Uranium hexaflouride (UF6) has a triple point at (T, p) = (337 K,152 kPa). Suppose you have a (gaseous) sample of UF6 at atmospheric pressure and room temperature. If you keep cooling your sample, will it undergo a phase transition from gas → liquid or from gas → solid?

Answer: 6kJ/mol

Explanation:

The full eclipse stereoisomer of ethanol has two H-H overlapped and one H-OH overlapped. The energy cost for H-H eclipse is 4kJ/mol or 1kcal/mol.

Total energy= 14kJ/mol

2(H-H eclipsed) = 2(4kJ/mol) = 8kJ/mol

Total Energy = H-H eclipsed + H-OH eclipsed

14kJ/mol = 8kJ/mol + H-OH eclipsed

14kJ/mol - 8kJ/mol = H-OH eclipsed

Therefore H-OH eclipsed = 6kJ/mol

Answer:

Explanation:

The solution has been attached

Answer:

a) P = 9.58 psi for  h=7.2 m

b) P=4.7 psi for h=5.94 m

Explanation:

Since the pressure Pon a static liquid level h is

P= p₀ + ρ*g*h

where p₀= initial pressure , ρ=density , g = gravity

then he variation of the liquid level Δh will produce a variation of pressure of

ΔP= ρ*g*Δh → ΔP/Δh =  ρ*g = ( 15 psi - 3 psi) /( 8.6 m - 5.5 m)  = 12/3.1 psi/m

if the liquid level is converted linearly

P = P₁ + ΔP/Δh*(h -h₁)

therefore choosing  P₁ = 3 psi and h₁= 5.5 m , for h=7.2 m

P = 3 psi  + 12/3.1 psi/m *(7.2 m -5.5 m) = 9.58 psi

then P = 9.58 psi for  h=7.2 m

for P=4.7 psi

4.7 psi = 3 psi  + 12/3.1 psi/m *(h -5.5 m)

h = (4.7 psi - 3 psi)/ (12/3.1 psi/m) + 5.5 m = 5.94 m

then P=4.7 psi for h=5.94 m

I hope this will help ;)

The structures for all constitutional isomers with the following molecular formulas a) CH₃-CH₃-CH₃-CH₃-C₂H₂ b) CH₃-CH₂-Cl  c) CH₃-CH₂-Cl d) CH₂-CH-Cl .

The isomers are those compounds which contain same molecular formula but different molecular structures but the physical and chemical properties will be same like melting and boiling points.

The isomers for the formula will be,

(a) C6H14 = CH₃-CH₃-CH₃-CH₃-C₂H₂

(b) C2H5Cl = CH₃-CH₂-Cl

(c) C2H4Cl2 = CH₃-CH₂-Cl

(d ) C2H3Cl3 = CH₂-CH-Cl .

The compounds are same in molecular formula but the representation of structure is different.

Therefore,  a) CH₃-CH₃-CH₃-CH₃-C₂H₂ b) CH₃-CH₂-Cl  c) CH₃-CH₂-Cl d) CH₂-CH-Cl . structures for all constitutional isomers with the following molecular formulas.

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Answer:

(a)

432 nm

(b)

287 nm

(c)

451 nm

Explanation:

[tex]Energy=\frac {h\times c}{\lambda}[/tex]

Where,  

h is Plank's constant having value [tex]6.626\times 10^{-34}\ Js[/tex]

c is the speed of light having value [tex]3\times 10^8\ m/s[/tex]

[tex]\lambda[/tex] is the wavelength of the light

(a)

Given that:- Energy = [tex]4.60\times 10^{-19}\ J[/tex]

[tex]4.60\times 10^{-19}=\frac{6.626\times 10^{-34}\times 3\times 10^8}{\lambda}[/tex]

[tex]4.6\times \:10^{26}\times \lambda=1.99\times 10^{20}[/tex]

[tex]\lambda=4.32\times 10^{-9}\ m[/tex] = 432 nm

(b)

Given that:- Energy = [tex]6.94\times 10^{-19}\ J[/tex]

[tex]6.94\times 10^{-19}=\frac{6.626\times 10^{-34}\times 3\times 10^8}{\lambda}[/tex]

[tex]6.94\times \:10^{26}\times \lambda=1.99\times 10^{20}[/tex]

[tex]\lambda=2.87\times 10^{-9}\ m[/tex] = 287 nm

(c)

Given that:- Energy = [tex]4.41\times 10^{-19}\ J[/tex]

[tex]4.41\times 10^{-19}=\frac{6.626\times 10^{-34}\times 3\times 10^8}{\lambda}[/tex]

[tex]4.41\times \:10^{26}\times \lambda=1.99\times 10^{20}[/tex]

[tex]\lambda=4.51\times 10^{-9}\ m[/tex] = 451 nm

Answer:

Diamagnetism in atom occurs whenever two electrons in an orbital paired equalises with a total spin of 0.

