Naturally occurring element X exists in three isotopic forms: X-28 (27.979 amu, 92.21% abundance), X-29 (28.976 amu 4.70% abundance), and X-30 (29.974 amu 3.09% abundance). Calculate the atomic weight of X.

a. 29.09 amu
b. 28.09 amu
c. 35.29 amu
d. 86.93 amu
e. 25.80 amu

Answer:

There are 0.00523 moles of oxygen in 0.150 grams of calcium sulfate crystal.

Explanation:

Mass of calcium sulfate crystal = m = 0.150 g

Molar mass of calcium sulfate crystal = M = 172 g/mol

Moles of magnesium nitrate = n

[tex]n=\frac{m}{M}[/tex]

[tex]n=\frac{0.150 g}{172 g/mol}=0.0008721 mol[/tex]

1 mole of calcium sulfate crystal has 6 moles of oxygen atoms. Then 0.004446 moles calcium sulfate crystal will have :

[tex]6\times 0.0008721 mol=0.0052326mol\approx 0.00523 mol[/tex]

There are 0.00523 moles of oxygen in 0.150 grams of calcium sulfate crystal.

In 0.150 g of CaSO₄·2H₂O, there are 3.15 × 10²¹ O atoms.

We want to calculate the number of oxygen atoms in 0.150 g of CaSO₄·2H₂O. We will need a series of conversion factors.

A conversion factor is an arithmetical multiplier for converting a quantity expressed in one set of units into an equivalent expressed in another.

We will need to consider the following conversion factors:

[tex]0.150 g CaSO_4.2H_2O \times \frac{1molCaSO_4.2H_2O}{172.17gCaSO_4.2H_2O} \times \frac{6.02\times 10^{23}moleculesCaSO_4.2H_2O }{1molCaSO_4.2H_2O} \times \frac{6\ O\ atoms}{1moleculeCaSO_4.2H_2O} = \\ = 3.15 \times 10^{21} O\ atoms[/tex]

In 0.150 g of CaSO₄·2H₂O, there are 3.15 × 10²¹ O atoms.

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Explanation:

Dehydrohalogenation reactions occurs as elimination reactions through the following mechanism:

Step 1: A strong base(usually KOH) removes a slightly acidic hydrogen proton from the alkyl halide.

Step 2: The electrons from the broken hydrogen‐carbon bond are attracted toward the slightly positive carbon (carbocation) atom attached to the chlorine atom. As these electrons approach the second carbon, the halogen atom breaks free.

However, elimination will be slower in the exit of Hydrogen atom at the C2 and C3 because of the steric hindrance by the methyl group.

Elimination of the hydrogen from the methyl group is easier.

Thus, the major product will A

Answer:

. 1. This amino acid has a positively charged R group: ARGININE

2. This amino acid has a negatively charged R group: ASPARTATE

3. This amino acid has a neutral polar R group: NONE

4. This amino acid has a nonpolar aliphatic R: VALINE

5. This amino acid has an aromatic R group: TRYTOPHAN

Explanation:

1) Arginine contains an extra amino group bearing a positive charge, in its chain which imparts basic properties to it

2) Aspartate contains an extra carboxyl group with a dissociable protron. Once the Protron is dissociated, it carries an extra negative charge in its side chain (R)

3) NONE of the amino acids given belong to this group because amino acids with neutral polar R groups contain functional groups that form hydrogen bonds with water. But, this is not the case with tryptophan aspartate valine or arginine

4) Valine has a R group that is hydrocarbon in nature and thus hydrophobic.

5) Trytophan has a benzene ring in its side chain

Amino acids are classified by the properties of their R groups at pH 7 to be arginine (positively charged), aspartate (negatively charged), valine (neutral polar and nonpolar aliphatic) and tryptophan (aromatic).

The chemical properties of an amino acid's side chain, or R group can be identified with their characteristics at pH 7. Here are each of the amino acids classified by the properties of their R groups at pH 7:

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Explanation:

Equation of the reaction:

2KMnO4(aq) + 16HCl(aq) --> 2MnCl2(aq) + 2KCl(aq) + 5Cl2(g) + 8H2O(aq)

To calculate the limiting reagent, we need to calculate the number of moles of the reactants :

KMnO4:

Molar mass = (39 + 55 + (16*4))

= 158 g/mol

Number of moles = mass/molar mass

= 229.19/158

= 1.4506 mol

HCl:

Molar mass = 1 + 35.5

= 36.5 g/mol

Number of moles = 526.64/36.5

= 14.428 mol

By stoichiometry, 2 moles of KMnO4 reacted with 16 moles of HCl

The limiting reagent :

14.428 moles of HCl * 2 moles of KMnO4/16moles of HCl

= 1.8035 moles of KMnO4 is required to react with 14.428 moles of HCl

But there's 1.4506 moles of KMnO4. Therefore, KMnO4 is the limiting reagent.

