For ionic compounds, the atoms are joined together by the ion-ion interactions, for metallic compounds, by metallic interactions, and, for molecular compounds, they can be attracted by London forces, dipole-dipole forces or by hydrogen bonds.

The London forces exist in nonpolar molecules, which are the ones formed by nonpolar bonds (elements with the same electronegativity), or form bonds with the same polarity that is opposite, and so are canceled. The dipole-dipole force exists in polar molecules, and so, the atoms have partial charges, and the interactions are stronger than in London forces. If the dipole-dipole exists with hydrogen and a high electronegative element (N, O, or F), the bond is even strong and is called a hydrogen bond.

So, let's analyze the statements:

I. As said above, dipole-dipole occurs in polar molecules, so they may have polar bonds, and the statement is correct;

II. Because the polarity of the bonds is a vector when the shape is symmetric, is more likely to the polarities be canceled, so it's usual to the polar molecules be asymmetric, and the statement is correct;

III. Because the dipole-dipole is a strong force, it's difficult to break it and the substance needs more energy to change phase, so they have a high boiling point, and the statement is incorrect;

IV. Because the shape of the molecule is asymmetric, the surface area intends to be small, the atoms are distributed in a small place. If the molecule is linear, for example, then the atoms are distributed in large spaces, so the statement is incorrect.

Answer:

a. Protons

b. Electrons

Explanation:

The atomic number of an element is also the proton number. Since the atom is neutral, the proton number and the electron number are equal. But this is different for an ion in that the element has either gain or loss an electron and so, the electron number differ from the proton number.

Answer:

The temperature of water will be 60 degree Celsius.

Explanation:

For every chemical there present a particular solubility temperature. The phenomenon occurs as the positive charge is attracted towards the negative and hence gives rise to a cohesive structure. When the polar compounds as well as the ions gets added to the water, they start getting into smaller parts and hence dissolve and becomes solution. The partial charge of water starts attracting different parts of the compound and makes them soluble in water.

Final answer:

To dissolve 91.6 g of KCl in 200 g of water, a temperature higher than 25°C will likely be necessary, but an exact temperature cannot be provided without specific solubility data for KCl.

Explanation:

To determine the temperature at which 91.6 g of KCl (potassium chloride) will dissolve in 200 g of water, we need to refer to the solubility data for KCl. Solubility tables or graphs provide this information and show the amount of a solute that can dissolve in a solvent at various temperatures. Generally, the solubility of ionic compounds like KCl increases with temperature. Without specific solubility data for KCl at different temperatures, an exact answer cannot be provided. However, it is known that at 25°C (77°F), approximately 34 g of KCl will dissolve in 100 mL of water. Therefore, for 91.6 g of KCl, a temperature higher than 25°C will likely be necessary to dissolve the entire amount in 200 g of water. More accurate determination requires a solubility chart or experimental data for KCl.

Answer:

Here is a similar question ; Identify the numbers from the statements above as exact or measured Sort these numbers into the proper categories. 93 g of silver, One minute, 2.20 lb ,One kilogram ,27 miles ,One yard, 100 g of sterling silver ,60 s, 1gal ,3 ft Exact Measured

Explanation:

Here is a similar question ; Identify the numbers from the statements above as exact or measured Sort these numbers into the proper categories. 93 g of silver, One minute, 2.20 lb ,One kilogram ,27 miles ,One yard, 100 g of sterling silver ,60 s, 1gal ,3 ft Exact Measured

Exact number are known from scientific research that has been proven to be true and widely acceptable or are known from true definition. for example 0 degree celsius = 273k, 1atm = 101.325 kPa

while measured numbers are not widely acceptable, they are as a result of individual measurement and discretion. A may measure the mass of a stone to be 3.5g, another person B may measure the same stone and arrive at 3.4g.

The enthalpy change for the sublimation of iodine, represented by I₂(s) → I₂(g), is 31 kJ/mol.