Paramagnetism in atom occurs whenever at least one orbital of an atom has a net spin of electron. That is a paramagnetic electron is just an unpaired electron in the atom.

Here is a twist even if an atom have ten diamagnetic electrons, the presence of at least one paramagnetic electron, makes it to be considered as a paramagnetic atom.

Simply put paramagnetic elements are one that have unpaired electrons, whereas diamagnetic elements do have paired electron.

The atomic orbital and radius increases by gaining electron linearly so even electron numbered atoms are diamagnetic while the odd electron numbered atoms are paramagnetic.

Running through the first 18 elements one can observe that there is an alternative odd number of electrons and an even number proofing that that half of the first 18 elements shows paramagnetism and diamagnetism respectively.

Explanation:

Answer:

The concentration of the Cu(II) ion is 0.777M

Explanation:

Step 1: Data given

Volume of  1.70 M CuCl2 = 48.0 mL = 0.0480 L

Volume of 0.800 M (NH4)2S = 57.0 mL = 0.0570 L

Step 2: The balanced equation

CuCl2 (aq) + (NH4)2S (aq) → 2 NH4Cl (aq) + CuS (s)

Step 3: Calculate moles CuCl2

moles CuCl2 = 0.0480 L * 1.70 M=0.0816 moles

Step 4: Calculate moles (NH4)2S

moles (NH4)2S = 0.0570 L * 0.800 M = 0.0456 moles

Step 5: Calculate the limiting reactant

The ratio between CuCl2 and (NH4)2S is 1 : 1 so (NH4)2S is the limiting reactant . IT will completely be consumed (0.0456 moles).

CuCl2 is in excess. There will remain 0.0816 - 0.0456 = 0.0360 moles

Step 6: Calculate moles of CuS

For 1 mol CuCl2 we need 1 mol (NH4)2S to produce 2 moles of NH4Cl and 1 mol CuS

For 0.0456 moles we'll produce 0.0456 moles CuS

Step 7: Calculate moles of Cu(II)ion

There remains 0.0360 moles CuCl2.

There will be 0.0456 moles CuS produced

Total moles Cu^2+ = 0.0816 moles

Step 8: Calculate concentration of Cu(II) ion

Concentration = moles / volume

Concentration = 0.0816 moles / (0.048+0.057)

Concentration = 0.777 M

The concentration of the Cu(II) ion is 0.777M

Answer: 2.5g

Explanation:

5 % sucrose solution means that 5 % of the weight of the solution is sucrose.

If 1 liter of water weighs 1000 grams. To prepare 5% sucrose solution 5/100 x 1000 = 50 grams. Since 1 liter equals 1000ml, thus a 5 % solution has 50 grams of solute dissolved in one liter.

To prepare 5% sucrose solution in 50mls

=5/100 x 50

= 0.05 x 50

= 2.5g

Therefore to prepare 5% sucrose solution in 50mls we dissolve 2.5g of sucrose in 50ml of water

Final answer:

To prepare a 5% by mass sucrose solution with 50.0 ml of distilled water, approximately 2.63 g of sucrose is required, which is calculated using the percent by mass formula and assumes that water's density is 1 g/ml.

Explanation:

The student's question asks to calculate the mass of sucrose necessary to prepare a 5% by mass sucrose solution using 50.0 ml of distilled water. The concept involved here is the percent by mass calculation which is used in preparing solutions in chemistry. To calculate the mass of the sucrose needed, one must use the formula:

Percent by mass = (mass of solute / mass of solution) × 100%

Given that we want a 5% sucrose solution, we can rearrange the formula to solve for the mass of sucrose:

Mass of sucrose = (Percent by mass × mass of solution) / 100%

First we need to convert the volume of water to mass, assuming the density of water is approximately 1 g/ml:

Mass of water (solvent) = 50.0 ml × 1 g/ml = 50.0 g

The total mass of the solution will be the mass of the water plus the mass of the sucrose, which we can call 'x':

Mass of solution = mass of water + x

Now, plugging in the known values and solving for 'x' gives us:

x = (5% × (50.0 g + x)) / 100%

Solving this equation for 'x' yields:

0.05 × 50.0 g + 0.05x = x

2.5 g + 0.05x = x

2.5 g = x - 0.05x

2.5 g = 0.95x

x = 2.5 g / 0.95

x = 2.6316 g

Therefore, the mass of sucrose necessary to make a 5% sucrose solution with 50.0 ml of distilled water is approximately 2.63 g.