Mass of the products:

KCl:

2 moles of KMnO4 will produce 2 moles of KCl

Moles of KCl = 1 * 1.4506 mol

= 1.4506 mol

Molar mass = 39 + 35.5 = 74.5 g/mol

Mass of KCl = 74.5 * 1.4506

= 108.07 g

MnCl2:

2 moles of KMnO4 will produce 2 moles of MnCl2

Number of moles of MnCl2 = 1 * 1.4506

= 1.4506 mol

Molar mass = 55 + (35.5*2)

= 126 g/mol

Mass of MnCl2= 1.4506 * 126

= 182.78 g

Cl2:

2 moles of KMnO4 will produce 5 moles of Cl2

Number of moles of Cl2 = 5/2 * 1.4506

= 3.6265 mol

Molar mass of Cl2 = 35.5*2

= 71 g/mol

Mass of Cl2 = 71* 3.6265

= 257.4815 g

H2O:

2 moles of KMnO4 will produce 8 moles of H2O

Number of moles of H2O = 8/2 * 1.4506

= 5.80 mol

Molar mass of H2O =( 1*2) + 16

= 18 g/mol

Mass of H2O = 18*5.80

= 104.44 g

Answer:

(1) mass of KCl =108.025g

(2) mass of MnCl2 =182.7 g

(3) mass of Cl2 =257.37 g

(4) mass of H20(water) =104.4g

Explanation:

we start with determining the limiting factor

molar mass of KMnO4 = 158g/mol

molar mass of HCl = 36.5g/mol

hence,

number of moles of KMnO4= mass /molar mass

number of moles of KMnO4= 229.19/158 = 1.45moles

number of moles of HCl = 526.64/36.5 = 14.43 moles

we chose the lowest number of moles from the reactants as the limiting factor

hence the limiting factor is KMnO4.

calculating the mass of the products:

for KCl

2moles of KMnO4 reacts to give 2moles of KCl

there for KCl contains 1.45moles

(1) mass of KCl = number of moles of KCl x molar mass of KCl = 1.45 x 74.5 = 108.025g

(2) from mole ratio 2 mole of KMnO4 gave 2 moles of MnCl2

therefore 1.45moles KMnO4 will give 1.45 MnCl2

mass of MnCl2 = 1.45moles x molar mass MnCl2 = 1.45 x 126 = 182.7 g

(3) from mole ratio

2moles KMnO4 gave 5moles Cl2

1.45 moles will give (5 x1.45)/2 moles of Cl2

mass of Cl2 = number of moles of Cl2 x molar mass Cl2 = 3.625 x 71 = 257.37 g

(4)from mole ratio

2moles KMnO4 gave 8moles water(H2O)

1.45 moles will give (1.45x 8)/2 moles of H20

mass of H2O produced = number of moles of H20 x the molar mass of H20 = 5.8 x 18 = 104.4g

Answer:

The correct answer is d.

Incorrect Answers:

Options a, b and c are incorrect answers because during folding of protein, surface area decreases and so there are less water molecules involved.The water molecules are free.The protein folding decreases entropy

Answer:

Molar mass of unknown gas = 41.45 g/mol

Explanation:

Time taken for a gas to effuse/diffuse ∝√(Molar Mass of the gas)

Let T₁ be the time taken for the Unknown gas to effuse. T₁ = 7.3 min = 438s

Let T₂ be the time taken for the Nitrogen gas to effuse. T₁ = 6 min = 360s

(T₁/T₂) = √(Molar mass of unknown gas/Molar mass of Nitrogen)

Molar mass of Nitrogen = 2×14 = 28 g/mol

(438/360) = √(Molar Mass of unknown gas/28)

√(Molar mass of unknown gas/28) = 1.21667

molar mass of unknown gas/28 = 1.21667² = 1.4803

Molar mass of unknown gas = 1.4803 × 28 = 41.45 g/mol

Hope this helps!

Answer:

Explanation:

Equation of the reaction:

NaOH(aq) + KHC8H4O4(aq) --> KNaC8H4O4(aq) + H2O(l)

A.

Number of moles = mass/molar mass

= 0.20/204.22

= 9.79 x 10-4 mol KHP

By stoichiometry, 1 mole of NaOH reacts with 1 mole of KHP

= 6.267 x 10-4 mol NaOH

Molar concentration = number of moles/volume

= 9.79 x 10-4/0.00953

= 0.103 M NaOH

B.