The enthalpy change for the sublimation of iodine, represented by I₂(s) → I₂(g), can be determined by comparing the enthalpy changes of the two given reactions. Since the equation ½ H₂(g) + ½ I₂(g) → HI(g) has a lower enthalpy change (-5.0 kJ/mol) compared to the equation ½ H₂(g) + ½ I₂(s) → HI(g) (26 kJ/mol), it means that the phase change from solid iodine to gaseous iodine requires an additional amount of energy. Thus, the enthalpy change for the sublimation of iodine is the difference between the two reaction enthalpies, which is 31 kJ/mol.

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The percent by mass of acetyl bromide in the given solution can be calculated by dividing its mass by the total mass of the solution and then multiplying by 100%, leading to a result of 43.5%.

To calculate the percent by mass of each component in the solution, we first need to find the total mass of the solution. The total mass is the sum of the mass of chloroform, acetone, and acetyl bromide, which is 96.2 g + 31.2 g + 98.1 g = 225.5 g.

Next, we calculate the percent by mass for each substance by dividing the mass of each substance by the total mass and then multiplying it by 100%.

For acetyl bromide, it's (98.1 g / 225.5 g) * 100% = 43.5%.

Please note that the number of significant digits in the answer should match the lowest number of significant digits in the input, that's why it's rounded off to three significant digits.

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Answer:

Polar/Hydrophilic

Explanation:

Fluorine, Nitrogen and Oxygen are strong electronegative atoms and by definition, Electronegativity is the amount of pull or the high affinity of an atom to electrons.

Polar bond occurs when there is a high difference between the electronegativity value of both atoms that take part in the bond.

A polar molecule has a net dipole from the distribution of its positive and negayive charges. Hydrophobic and Hydrophilic (in chemistry, Polar) are terms dependent on the overall distribution of charge in its molecule.

Therefore, bonds between C-N, C-O and C-Cl are polar covalent bonds a d this is because of the jigh electronegativity possessed by Nitrogen, Oxygen and Chlorine.

Answer:

19.91 J/K

Explanation:

The entropy is a measure of the randomness of the system, and it intends to increase in nature, thus for a spontaneous reaction ΔS > 0.

The entropy variation can be found by:

ΔS = ∑n*S° products - ∑n*S° reactants

Where n is the coefficient of the substance. The value of S° (standard molar entropy) can be found at a thermodynamic table.

S°, Cl(g) = 165.20 J/mol.K

S°, O3(g) = 238.93 J/mol.K

S°, O2(g) = 205.138 J/mol.K

So:

ΔS = (1*205.138 + 1*218.9) - (1*165.20 + 1*238.93)

ΔS = 19.91 J/K

Answer:

The work done is 123.5 J

Explanation:

Given that:-

Temperature = 260 K

The expression for the work done is:

[tex]W=-nRT \ln \left( \dfrac{V_2}{V_1} \right)[/tex]

Where,  

n is the number of moles = 52.0 mmol = [tex]52.0\times 10^{-3}\ moles[/tex]

W is the amount of work done by the gas

R is Gas constant having value = 8.314 J / K mol  

T is the temperature  

V₁ is the initial volume  = 300 cm³

V₂ is the final volume  = 100 cm³

Applying in the equation as:

[tex]W=-52.0\times 10^{-3}\ moles\times 8.314\ J/Kmol\times 260\ K \ln \left( \dfrac{100\ cm^3}{300\ cm^3} \right)[/tex]

[tex]W=-52.0\times 10^{-3}\times 8.314\times 260 \ln \left( \dfrac{100}{300} \right)\ J=123.5\ J[/tex]

The work done is 123.5 J

The work done in the isothermal reversible compression of an ideal gas, where the volume of gas was reduced from 300 cm3 to 100 cm3 is about 284 J.

In the given problem, we're dealing with isothermal reversible compression of an ideal gas. Under these conditions, the work done (w) can be calculated using the formula: w = -nRT ln(Vf/Vi), where n is the number of moles, R is the gas constant, T is the temperature, Vi is the initial volume, and Vf is the final volume.

Here, n = 52.0 mmol = 0.052 mol, R = 8.314 J / (mol·K) (in these units to match those of the problem), T = 260 K, Vi = 300 cm3 = 0.3 L, and Vf = 100 cm3 = 0.1 L.

