Lithium ions in Lithium selenide (Li2Se) have an atomic radius of 73 pm whereas the selenium ion is 184 pm. This compound is most likely to adopt a:Select the correct answer below:
a. closest-packed array with lithium ions occupying tetrahedral holes
b. closest-packed array with lithium ions occupying octahedral holes
c. body-centered cubic array with lithium ions occupying cubic holes
d. none of the above

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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Answer:

The mass of AlBr3 is 24.5 grams

Explanation:

Step 1: Data given

Mass of aluminium = 5.0 grams

Mass of bromine = 22.0 grams

Molar mass of aluminium = 26.98 g/mol

Molar mass of br2= 159.8 g/mol

Step 2: The balancced equation

2 Al + 3 Br2 → 2 AlBr3

Step 3: Calculate moles Al

Moles Al = mass Al / molar mass Al

Moles Al = 5.0 grams / 26.98 g/mol

Moles Al = 0.185 moles

Step 4: Calculate moles Br

Moles Br = 22.0 grams / 159.8 g/mol

Moles Br = 0.138 moles

Step 5: Calculate limiting reactant

For 2 moles Al we need 3 moles Br2 to produce 2 moles AlBr3

Br2 is the limiting reactant. It will completely be consumed (0.0313 moles)

Al is in excess. There will react 2/3*0.138 = 0.092 moles

There will remain 0.185- 0.092 = 0.093 moles Al

Step 6: Calculate moles AlBr3

For 2 moles Al we need 3 moles Br2 to produce 2 moles AlBr3

For 0.0313 moles Br2 we'll have 2/3*0.138 = 0.092 moles AlBr3

Step 7: Calculate mass AlBr3

Mass AlBr3 = moles * molar mass

Mass AlBr3 = 0.092 moles * 266.69 g/mol

Mass AlBr3 = 24.5 grams

The mass of AlBr3 is 24.5 grams

Final answer:

To find out the mass of aluminum bromide produced, we calculate the moles of aluminum and bromine, identify the limiting reactant, and then calculate the mass of the product from the moles of the limiting reactant using the stoichiometry of the reaction.

Explanation:

The question involves a stoichiometry calculation to determine the amount of product formed from a chemical reaction between aluminum and bromine (2 Al + 3 Br2 → 2 AlBr3). To solve this, we need to identify the limiting reactant and use it to calculate the mass of the product formed. First, the moles of each reactant is determined using their molar masses. Next, the stoichiometry of the balanced chemical equation is used to find the moles of product that can be formed from the limiting reactant.

Then, the moles of product are converted to grams using the molar mass of the product. This step-by-step process will lead us to find the mass of aluminum bromide (AlBr3) that can be produced.

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: 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:

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 :

(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).

The osmotic pressure of this isotonic saline solution at 25 °C, calculated using the formula II = MRT and performing the necessary conversions for temperature and molarity, is approximately 3.86 atm.

The osmotic pressure of an isotonic solution can be calculated using the formula II = MRT. In this case, 'M' represents the molarity of the solution, 'R' is the universal gas constant (0.08206 L atm/mol K), and 'T' is the temperature in Kelvin.

To calculate the osmotic pressure, we need to first convert the temperature from Celsius to Kelvin: 25 degrees Celsius = 298.15 Kelvin. Then, we calculate the molarity of the solution. Here, we know that 0.923 g of NaCl is dissolved in 100 mL of water. The molar mass of NaCl is approximately 58.44 g/mol, so the number of moles of NaCl in the solution is 0.923g / 58.44g/mol = 0.0158 mol. As the volume of the solution is 100 ml or 0.1 L, the molarity 'M' is 0.0158 mol / 0.1 L = 0.158 mol/L.

Now we substitute these values into the formula II = MRT to calculate osmotic pressure. Thus, II = 0.158 mol/L * 0.08206 L atm/mol K * 298.15 K = 3.86 atm.

So, the osmotic pressure of this isotonic saline solution at 25 °C is approximately 3.86 atm.

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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.

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:

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:

(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:

(d) Cl, Br, Se

Explanation:

Ionisation energy is defined as the minimum energy required to remove the loosely bound valence electron of a neutral gaseous atom or molecule. It is usually endothermic process.

Accorfing to the trends down the group and across the period:

Ionisation energy generally increases as one moves from left to right in a given period while ionisation generally decreases as one moves from top to bottom in a given group. This is of the magnitude of the effective Nuclear charge and the Number of electron shells.

From the elements above:

Ionisation energy decreases down the group while it increases across the period.

F > C > Be

Li > Na > K

Ar > Cl > Na

Cl > Br > Se

Final answer:

The first ionization energy (IE₁) is the energy required to remove one electron from an atom. The arrangement in decreasing IE₁ for the given sets of elements are: K, Na, Li; Be, C, F; Cl, Na, Ar; and Cl, Br, Se.