Number of moles of kool aid = mass/molar mass

= 3.607/342

= 0.01055 mol

Concentration = 0.01055/0.005

= 2.1 M

C1 * V1 = C2 * V2

Concentration of NaOH = 2.1 * 0.005/0.01079

= 0.977 M

Number of moles of NaOH = 0.0105 mol.

Answer:

(a) Cl

[tex]1s^22s^22p^63s^23p^5[/tex]

(b) Si

[tex]1s^22s^22p^63s^23p^2[/tex]

(c) Sr

[tex]1s^22s^22p^63s^23p^63d^{10}4s^24p^65s^2[/tex]

Explanation:

(a) Cl

Atomic number = 17

The electronic configuration is -  

[tex]1s^22s^22p^63s^23p^5[/tex]

(b) Si

Atomic number = 14

The electronic configuration is -  

[tex]1s^22s^22p^63s^23p^2[/tex]

(c) Sr

Atomic number- 38

The electronic configuration is -  

[tex]1s^22s^22p^63s^23p^63d^{10}4s^24p^65s^2[/tex]

The energy of attraction between a cation with a valence of 1 and an anion with a valence of 3 can be calculated using Coulomb's law.

The energy of attraction between a cation and an anion can be calculated using Coulomb's law. Coulomb's law states that the energy of attraction is directly proportional to the product of the charges and inversely proportional to the distance between them. In this case, the cation has a valence of +1 and the anion has a valence of -3. The distance between their centers is given as 7.5 nm. The equation to calculate the energy of attraction is:

E = k * (q1 * q2) / r

Where E is the energy of attraction, k is Coulomb's constant (9 x 10^9 Nm^2/C^2), q1 and q2 are the charges of the cation and anion respectively, and r is the distance between their centers.

Substituting the values:

E = (9 x 10^9 Nm^2/C^2) * ((+1) * (-3)) / (7.5 x 10^-9 m)

Simplifying the equation gives:

E = -36 Nm

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Answer:

1 ClO-(aq) + 1 I-(aq) ---> 1 Cl -(aq) + 1 IO-(aq).

Explanation:

1. ClO-(aq) + H2O(l) <=> HClO(aq) + OH-(aq)

2. I-(aq) + HClO(aq) <=> HIO(aq) + Cl-(aq)

3. OH-(aq) + HIO(aq) => H2O(l) + IO-(aq)

Adding all the 3 equations together gives and it gives:

ClO-(aq) + H2O(l) + I-(aq) + HClO(aq) + OH -(aq) + HIO(aq)

---> HClO(aq) + OH-(aq) + HIO(aq) + Cl -(aq) + H2O(l) + IO-(aq)

Deleting the same species on both sides of the equation gives:

1 ClO-(aq) + 1 I-(aq) ---> 1 Cl -(aq) + 1 IO-(aq)

The overall equation:

1 ClO-(aq) + 1 I-(aq) ---> 1 Cl -(aq) + 1 IO-(aq)

The overall chemical reaction is ClO- (aq) + I- (aq) --> IO- (aq) + Cl- (aq). The intermediates, compounds produced and then consumed in later reaction steps, are HClO and HIO.

The overall reaction occurring is the sum of the provided stepwise reactions, where intermediates, or species that are formed in one step and consumed in another, are cancelled out. The chemical species that are intermediates in this case are HClO and HIO.

Adding all three reactions together and cancelling out the intermediates, we get:

ClO-(aq) + I-(aq) --> IO-(aq) + Cl-(aq)

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Answer:The concentration of 2-phosphoglycerate is 0.415 mM

Explanation:

[tex]3-phosphoglycerate\rightleftharpoons 2-phosphoglycerate[/tex]

Relation between standard Gibbs free energy and equilibrium constant follows:

[tex]\Delta G^o=-RT\ln K[/tex]

where,

[tex]\Delta G^o[/tex] = Standard Gibbs free energy = +4.40  kJ/mol = 4400 J/mol  (Conversion factor: 1kJ = 1000J)

R = Gas constant = [tex]8.314J/K mol[/tex]

T = temperature = [tex]25^0C=(25+273)K=298 K[/tex]

Putting values in above equation, we get:

[tex]4400J/mol=-(8.314J/Kmol)\times 298K\times \ln K[/tex]

[tex]\ln K=-1.776[/tex]

[tex]K=0.169[/tex]

[tex]K=\frac{ 2-phosphoglycerate}{3-phosphoglycerate}[/tex]

[tex]0.169=\frac{ 2-phosphoglycerate}{2.45mM}[/tex]

[tex]2-phosphoglycerate}=0.415mM[/tex]

Thus the concentration of 2-phosphoglycerate is 0.415 mM

Answer:

0.104 M

Explanation:

A saline solution contains 0.770 g of NaCl (molar mass = 58.55 g/mol) in 133 mL.