Plugging these values into the formula, we get:

w = -0.052 mol * 8.314 J / (mol·K) * 260 K * ln(0.1/0.3)

Therefore, the work done in this isothermal reversible compression where the volume of gas was reduced from 300 cm3 to 100 cm3 is approximately 284 J.

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Answer:

1.05 × 10²⁵ atoms H₂

General Formulas and Concepts:

Atomic Structure

Stoichiometry

Explanation:

Step 1: Define

[Given] 35.0 g H₂

[Solve] atoms H₂

Step 2: Identify Conversions

Avogadro's Number - 6.022 × 10²³ atoms, molecules, formula units, etc.

[PT] Molar Mass of H - 1.01 g/mol

Molar Mass of H₂: 2(1.01) = 2.02 g/mol

Step 3: Convert

Step 4: Check

Follow sig fig rules and round. We are given 3 sig figs.

1.04342 × 10²⁵ atoms H₂ ≈ 1.04 × 10²⁵ atoms H₂

Topic: AP Chemistry

Unit: Atomic Structure

Final answer:

To find the number of hydrogen atoms in 35.0 grams of hydrogen gas, we calculated the moles of hydrogen gas and used Avogadro's number. We determined that there are 2.09 × 1025 hydrogen atoms in 35.0 grams of hydrogen gas.

Explanation:

To determine how many hydrogen atoms are in 35.0 grams of hydrogen gas, we use Avogadro's number and the concept of molar mass. The molar mass of hydrogen gas (H2) is 2.0158 grams per mole. First, we calculate the number of moles in 35.0 grams of hydrogen gas:
(35.0 grams H2) ÷ (2.0158 grams/mol) = 17.37 moles H2

Since each mole of hydrogen gas contains two hydrogen atoms, we need to multiply the number of moles by Avogadro's number, which is 6.022 × 1023 atoms/mol:
(17.37 moles H2) × (6.022 × 1023 molecules H2/mol) × (2 atoms H/molecule H2)

After calculating, we find that there are 2.09 × 1025 hydrogen atoms in 35.0 grams of hydrogen gas. Therefore, the correct answer is 2.09 × 1025.

Answer:

1.472 x 10^-17 J energy of the photon is released (exothermic).

Explanation:

E = –kZ^2 / n^2

k = 2.18 x10^-18 J

Z = atomic number

Li2+ atomic number = 3

Ei = –(2.18 x 10^-18 J) x 3^2 / 2^2 = –4.905 x 10^-18 J

Ef = –(2.18 x 10^-18 J) x 3^2 / 1^2 = –1.962 x 10^-17 J

ΔE = |Ef – Ei| = |(–1.962 x 10^-17 J) – (–4.905 x 10^-18 J)| = 1.472x10^-17  

1.472 x 10-17 J energy of the photon is released (exothermic).

Using the Bohr model, [tex]\rm 1.472\times 10^{-17 } J[/tex] is the energy in joules of the photon produced when an electron in a Li2+ ion moves from the orbit with n = 2 to the orbit with n = 1.

A photon is the basic unit of light and electromagnetic radiation. It is a quantum of energy and a carrier of electromagnetic force. Light was assumed to be a continuous wave in classical physics, but scientists found in the early twentieth century that light and other kinds of electromagnetic radiation behave as both waves and particles. This dual nature is a key concept in quantum mechanics.

E =[tex]\rm -kZ^2 / n^2[/tex]

k =[tex]\rm 2.18 \times10^{-18 }[/tex]J

Z = atomic number

[tex]\rm Li^{2+}[/tex] atomic number = 3

Ei =[tex]\rm -(2.18 \times 10^{-18} J) \times 3^2 / 2^2[/tex]

  = [tex]\rm -4.905 \times 10^{-18}[/tex] J

Ef = [tex]\rm -(2.18 \times 10^{-18} J) \times 3^2 / 1^2[/tex]

   =[tex]\rm -1.962 \times 10^{-17[/tex] J

ΔE = |Ef – Ei|

    = |([tex]\rm -1.962 \times 10^{-17}[/tex] J) – ([tex]\rm -4.905 \times 10^{-18 } J[/tex])|

    = [tex]\rm 1.472\times 10^{-17 }[/tex] J

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The periodic properties of elements are chiefly due to the numbers of electrons in each element's atoms and the distribution of these electrons. These properties don't significantly relate to the neutrons in the atoms. Thus, the correct answer is h) I and II only.