Answer: The precipitate will not be formed in the above solution.

Explanation:

The chemical equation for the reaction of potassium bromide and lead acetate follows:

[tex]2KBr(aq.)+Pb(CH_3COO)_2(aq.)\rightarrow PbBr_2(s)+2CH_3COOK(aq.)[/tex]

We are given:

Concentration of KBr = 0.013 M

Concentration of lead acetate = 0.0035 M

1 mole of KBr produces 1 mole of potassium ions and 1 mole of bromide ions

So, concentration of bromide ions = 0.013 M

1 mole of lead (II) acetate produces 1 mole of lead (II) ions and 2 moles of acetate ions

So, concentration of lead (II) ions = 0.0035 M

The salt produced is lead (II) bromide

The equation follows:

[tex]PbBr_2(s)\rightleftharpoons Pb^{2+}(aq.)+2Br^-(aq.)[/tex]

The expression of [tex]Q_{sp}[/tex] for above equation follows:

[tex]Q_{sp}=[Pb^{2+}]\times [Br^-]^2[/tex]

Putting values of the concentrations in above expression, we get:

[tex]Q_{sp}=(0.0035)\times (0.013)^2\\\\Q_{sp}=5.92\times 10^{-7}[/tex]

We know that:

[tex]K_{sp}[/tex] for lead (II) bromide = [tex]4.67\times 10^{-6}[/tex]

There are 3 conditions:

As, the [tex]Q_{sp}[/tex] is less than [tex]K_{sp}[/tex]. The above reaction is product favored. This means that no salt or precipitate will be formed.

Hence, the precipitate will not be formed in the above solution.

Answer:

0.327 M is the concentration of hydrochloric acid in the final solution.

Explanation:

Case 1:

716.7 ml of 3.86 M HCl. Using a volumetric pipet, you take 119.56 ml of that solution and dilute it to 969.88 ml in a volumetric flask.

Before dilution :

[tex]M_1=3.86 M, V_1=119.56 ml[/tex]

After dilution :

[tex]M_2=?, V_2=969.88 ml[/tex]

[tex]M_1V_1=M_2V_2[/tex] ( dilution )

[tex]M_2=\frac{M_1V_1}{V_2}=\frac{3.86 M\times 119.56 mL}{969.88 mL}[/tex]

[tex]M_2=0.4758 M[/tex]

Case 2:

Now you take 100.00 ml of case 1 solution and dilute it to 145.62 ml in a volumetric flask

Before diluting it further  :

[tex]M_2=0.4758 M, V_1=100.00 ml[/tex]

After dilution :

[tex]M_3=?, V_3=145.62 ml[/tex]

[tex]M_3V_3=M_2V_2[/tex] ( dilution )

[tex]M_3=\frac{M_2V_2}{V_3}=\frac{0.4758 M\times 100.00 mL}{145.62 mL}[/tex]

[tex]M_3=0.327 M[/tex]

0.327 M is the concentration of hydrochloric acid in the final solution.

Answer:

ΔSmix,m = 5.7628 J/K.mol

Explanation:

mix: A + B

∴ nA = x mol A

∴ nB = y mol B

⇒ n mix = x + y = 1 mol

∴ P total = 1 bar

∴ T: constant

entropy of gases when mixing:

∴ XA = x/1 = x

∴ XB = y/1 = y

⇒ ΔSmix = - x*R*Lnx - y*R*Lny

assuming: x = y = 0.5 mol

⇒ ΔSmix = - (0.5)(R)(- 0.693) - (0.5)(R)(- 0.693)

⇒ ΔSmix = (0.3465)(R) + (0.3465)(R)

⇒ ΔSmix = (0.6931)(R)

∴ R = 8.314 J/K.mol

⇒ ΔSmix,m = (0.6931)(8.314 J/K.mol)

⇒ ΔSmix,m = 5.7628 J/K.mol

Answer:

n g ng

Explanation:

chn gcfn h

Answer: (a) m=7.64kg (b)m=7.64kg

Explanation: (a) Calculating the mass of nitrogen using the ideal gas model:

P.V=n.R.T

Note: as pressure is in Pascal and volume is in m³, the constant R will be 8,31J/K.mol.