The molar mass of the solute (NaCl) is 58.55 g/mol. The moles corresponding to 0.770 g are:

0.770 g × (1 mol/55.85 g) = 0.0138 mol

The volume of solution is 133 mL. In liters,

133 mL × (1 L/1000 mL) = 0.133 L

The molarity of NaCl is:

M = moles of solute / liters of solution

M = 0.0138 mol / 0.133 L

M = 0.104 M

Final answer:

The overall reaction for converting coal to methane is 2C(s) + 2H2O(g) → CH4(g) + CO2(g). The ΔH°rxn for the overall process is -117.3 kJ. Using the molar mass of coal and the heat of combustion for methane, the total heat for gasifying 6.28 kg of coal and burning the resulting methane is calculated.

Explanation:

To answer the student's question regarding the production of methane from coal via gasification, we combine the given reactions to find the overall reaction:

The H2O(g) and CO(g) appear on both sides of the above equations and can be cancelled out when added together. The overall reaction is:

2C(s) + 2H2O(g) → CH4(g) + CO2(g)

To calculate ΔH°rxn for the overall reaction, we sum the enthalpies of the individual steps:

ΔH°rxn (overall) = 129.7 kJ + (-41 kJ) + (-206 kJ) = -117.3 kJ

Using the heat of combustion for methane, 890.4 kJ/mol, and the assumption that coal is approximated as pure carbon (12.00 g/mol), we can calculate the total heat for gasifying 6.28 kg of coal and then burning the methane produced:

Moles of C in 6.28 kg = (6280 g) / (12.00 g/mol) = 523.33 mol

Total heat from gasification = 523.33 mol × -117.3 kJ/mol = -61411.3 kJ

Total heat from combustion = 523.33 mol × 890.4 kJ/mol = 465774.9 kJ

The total heat for the entire process is the sum of heat from gasification and combustion.

Answer:

Al(NO₃)₃ →  Al³⁺ (aq)  +  3NO₃⁻ (aq)

FeCl₂ → Fe²⁺ (aq) +  2Cl⁻ (aq)

K₂S → 2K⁺ (aq) +  S⁻²(aq)

Mg(CH3COO)₂ → Mg²⁺ (aq) + 2CH3COO⁻ (aq)

(NH₄)₃PO₄ → 3NH₄⁺ (aq) + PO₄⁻³(aq)

Explanation:

Al(NO₃)₃ → Aluminum nitrate

FeCl₂ → Iron (II) chloride

K₂S → Potassium sulfide

Mg(CH3COO)₂ → Magnesium acetate

(NH₄)₃PO₄ → Ammonium phosphate

Answer:

Hello, the above question is not complete, nonetheless let us check somethings out.

Explanation:

Paleothermometer definition is from two words, that is "Paleo" which means something that is old and ''thermometer" which is an instrument for measuring temperature. So, if we add this up, Paleothermometer is an instrument for measuring "old" temperature, that is temperature. One of the Paleothermometer that is been used is the δ18O which is the one in the question that has isotopic ratio of 18O/16O, and it deals with the measurement of 18O to 16O. The others include Alkenones Paleothermometer, Mg/Ca Paleothermometer, Leaf physiognomy and so on.

If the values of the isotopic ratio that is 18O/16O ratio is low, then the temperature is high. To Calculate the 18O/16O ratio for ancient ocean then we will be using the equation below;

δ18O = (z - 1) × 1000. Where z= [(18O/16O)/( 18O/16O)sm. And sm= standard mean.

Answer:

[tex]2.023 m^3[/tex] is the total displacement (volume) of the submarine.

Explanation:

Mass of water carried by submarine at 1000 ft depth = m = 2100 kg

The density of seawater at 1000 ft depth = d = [tex]1033 kg/m^3[/tex]

Volume of the water displaced = V= ?

Total displacement of the submarine = Volume of the water displaced = V

[tex]Density=\frac{Mass}{Volume}[/tex]

[tex]V=\frac{m}{d}=\frac{2100 kg}{1033 kg/m^3}=2.023 m^3[/tex]

[tex]2.023 m^3[/tex] is the total displacement (volume) of the submarine.