The periodicity of the properties of elements is chiefly due to the numbers of electrons in the atoms of the elements and the distribution of electrons in the atoms of the elements. In the periodic table, these properties are mainly arranged by the atomic number of the elements, which determines the number of electrons, and their electronic configuration. More specifically, the arrangement is based on the number of protons (nuclear charge) and how the electrons are distributed in energy levels/shells around the nucleus.

So, the answer is h) I and II only.

The number of neutrons does not significantly affect periodic properties as the nature of an atom is primarily defined by its number of protons and electrons.

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Final answer:

The answer is (h) I and II only, where periodicity is primarily due to the number and distribution of electrons, which are related to the atomic number. The number of neutrons does not impact periodicity.

Explanation:

The periodicity of the properties of elements is chiefly due to the distribution of electrons in the atoms of the elements and the overall number of electrons, which correspond to their atomic number. The correct answer to the question is hence, option (h) I and II only. The number of protons in an atom (which is equivalent to the number of electrons in a neutral atom) and how these electrons are distributed among the different energy levels, or shells, determine the chemical and physical properties of an element.

Various properties such as atomic radius, ionization energy, and electron affinity can be understood through the element's position on the periodic table. Neutrons do not affect the chemical properties of elements and are thus irrelevant to periodicity. The specific arrangement of electrons, particularly in the outermost shells or valence shells, influences how an element reacts chemically and thus its placement in the periodic table.

Answer: AIBN is good radical initiator

Explanation:

A radical in organic chemistry refers to any specie having a single unshared electron. An initiator is any specie capable of producing radicals thus starting up a chain reaction. Azobisisobutyronitrile is a good initiator basically owing to its structure. It forms radicals by breaking up to release nitrogen gas as shown in the image attached. The two radicals formed both contain the -CN group which stabilizes the radical.

Answer:

Explanation:

1 - Octyne is a member of the Alkyne family. Alkyne has a general formular is CnH2n-2 which is a homologous series with a least one carbon - carbon triple bond.

Where n is the number of carbon atoms needed in the structure.

The "1" prefix is an indicator for where the triple bond of the compound will be found which is at carbon -1. Because of its triple bond it is termed an unsaturated hydrocarbon. Their properties include:

1. Highly reactivity

2. Hydrophobic compound.

Octa- decribes the number of carbon atoms present in the compound which is 8. Also note that, it is not a cycloalkyne but a straight chain compound with a triple bond at one end.

So using the general formular, CnH2n-2

For n = 8,

1-Octyne is C8H14

Below in the attac is the structural formula of 1 - Octyne

Answer:

Explanation:

1-Octyne is a member of the Alkyne series, Alkyne is a homologous series with the general molecular formula;

CnH2n-2

n is a positive whole number which equal to or greater than 2.

i.e n= number of carbon given  

In the case of the compound given (1-Octyne), the prefix “Oct” means 8, which is the number of  

carbon atoms present in the compond while the suffix “yne” is

coined from the family name “Alkyne”

Therefore, C8H(2x 8) – 2 = C8H16-2 = C8H14

It is an unsaturated hydrocarbon, each alkyne molecule contains four hydrogen atoms

corresponding to alkane and two hydrogen atom less than the corresponding alkene. This is because

each alkyne molecule contains a carbon-carbon triple bond, where two  carbon atoms are  bonded  

to each other by the sharing of three pairs of electrons.

The Alkyne family show a higher degree of unsaturation than the alkenes, hence they are chemically  

reactive than alkane and alkene .

Answer: option E. None because in all the reactions O2 is in excess

Explanation:

In the above reactions, only Reaction A and Reaction B have oxygen (O2) as the limiting reactant. For the rest of the reactions, the other compound will be used up before all the O2 is spent.