Totransform kPa in Pa: P=1356kPa → P=1356.10³ Pa

To use the temperature, it has to be in Kelvin: T=2718°C → T = 2718 + 273 = 2991K

P.V=n.R.T  

n=(P.V) / (R.T)

n= (1356.10³. 0.5) / (8.31.2991)

n= 27,28 mols

For nitrogen, 1 mol = 28,014 g so m=764,22g or, in this case, m=764220kg.

(b) To calculate the mass with the compressibility chart, use a compatibility factor Z, which is Z=(p.v) / (R.T). If the factor Z=1, the gas behaves ideally.

This means that, in general, gases behaves almost ideally. So, in this case, calculating the mass of nitrogen by the ideal gas law or the Z factor will result in the same quantity.

Explanation:

A solution is red in color means it has absorbed other color wavelengths and reflects only red color wavelength. A red color wavelength would absorb wavelength corresponding to violet color. Hence, the wavelength should be fixed in the range of 400 nm to measure the absorbance. And not red light range ( 700 nm) to measure its absorbance."

Answer:

The concentration the student should write down in her lab is 2.2 mol/L

Explanation:

Atomic mass of the elements are:

Na: 22.989 u

S: 32.065 u

O: 15.999 u

Molar mass of sodium thiosulfate, Na2S2O3 = (2*22.989 + 2*32.065 + 3*15.999) g/mol = 158.105 g/mol.

Mass of Na2S2O3 taken = (19.440 - 2.2) g = 17.240 g.

For mole(s) of Na2S2O3 = (mass taken)/(molar mass)

= (17.240 g)/(158.105 g/mol) = 0.1090 mole.

Volume of the solution = 50.29 mL = (50.29 mL)*(1 L)/(1000 mL)

= 0.05029 L.

To find the molar concentration of the sodium thiosulfate solution prepared we use the formula:

= (moles of sodium thiosulfate)/(volume of solution in L)

= (0.1090 mole)/(0.05029 L)

= 2.1674 mol/L

Answer:

1. LiOH : strong base

2. HF : weak acid

3. HCl : strong acid

4. NH3 : weak base

Ka expression: Ka = [A- ] * [H+] / [HA]

Explanation:

The Acidity Constant:

The acid dissociation constant, Ka, (or acidity constant, or acid ionization constant) is a measure of the strength of a weak acid (which is not completely dissociated):

HA ↔ A- + H +

HA is a generic acid that dissociates into A- (the conjugate base of the acid), and the hydrogen or proton ion, H +.

The dissociation constant Ka is written as the ratio of equilibrium concentrations (in mol / L):

Ka = [A- ] * [H+] / [HA]

When we write in square brackets, we refer to the concentration of that element.

Final answer:

LiOH is a strong base, HF is a weak acid with a Kₐ expression of Kₐ = [H⁺][F⁻]/[HF], HCl is a strong acid, and NH₃ is a weak base with a Kₐ expression for its conjugate acid NH₄⁺ as Kₐ = [NH₄⁺][OH⁻]/[NH₃].

Explanation:

Compounds can be identified as strong acids, weak acids, strong bases, or weak bases depending on their ability to dissociate in solution. The classification is based on the strength of the acids and bases, which is a measure of their tendency to donate or accept protons.

Since, the options are not given the question is incomplete. The complete question is as follows:

A molecule from a new organism contains adenine, cytosine, guanine, and thymine. What is this unknown molecule?

DNA

lipid

carbohydrate

RNA

protein

Answer: DNA

Explanation:

The DNA short for deoxyribonucleic acid is a genetic material that can be found in majority of the living beings on earth. It is composed of two strands that are coiled around each other. Each strand consists of nucleotides. Each nucleotide is composed of four nitrogen exhibiting nucleobases like guanine, cytosine, adenine and thymine along with the deoxyribose sugar and phosphate group. In DNA there are two groups of nitrogenous bases these includes the pyrimidines and purines. The pyrimidines are cytosine and thymine and the purines are guanine and adenine.  

According to the given situation, a molecule from a new organism consists of adenine, cytosine thymine and guanine these all are nitrogenous bases which can be found in DNA.