Explanation:

The first ionization energy (IE₁) refers to the energy required to remove one electron from an atom in its gaseous state. It is generally observed that ionization energy increases across a period and decreases down a group in the periodic table.

(a) Na, Li, K: These elements are in the same group of alkali metals. Since potassium (K) is located below sodium (Na) and lithium (Li) in the periodic table, K has the lowest ionization energy, followed by Na, and then Li. Therefore, the arrangement in decreasing IE₁ would be: K, Na, Li.

(b) Be, F, C: Beryllium (Be) has a higher ionization energy compared to carbon (C) and fluorine (F) because Be has a smaller atomic radius. Thus, the arrangement in decreasing IE₁ would be: Be, C, F.

(c) Cl, Ar, Na: Chlorine (Cl) has a higher ionization energy compared to argon (Ar) and sodium (Na) because Cl has fewer energy levels and a higher effective nuclear charge. Therefore, the arrangement in decreasing IE₁ would be: Cl, Na, Ar.

(d) Cl, Br, Se: Chlorine (Cl) has the highest ionization energy among the three elements because it requires the most energy to remove an electron. Bromine (Br) has a lower ionization energy compared to chlorine, and selenium (Se) has the lowest ionization energy of the three elements. Therefore, the arrangement in decreasing IE₁ would be: Cl, Br, Se.

Final answer:

The general expression for the ionization energy of any one-electron species is IE = -13.6/n^2 eV. The ionization energy of B⁴⁺ is approximately -13.6/2^2 eV. The minimum wavelength required to remove the electron from the n = 3 level of He⁺ and from the n = 2 level of Be³⁺ can be calculated using the equation E = hc/λ.

Explanation:

(a) The general expression for the ionization energy of any one-electron species can be written as: IE = -13.6/n^2 eV, where n is the principal quantum number. This expression indicates that the ionization energy decreases as the principal quantum number increases.

(b) To calculate the ionization energy of B⁴⁺, we need to find the value of n in the expression IE = -13.6/n^2 eV. Since B⁴⁺ has 5 protons in its nucleus, we can use the formula Z = 1.33n^2 to determine the value of n for B⁴⁺. Plugging in Z = 5, we get n ≈ 2.14. Therefore, the ionization energy of B⁴⁺ is approximately -13.6/2^2 eV.

(c) The minimum wavelength required to remove the electron from the n = 3 level of He⁺ can be found using the equation E = hc/λ, where E is the energy, h is Planck's constant, c is the speed of light, and λ is the wavelength. We can calculate the energy using the ionization energy of He⁺ from the given information, and then rearrange the equation to solve for λ.

(d) Similarly, we can use the same equation and rearrange it to solve for λ in order to find the minimum wavelength required to remove the electron from the n = 2 level of Be³⁺.

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: 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

Answer:

NO, they are not the same compound

Explanation:

Given that;

Compound A melts at  220.5 °C - 222.1 °C;      &

Compound B melts at   221.2 °C - 223.4 °C

It is seen from above that there is little difference in the melting point of Compound A and B. This little difference can be as a result of factors associated when carrying the melting process or because different methods were employed in the establishing their melting points.

Also, we were told that when they were both mixed together ,  the mixture of compound A and B melts at 216.4 °C  - 224.6 °C.

This statement has largely indicated that both compounds are not the same at all, because if they were, the mixture of compound A and B melting point must be identical to one of the individual compound's melting point either from compound A or from compound B.

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. 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)

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:

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.

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:

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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Explanation:

Chlorine atom has an atomic number of 17

Electronic configuration

Cl = 1s2 2s2 2p6 3s2 3p5

Cl- = 1s2 2s2 2p6 3s2 3p6

Soduim atom has an atomic number of 11

Electronic configuration

Na = 1s2 2s2 2p6 3s1

Na+ = 1s2 2s2 2p6

Calcium atom has an atomic number of 20

Electronic configuration

Ca = 1s2 2s2 2p6 3s2 3p6 4s2

Ca2+ = 1s2 2s2 2p6 3s2 3p6

The monatomic ions most likely formed by Cl, Na, and Ca are Cl-, Na+, and Ca2+, respectively.

(a) The most likely monatomic ion formed by Chlorine (Cl) is Cl-. The electronic configuration of Cl is 1s2 2s2 2p6 3s2 3p5. By gaining one electron, Cl achieves a stable octet and becomes a chloride ion (Cl-).

(b) The most likely monatomic ion formed by Sodium (Na) is Na+. The electronic configuration of Na is 1s2 2s2 2p6 3s1. By losing one electron, Na achieves a stable octet and becomes a sodium ion (Na+).

(c) The most likely monatomic ion formed by Calcium (Ca) is Ca2+. The electronic configuration of Ca is 1s2 2s2 2p6 3s2 3p6 4s2. By losing two electrons, Ca achieves a stable octet and becomes a calcium ion (Ca2+).

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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

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: 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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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:

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: 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