In every reaction, you have 5.73 moles of O2 and 1.70 moles of the other reactants. You can determine which of the two reactants are limiting (which will be consumed first) by comparing the number of moles of that reactant divided by its stoichiometric coefficient (the numbers in front of the chemicals in the reaction) to the same calculation for the other reactant.

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Final answer:

A pound-mole of a substance contains approximately 6.022 × 10²³ × 453.592 atoms, as Avogadro's number is used in the definition of a mole, which is based on grams, and needs to be converted using the pound-to-gram conversion factor.

Explanation:

The concept of mole is fundamental in chemistry for measuring quantities of substances. When referring to a pound-mole is equivalent to the number of atoms in a substance weighing one pound. However, Avogadro's number, which is 6.022 × 10²³, is defined based on the international standard mole unit, which is mass in grams. To convert between the pound-mole and the gram-mole, one must use the conversion factor between pounds and grams. Since 1 pound is equal to approximately 453.592 grams, a pound-mole will contain Avogadro's number multiplied by the ratio between the pound and grams (453.592). Therefore, there are approximately 6.022 × 10²³ × 453.592 atoms in a pound-mole of a substance.

Final answer:

The sulfur-containing product of the oxidation of two 2-methyl-1-propanethiol molecules is a disulfide, where the sulfur atoms form a bond between the two original thiol molecules, resulting in a dimer.

Explanation:

The student has asked for the sulfur-containing product of the oxidation reaction between two 2-methyl-1-propanethiol molecules. In the presence of mild oxidizing agents such as oxygen, sulfur atoms can be oxidized from the sulfhydryl (S-H) groups present in 2-methyl-1-propanethiol. When two such sulfur atoms, each with an unpaired electron, come into contact, they can form a sulfur-sulfur bond (disulfide bond), resulting in the dimerization of the two thiol molecules.

The chemical formula of 2-methyl-1-propanethiol is CH3CH2CH(SH)C(CH3)2. Upon oxidation, the sulfur atoms from two of these molecules will bond together to form a disulfide. The resulting molecule will be CH3CH2CH(S-S)CH2C(CH3)2.

This question does not contain the structures of the molecules. The structures in Daylight SMILES format are:

I. C1=CC=CC=C1C(=O)C

II. C1=CC=CC=C1CC=O

III. C1=CC(C)=CC=C1C=O

IV. C1=CC=CC=C1CCC

V. C1=CC=CC=C1C(C)C

The structures are also attached

Answer:

The structure of compound IV is consistent with the information obtained analysis

Proposed structures for the ions with m/z values of 120, 105,77 and 43 are (also attached):

C1=CC=CC=C1C(=[OH0+])C |^1:7|

C1C([CH0+]=O)=CC=CC=1

C1[CH0+]=CC=CC=1

C(#[OH0+])C

respectively

Explanation:

The IR peak at 1687 cm⁻¹ is indicative of an α unsaturated carbonyl carbon. While the 1H NMR singlet is of the methyl group next to carbonyl and the multiplet near 7.1 ppm is a characteristic peak of benzene. This data shows points towards structure I.

Mass spectrum peak at 120 m/z is of molecular ion peak. In the case of carbonyl-containing molecule, this peak is observable. The signal at 105 shows the loss of a methyl group next to the carbonyl. m/z value of 77 is the characteristic cationic peak of benzene, while the peak at 43 infers the formation of acylium ion (RCO+) due to α-cleavage. All this data agrees with the structure of acetophenone (Structure 1)

Answer:

2 8 5

2 8

2 8, 6

2 8 2

2 8 3

2 7

Explanation:

This is your question :

Write the electron arrangement for each of the following atoms:(Example sodium is 2,8,1)

a. phosphorus

b. neon

c. sulfur

d. magnesium

e. aluminum

f. fluorine

Phosphorus is an element in period 3. phosphorus has an atomic number of  15 and it contains 15 electrons. The electron arrangement can be computed as follow ;

2 8 5

This means is has 2 electrons at the first energy level, 8 electron at the second energy level and 5 electron at the third energy level. Base on it configuration phosphorus requires 3 electrons to fulfill it octet rule .