Answer:

In comparison to Part 1 of this experiment, we observed similar reactions when determining the make up of our unknown. When testing for Mn2+ we observed a color change that resulted in a darker brown/red color, when testing for Co2+ we observed the formation of foamy bubbles but we could not conclude that a gas had formed, when testing for Fe3+ the result was a liquid red in color, when testing for Cr3+ we observed no change, when testing for Zn2+ we observed the formation of a pink/red liquid, when testing for K+ we observed the formation of a precipitate, when testing for Ca2+ we observe the formation of a precipitate. Sources of error may have occurred when observing whether or not an actual reaction had taken place or not, using glassware that wasn't fully cleaned, or the accidental mix of various other liquids in the lab

Explanation:

Final answer:

When a hydrogen atom in the ground state absorbs a photon of wavelength 97.20 nm, it moves to energy level 2.

Explanation:

When a hydrogen atom in the ground state absorbs a photon of wavelength 97.20 nm, it moves to a higher energy level. To determine the energy level, we can reference the Lyman series of photons, which have energies capable of exciting the hydrogen atom from the ground state to higher energy levels. By comparing the wavelength of the absorbed photon to the wavelengths of the Lyman series, we can find the corresponding energy level.

Based on the information given, the first five wavelengths in the Lyman series are 121.6 nm, 102.6 nm, 97.3 nm, 95.0 nm, and 93.8 nm. The ground state energy of hydrogen is -13.6 eV. By calculating the energies corresponding to these wavelengths, we find that the absorbed photon with a wavelength of 97.20 nm corresponds to the energy level 2. Thus, the electron in the ground state hydrogen atom moves to energy level 2 when absorbing a photon with a wavelength of 97.20 nm.

Answer:

1. When observing a positive test for the jones reagent and negative for the Lucas test, it indicates that it is in the presence of a primary alcohol.

Jones reagent behaves like a strong oxidant, where it transforms the primary alcohols into carboxylic acids and the secondary alcohols into ketones. Tertiary alcohols do not react.

With the Lucas test, tertiary alcohols react immediately producing turbidity, while secondary alcohols do so in five minutes. Primary alcohols do not react significantly with Lucas reagent at room temperature.

2. No reaction (See the attached drawing)

3. (see the attached drawing)

The Justification are:

(a) has 7 electrons in its outermost shell, making it a halogen.

(b) has a full s subshell and belongs to the alkaline earth metals.

(c) has a complete electron shell and is an inert gas.

(d) has 1 electron in its outermost shell, classifying it as an alkali metal.

(e) has a partially filled d subshell, indicating a transition metal.

(f) has a full s subshell and is categorized as an alkaline earth metal.

(a) 1s² 2s² 2p⁶ 3s² 3p⁵- This configuration corresponds to the element Chlorine (Cl), which is a halogen.

(b) 1s² 2s² 2p⁶ 3s² 3p⁶ 3d¹⁰ 4s²- This configuration corresponds to the element Calcium (Ca), which is an alkaline earth metal.

(c) 1s² 2s² 2p⁶ 3s² 3p⁶ 3d¹⁰ 4s² 4p⁶- This configuration corresponds to the noble gas Krypton (Kr), which is an inert gas.

(d) 1s² 2s² 2p⁶ 3s² 3p⁶ 4s¹ -This configuration corresponds to the element Potassium (K), which is an alkali metal.

(e) 1s² 2s² 2p⁶ 3s² 3p⁶ 3d¹⁰ 4s² 4p⁶ 4d⁵ 5s² - This configuration corresponds to the element Yttrium (Y), which is a transition metal.

(f) 1s² 2s² 2p⁶ 3s² -This configuration corresponds to the element Magnesium (Mg), which is an alkaline earth

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See text below

Determine whether each of the following electron configurations is an inert gas, a halogen, an alkali metal, an alkaline earth metal, or a transition metal. Justify your choices.(a) 1s22s22p63s23p5(b) 1s22s22p63s23p63d74s2(c) 1s22s22p63s23p63d104s24p6(d) 1s22s22p63s23p64s1(e) 1s22s22p63s23p63d104s24p64d55s2(f) 1s22s22p63s2

The electron configurations correspond to different types of elements: (a) is a halogen, (b) is a transition metal, (c) is an inert gas, (d) is an alkali metal, (e) is an alkaline earth metal, and (f) is also an alkaline earth metal.

This question is about determining the type of atom from their electron configurations. Just by looking at the electron configurations, we can deduce what type of element we’re dealing with.