Neon is grouped as a noble gas or inert gas because of it less reactive nature . The atomic number of Neon is 10. The electron arrangement can be computed as follows;

2 8

This means it has 2 electrons on the first energy level and 8 electrons on the second energy level. Neon is in period 2 on the periodic table. Neon electron configuration shows it has fulfilled it octet rule, this is why it rarely goes into a reaction with other elements.

Sulfur

Sulfur has an atomic number of 16 . The electronic configuration is as follows ;

2 8, 6

Sulfur is in period 3 base on it configuration . it possess three energy level with 2 electrons at the first energy level , 8 electrons at the second energy level and 6 at the third energy level.  

Magnesium

Magnesium has an atomic number of 12 . Magnesium is in period 3 in the periodic table and are also known as the alkali earth metals. The electronic configuration is as follows ;

2 8 2

The three energy level has 2 electrons, 8 electrons and finally 2 electrons. Magnesium is reactive.

Aluminium

Aluminium has an atomic number of 13 . Aluminium is in period 3 . The electronic configuration is as follows;

2 8 3

The three energy level has 2 electrons, 8 electrons and 3 electrons each .

Florine

Florine has an atomic number of 9. Florine is in period 2 on the periodic table. The electronic configurations is as follows ;

2 7

The two energy level has 2 electrons and 6 electrons respectively .

Answer:

Lowering the temperature typically reduces the significance of the decrease in entropy. That makes the Gibbs Free energy of the reaction more negative. As a result, the reaction becomes more favorable overall.  

Explanation:

In an addition reaction there's a decrease in the number of particles. Consider the hydrogenation of ethene as an example.

[tex]\rm H_2C\text{=}CH_2\; (g) + H_2\; (g) \stackrel{\text{Ni}^\ast}{\to} H_3C\text{-}CH_3\; (g)[/tex].

When [tex]\rm H_2[/tex] is added to [tex]\rm H_2C\text{=}CH_2[/tex] (ethene) under heat and with the presence of a catalyst, [tex]\rm H_3C\text{-}CH3[/tex] (ethane) would be produced.

Note that on the left-hand side of the equation, there are two gaseous molecules. However, on the right-hand side there's only one gaseous molecule. That's a significant decrease in entropy. In other words, [tex]\Delta S < 0[/tex].

The equation for the change in Gibbs Free Energy for a particular reaction is:

[tex]\Delta G = \Delta H + (\underbrace{- T \, \Delta S}_{\text{entropy}\atop \text{term}})[/tex].

For a particular reaction, the more negative [tex]\Delta G[/tex] is, the more spontaneous ("favorable") the reaction would be.

Since typically [tex]\Delta S < 0[/tex] for addition reactions, the "entropy term" of it would be positive. That's not very helpful if the reaction needs to be favorable.

[tex]T[/tex] (absolute temperature) is always nonnegative. However, lowering the temperature could help bring the value of

The number of alpha particles emitted from a 1.0ng sample of Polonium-209 in 1.0s depends on the specifics of the isotope's half-life and the initial amount of the isotope. The emissions of these alpha particles are part of the isotope's decay process.

The half-life concept is key in understanding the decay of radioactive isotopes like polonium-209. Given that polonium-209 is an alpha emitter, it releases alpha particles during its decay process. These alpha particles are essentially helium-4 nuclei (two protons and two neutrons).

Understanding the number of alpha particles emitted in a certain time requires the understanding of the isotope's half-life and atomic decay. However, without knowing the specific numbers of atoms present in the given 1.0 ng sample of 209Po, it is difficult to provide a precise number of alpha particles emitted in 1.0 s. You would need to make use of the radioactive decay formula N = N0e^-λt where N0 represents the initial amount of the isotope, N represents the remaining amount of the isotope, λ represents the decay constant, and t represents time.

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Answer:

a.Carbon :proton number =6

             electron number =6

b. Fluorine: proton number=9

                  electron number =9

c. tin:  proton number=50

           electron number =50

d. nickel:proton number=28

           electron number =28

Explanation:

In the neutral state of an atom the number of protons is always equal to the number of electrons and that whats makes it electrically neutral as the positive charges of protons balances the negative charges of electrons.

The correct answer is "0.000035 M/s". The rate of disappearance of NO₂ in the reaction is 0.000035 M/s, found by dividing the change in concentration of NO₂ (-0.00350 M) by the time period (100 s).

The rate of disappearance of NO₂ in the reaction 2NO₂
ightarrow 2NO₂ + NO₂ can be calculated using the change in concentration of NO₂ over time. We are given that the concentration of NO₂ drops from 0.0100 to 0.00650 M in 100 seconds.

First, we calculate the change in concentration (also known as the concentration difference):
Concentration difference = Final concentration - Initial concentration = 0.00650 M - 0.0100 M = -0.00350 M (a negative sign indicates disappearance).

To find the rate of disappearance, we divide the concentration difference by the time period:
Rate of disappearance = Concentration difference / Time period
Rate of disappearance = -0.00350 M / 100 s = -0.000035 M/s

Since we are interested in the rate of disappearance, we take the absolute value of the result, making it 0.000035 M/s.

The rate of disappearance of NO2 for the given period is [tex]\( 3.5 \times 10^{-5} \) M/s.[/tex]

To find the rate of disappearance of NO2, we need to calculate the change in concentration of NO2 over time. The reaction given is:

[tex]\[ 2NO_2 \rightarrow 2NO + O_2 \][/tex]

From the problem, we have the initial concentration of NO2 as 0.0100 M and the final concentration as 0.00650 M. The time taken for this change in concentration is 100 seconds.

The rate of disappearance of NO2 can be calculated using the formula:

[tex]\[ \text{Rate} = -\frac{\Delta [NO_2]}{\Delta t} \][/tex]

where [tex]\( \Delta [NO_2] \)[/tex] is the change in concentration of NO2, and [tex]\( \Delta t \)[/tex] is the change in time.

The change in concentration of NO2 is:

[tex]\[ \Delta [NO_2] = [NO_2]_{\text{initial}} - [NO_2]_{\text{final}} \][/tex]

[tex]\[ \Delta [NO_2] = 0.0100 \, \text{M} - 0.00650 \, \text{M} = 0.00350 \, \text{M} \][/tex]

The change in time is:

[tex]\[ \Delta t = t_{\text{final}} - t_{\text{initial}} \][/tex]

[tex]\[ \Delta t = 100 \, \text{s} - 0 \, \text{s} = 100 \, \text{s} \][/tex]

Now, we can calculate the rate:

[tex]\[ \text{Rate} = -\frac{0.00350 \, \text{M}}{100 \, \text{s}} \][/tex]

[tex]\[ \text{Rate} = -3.5 \times 10^{-5} \, \text{M/s} \][/tex]

The negative sign indicates that the concentration of NO2 is decreasing over time. However, the rate of disappearance is typically reported as a positive value, so we take the absolute value:

[tex]\[ \text{Rate} = 3.5 \times 10^{-5} \, \text{M/s} \][/tex]

Answer: option A. Lab.

Explanation:

Answer:

206 mL

Explanation:

In the annexed picture you can see your same question, just in another format.

First we calculate the total moles of CuF₂ that are required in the working solution:

1967 μM ⇒ 1967 / 10⁶ = 1.967 x10⁻³M

1.967 x10⁻³M * 0.275 L = 5.409x10⁻⁴ mol

Now we divide those moles by the concentration of the stock solution, to calculate the volume:

5.409x10⁻⁴ mol ⇒ 5.409x10⁻⁴ * 1000 = 0.5409 mmol

0.5409 mmol ÷ (2.63 mmol/L) = 0.206 L

0.206 L ⇒ 0.206 * 1000 = 206 mL

Answer:1. False, 2. False, 3. True, 4. True

Explanation:

1. FALSE. Pure crystalline solids have sharp melting points, while the present of impurity lower the melting point. This is True, but the exception is a eutectic mixture. A properly mixed eutectic mixture have a sharp melting point also.