(a) 1s2 2s2 2p6 3s2 3p5: atoms with 5 valence electrons in their outer shell are in Group 17 of the periodic table, which are halogens.

(b) 1s2 2s2 2p6 3s2 3p6 3d7 4s2: transition metals have incomplete d-subshells in their electron configurations, therefore, this is a transition metal.

(c) 1s2 2s2 2p6 3s2 3p6 3d10 4s2 4p6: This configuration ends with a filled p6 subshell. This indicates that the element in question is an inert gas, which are known for their stability due to a complete set of electrons in their outermost energy level.

(d) 1s2 2s2 2p6 3s2 3p6 4s1: atoms that have one electron in their highest energy level (s1) are in Group 1 of the periodic table, which are alkali metals.

(e) 1s2 2s2 2p6 3s2 3p6 3d10 4s2 4p6 5s2: This is an alkaline earth metal since it ends with an s2 orbiltal.

(f) 1s2 2s2 2p6 3s2: atoms with two valence electrons in their outer shell are in Group 2 of the periodic table, which are alkaline earth metals.

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Answer: Least Acidic:  A

Moderate Acidic: C

Most Acidic: B

Explanation:

Given compounds' acidity cannot be determined as the molecular formulas seem incorrect. Normally, acidity is determined by factors such as the presence of hydrogen atoms and their ability to be donated influenced by bond polarity and molecular structure.

The acidity of the given compounds cannot be determined as the given molecular formulas (h5ch17p19b1, h5ch17p19a1, h5ch17p19c1) appear to be incorrect or non-standard. Typically, the acidity of a compound is influenced by factors like the presence of hydrogen atoms, how easily these can be donated (as determined by bond polarity and structure of the molecule), and the stability of the conjugate base after a hydrogen atom has been donated.

In common terminology, acidity refers to the ability of a substance to donate a proton (H+) in a chemical reaction. The traditional scale for measuring acidity is the pH scale, where lower pH values indicate higher acidity.

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Answer :

(a) The symbol, group number, and period number of the element is, O, 16 and 2 respectively.

(b) The symbol, group number, and period number of the element is, P, 15 and 3 respectively.

Explanation :

Electronic configuration : It is defined as the representation of electrons around the nucleus of an atom.

Number of electrons in an atom are determined by the electronic configuration.

To identify the block of the element after you get its electronic configuration:

(i) If the element belongs to s-block.

Group number = Number of valence electrons (or outermost shell electrons).

(ii) If the element belongs to p- block.

Group number = No. of valence electrons + 10 .i.e., 10 + np electrons + ns electrons.

(iii) If the element belongs to d-block.

Group no. = no. of electrons in (n-1) d subshell + no. of electron/s in ns shell.

(iv) If the element belongs to f-block, Group no. = 3.

(a) [He] 2s²2p⁴

From the given electronic configuration, we conclude that it has 6 valence electrons and belongs to p-block. So, it belongs to group number 16 (6+10).

The highest energy level in the electronic configuration shows the period number. In this, highest energy level is (n=2). So, the period number is, 2.

Thus, the element which is present in 2nd period and 16 group number is, oxygen (O).

(b) [Ne] 3s²3p³

From the given electronic configuration, we conclude that it has 5 valence electrons and belongs to p-block. So, it belongs to group number 15 (5+10).

The highest energy level in the electronic configuration shows the period number. In this, highest energy level is (n=3). So, the period number is, 3.

Thus, the element which is present in 3rd period and 15 group number is, phosphorous (P).

Answer: The mass of estrogen that must be added is 2.83 grams

Explanation:

The equation used to calculate relative lowering of vapor pressure follows:

[tex]\frac{p^o-p_s}{p^o}=i\times \chi_{solute}[/tex]

where,

[tex]\frac{p^o-p_s}{p^o}[/tex] = relative lowering in vapor pressure

i = Van't Hoff factor = 1 (for non electrolytes)

[tex]\chi_{solute}[/tex] = mole fraction of solute = ?