2. FALSE, although the present of impurity lower the melting point of a organic compound, but if a molecular size impurity with a significantly higher melting point and is present in large quantities the melting point of the organic compound will not be lowered.

3. TRUE, two compounds with same melting point, when mixed together will not change the melting point of either compounds. But if A has higher MP than B and the addition of A to B creates mixture with same melting point as pure A then the statement will be false.

4. TRUE, a properly mixed eutectic mixture will have a sharp melting point.

Melting point and freezing point of a pure compound will be identical because the liquid form of a pure substance when cooled it forms a stable arrangement that supercooling is prevented, freezing occur at the same temperature at which the pure solid phase melts

Answer: 8.2 grams of potassium dichromate the chemist has been added to the flask.

Explanation:

Molarity is defined as the number of moles of solute dissolved per liter of the solution.

To calculate the number of moles for given molarity, we use the equation:

[tex]\text{Molarity of the solution}=\frac{\text{Moles of solute}}{\text{Volume of solution (in L)}}[/tex]     .....(1)

Molarity of potassium dichromate solution = 0.11 M

Volume of solution = 0.25 L

Putting values in equation 1, we get:

[tex]0.11M=\frac{\text{Moles of potassium dichromate}}{0.25L}\\\\{\text{Moles of potassium dichromate}}={0.11mol/L\times 0.25L}=0.028mol[/tex]

Mass of potassium dichromate =[tex]moles\times {\text {molar mass}}=0.028mol\times 294g/mol=8.2g[/tex]

Thus mass in grams of potassium dichromate the chemist has added to the flask is 8.2.

To calculate the mass of potassium dichromate, use the equation moles = volume (L) × concentration (mol/L), and then multiply the moles by the molar mass of K₂Cr2O7. Round your answer to the correct number of significant digits.

To calculate the mass of potassium dichromate that the chemist added to the flask, we can use the equation:

moles = volume (L) × concentration (mol/L)

First, convert the volume to liters by dividing it by 1000. Then, use the equation with the given volume and concentration to find the moles of potassium dichromate. Finally, multiply the moles by the molar mass of potassium dichromate to get the mass in grams. The molar mass of K₂Cr2O7 is 294.185 g/mol.

mass = moles × molar mass

Make sure to round your answer to the correct number of significant digits based on the given information.

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Answer:

This is as a result of their property type

ΔG is extensive and E is Intensive. The explanation is as given below

Explanation:

Basically both ΔG and the cell potential or the electromotive force (E.M.F) has some disparity especially in their spontaneity, for spontaneous reaction ΔG = -ve while E = +ve and vice versa. But the most important disparity is their state function i.e one is intensive and the other is extensive property.

ΔG is an example of an extensive property, they are properties whose value is dependent on the volume or the size of the system. other examples are mass, volume etc.

E on the other hand is an intensive property, they are properties whose value is not dependent on the size of the system. As such, this differences explains why ΔG for a reaction scale with a reaction quantity and E does not.

The change in Gibbs free energy (ΔG) for a reaction scales with reaction quantity because it is an extensive property, while the standard electrode potential (E) does not scale with reaction quantity because it is an intensive property.

The change in Gibbs free energy (ΔG) for a reaction scales with reaction quantity because it is an extensive property, meaning it depends on the amount of substance involved in the reaction. On the other hand, the standard electrode potential (E) does not scale with reaction quantity because it is an intensive property, meaning it is independent of the amount of substance.

For example, in the combustion of hydrogen, the standard Gibbs free energy change (ΔG°) for the reaction is -237 kJ/mol for 1 mole of hydrogen and -474 kJ/mol for 2 moles of hydrogen. Since ΔG is an extensive property, it doubles when the quantity of hydrogen doubles. However, the standard electrode potential (E°) remains the same regardless of the quantity of hydrogen.

In summary, the scaling of ΔG with reaction quantity and the independence of E from reaction quantity are due to the differences in their nature as extensive and intensive properties, respectively.

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