[tex]p^o[/tex] = vapor pressure of pure ethanol = 54.68 mmHg

[tex]p_s[/tex] = vapor pressure of solution = 54.11 mmHg

Putting values in above equation, we get:

[tex]\frac{54.68-54.11}{54.68}=1\times \chi_{\text{estrogen}}\\\\\chi_{\text{estrogen}}=0.0104[/tex]

This means that 0.0104 moles of estrogen are present in the solution

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

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

Moles of estrogen = 0.0104 moles

Molar mass of estrogen = 272.4 g/mol

Putting values in above equation, we get:

[tex]0.0104mol=\frac{\text{Mass of estrogen}}{272.4g/mol}\\\\\text{Mass of estrogen}=(0.0104mol\times 272.4g/mol)=2.83g[/tex]

Hence, the mass of estrogen that must be added is 2.83 grams

Answer: The pH of the solution is 13.0

Explanation:

To calculate the molarity of solution, we use the equation:

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

Given mass of KOH = 716. mg = 0.716 g    (Conversion factor:  1 g = 1000 mg)

Molar mass of KOH = 56 g/mol

Volume of solution = 130 mL

Putting values in above equation, we get:

[tex]\text{Molarity of solution}=\frac{0.716\times 1000}{56g/mol\times 130}\\\\\text{Molarity of solution}=0.098M[/tex]

1 mole of KOH produces 1 mole of hydroxide ions and 1 mole of potassium ions

[tex]pOH=-\log[OH^-][/tex]

We are given:

[tex[[OH^-]=0.098M[/tex]

Putting values in above equation, we get:

[tex]pOH=-\log(0.098)\\\\pOH=1.00[/tex]

To calculate the pH of the solution, we use the equation:

pOH + pH = 14

So,  pH = 14 - 1.00 = 13.0

Hence, the pH of the solution is 13.0

The given configurations correspond to the elements Cadmium (Cd) and Nickel (Ni) respectively. Cd is in group 12, period 5 and Ni is in group 10, period 4.

The requested electron configurations correspond to specific elements in the periodic table.

(a) The configuration [Kr] 5s²4d¹⁰ corresponds to the element Cadmium (Cd). Its symbol is Cd, it is in group 12, and period 5. The partial valence-level orbital diagram is as follows:

(b) The configuration [Ar] 4s²3d⁸ corresponds to the element Nickel (Ni). Its symbol is Ni, it is in group 10, and period 4. The partial valence-level orbital diagram is as follows:

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The configurations [Kr] 5s²4d¹⁰ and [Ar] 4s²3d⁸ represent Cadmium and Nickel respectively. Cadmium is in the 12th group, 5th period, and Nickel is in the 10th group, 4th period. Both are transition metals with distinct chemical reactions.

The element with the electron configuration [Kr] 5s²4d¹⁰ is Cadmium (Cd). It belongs to the 12th group and 5th period. Its valence electron configuration diagram shows that there are 2 electrons in the 5s subshell and 10 electrons in the 4d subshell.

On the other hand, the element with the electron configuration [Ar] 4s²3d⁸ is Nickel (Ni). This element belongs to the 10th group and 4th period. Its valence electron configuration diagram shows it has 2 electrons occupying the 4s subshell and 8 electrons in the 3d subshell.

These identified elements, Cadmium and Nickel, represent their unique chemical and physical properties. For instance, they are both transition metals and behave similarly in many chemical reactions.

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Answer : The radii of the two ions Cl⁻ ion and Na⁺ ion is, 181 and 102 pm respectively.

Explanation :

As we are given that the Na⁺ radius is 56.4% of the Cl⁻ radius.

Let us assume that the radius of Cl⁻ be, (x) pm

So, the radius of Na⁺ = [tex]x\times \frac{56.4}{100}=(0.564x)pm[/tex]

In the crystal structure of NaCl, 2 Cl⁻ ions present at the corner and 1 Na⁺ ion present at the edge of lattice.

Thus, the edge length is equal to the sum of 2 radius of Cl⁻ ion and 2 radius of Na⁺ ion.

Given:

Distance between Na⁺ nuclei = 566 pm

Thus, the relation will be:

[tex]2\times \text{Radius of }Cl^-+2\times \text{Radius of }Na^+=\text{Distance between }Na^+\text{ nuclei}[/tex]

[tex]2\times x+2\times 0.564x=566[/tex]

[tex]2x+1.128x=566[/tex]

[tex]3.128x=566[/tex]

[tex]x=180.9\approx 181pm[/tex]

The radius of Cl⁻ ion = (x) pm = 181 pm

The radius of Na⁺ ion = (0.564x) pm = (0.564 × 181) pm =102.084 pm ≈ 102 pm

Thus, the radii of the two ions Cl⁻ ion and Na⁺ ion is, 181 and 102 pm respectively.

Answer:

Explanation:hiii

80 to 95 grams of carbohydrate should be consumed every hour for 4 hours post-exercise.

Carbohydrates are biomolecules that are made up of carbon, hydrogen, and oxygen atoms.

Examples of carbohydrates are starch, sugar, fiber.

Carbohydrate is the main component of our food which gives us energy.

If an athlete weighs 175 lb and has to exercise every four hours.

He will need a regular amount of carbohydrates to get energy.

Thus, the amount of carbohydrate required is 80 to 95 grams.

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Answer: The concentration of chloride ions in the solution obtained is 0.674 M

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]     .....(1)

Molarity of KCl solution = 0.675 M

Volume of solution = 297 mL

Putting values in equation 1, we get:

[tex]0.675=\frac{\text{Moles of KCl}\times 1000}{297}\\\\\text{Moles of KCl}=\frac{(0.675mol/L\times 297)}{1000}=0.200mol[/tex]

1 mole of KCl produces 1 mole of chloride ions and 1 mole of potassium ion

Moles of chloride ions in KCl = 0.200 moles

Molarity of magnesium chloride solution = 0.338 M

Volume of solution = 664 mL

Putting values in equation 1, we get:

[tex]0.338=\frac{\text{Moles of }MgCl_2\times 1000}{664}\\\\\text{Moles of }MgCl_2=\frac{(0.338mol/L\times 664)}{1000}=0.224mol[/tex]

1 mole of magnesium chloride produces 2 moles of chloride ions and 1 mole of magnesium ion

Moles of chloride ions in magnesium chloride = [tex](2\times 0.224)=0.448mol[/tex]

Calculating the chloride ion concentration, we use equation 1:

Total moles of chloride ions in the solution = (0.200 + 0.448) moles = 0.648 moles

Total volume of the solution = (297 + 664) mL = 961 mL

Putting values in equation 1, we get:

[tex]\text{Concentration of chloride ions}=\frac{0.648mol\times 1000}{961}\\\\\text{Concentration of chloride ions}=0.674M[/tex]

Hence, the concentration of chloride ions in the solution obtained is 0.674 M

Based on the concentrations of the given solutions, the concentration of Cl⁻ is 0.674 M.

Moles of a substance is calculated using the formula:

For 297 mL of 0.675 M KCl

297 mL = 0.297 L

moles of KCl = 0.675 * 0.297

moles of KCl = 0.200 moles

1 mole of KCl produces 1 mole of Cl⁻

Thus, moles of Cl⁻ = 0.200 moles

For 664 mL of 0.338 M MgCl₂

664 mL = 0.664 L

moles of MgCl₂ = 0.338 * 0.664

moles of MgCl₂ = 0.224 moles

1 mole of MgCl₂ produces 2 moles of Cl⁻

moles of Cl⁻ = 2 * 0.224

moles of Cl⁻ = 0.448 moles

Total moles of Cl⁻ = 0.448 + 0.200

moles of Cl⁻ = 0.648 moles

Volume of solution = 664 + 297

Volume of solution = 961 = 0.961 L

Molarity of Cl⁻ = 0.648 / 0.961

Molarity of Cl⁻ = 0.674 M

Therefore, the concentration of Cl⁻ is 0.674 M

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Answer:1s22s22p63s2- Magnesium

1s22s22p63s23p3- phosphorus

1s22s22p63s23p64s23d104p6- krypton

1s22s2- Beryllium

1s22s22p6- neon

1s22s22p3- nitrogen

Explanation:

The identity of an element is deducible from its electronic configuration by looking out for the outermost shell configuration and counting the total number of electrons present. for example; phosphorus 15 electrons and an outermost shell configuration of ns2 np3. Any electronic configuration written above which reflects these characteristics must belong to phosphorus.

The question is about identifying elements based on given electron configurations and then matching these configurations to known ones. The elements matching the given configurations have the same number of electrons in their outermost energy level, indicating the possibility of similar chemical properties.

The question is regarding the identification of elements based on their electron configurations and comparing them to known configurations. The electron configuration describes the distribution of electrons in an atom's electron shells and subshells.

For instance, the given first electron configuration (1) 1s22s22p63s2 corresponds to the element Neon. The element that would match this configuration is one whose electron configuration is 1s22s22p6, which is also for Neon in a neutral state.

For the second case (1s22s22p63s23p64s23d104p6), it corresponds to the element Krypton. The element that would match this configuration is also Krypton in a neutral state. These elements are said to have similar chemical properties because they have the same number of electrons in their outermost energy level, making their bonding patterns similar.

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