Answer:

C₄H₈O₄ is the molecular formula of the solid.

Explanation:

Let's apply the freezing point depression to solve this:

ΔT = Kf . m

where Δt = Freezing T° pure solvent - Freezing T° of solution

Kf, the cryoscopic constant

m = molalilty, moles of solute in 1kg of solvent.

We must determine the molecular weight to know the molecular formula of the solid

Let's replace the data.

1.6°C = 8 °C/m . m

We can determine molality, by this:

1.6 °C / 8°C/m = 0.2 m

mol of solute / 1kg of solvent  = 0.2

Let's convert the mass of solvent from g to kg, to determine the moles of solute.

mol of solute / 0.0281 kg = 0.2 mol/kg

28.1 g . 1kg / 1000 g = 0.0281 kg

mol of solute = 0.0281 kg . 0.2 mol/kg → 0.00562 moles

Molar mass (g/mol) → 0.673 g / 0.00562 mol = 120 g/mol

Now, we can apply the percent composition.

100 g of compound have ___ 40 g C ___6.7 g H ___ 53.3 g O

120 g of compound must have:

(120 . 40) / 100 = 48 g of C

(120 . 6.7) / 100 = 8 g of H

(120 . 53.3) / 100 = 64 g of O

Let's convert the mass of each elements to moles

48 g . 1 mol/12 g = 4 C

8 g . 1 mol /1g = 8 H

64 g . 1 mol / 16g = 4 O

To determine the molecular formula of the solid compound, calculate the empirical formula using the mass percent composition. Then, divide the molar mass of the compound by the molar mass of the empirical formula to find the number of empirical formula units in the compound.

To determine the molecular formula of the solid compound, we first need to calculate its empirical formula. We can assume a 100g sample of the compound, so we have 40g of C, 6.7g of H, and 53.3g of O. Converting the grams to moles, we find that we have approximately 3.33 moles of C, 6.65 moles of H, and 3.33 moles of O.

Next, we need to find the smallest whole number ratio between the elements. The ratio between C, H, and O is approximately 1:2:1. So, the empirical formula of the compound is CH2O.

To find the molecular formula, we need to know the molar mass of the compound. The molar mass of the empirical formula CH2O is approximately 30 g/mol. To determine the molecular formula, we need to divide the molar mass of the compound by the molar mass of the empirical formula. Assuming the molar mass of the compound is a multiple of 30 g/mol, we can divide it by 30 to find the number of empirical formula units in the compound.

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Answer:

1.126 grams

Explanation:

Given that:

Standard Solution = 600 mg O₂/L

The molecular weight of C₆H₁₂O₆ (glucose) = 180.156 g/mol

The mass of O₂  in 1 mole of C₆H₁₂O₆ can be determined  as:

C₆H₁₂O₆ = 6 × 16 g ( of one oxygen)

              = 96 g

∴ 96 g  of O₂ is available in 180.156 gram of C₆H₁₂O₆

Thus C₆H₁₂O₆  required for giving 600 mg = 0.60 g of O₂

∴ [tex]\frac{180.156}{96}*0.6[/tex]

= 1.876625 × 0.6

= 1.125975 g

≅ 1.126 grams

Hence, 1.126 grams of C₆H₁₂O₆ (glucose) will be added to one liter of distilled water in order to get 600mg O₂/L.

To calculate the concentration of a second-order reaction after a certain time, use the half-life equation and substitute the given values.

A second-order reaction follows the equation relating the half-life of the reaction to its rate constant and initial concentration:

t1/2 = 1 / (k * [A]₀)

To calculate the concentration of A in the vessel after a certain number of seconds, you can substitute the given values into this equation. For example, if t1/2 = 18 min, k = 0.0576 L mol-1 min-1, and [A]₀ = 0.200 mol L-1, you can solve for [A].

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The concentration in the vessel seconds later is ≈ 0.179 M which will be required for 0.200 M and a rate constant of 5.76 × 10-² L/mol/min over 10 minutes.

To solve this question, we can use the integrated rate law for second-order reactions:

[tex]\frac{1}{[A]t} = \frac{1}{[A]_0} + kt[/tex]

Substitute these values into the integrated rate law equation:

[tex]\frac{1}{[A]t} = \frac{1}{0.200} + (5.76 \times 10^{-2} L/mol/min)(10.0 min)[/tex]

Calculate the right-hand side:

[tex]\frac{1}{[A]t} = 5.00 + 0.576 \\\\\frac{1}{[A]t} = 5.576[/tex]

So, [tex][A]t = 1 / 5.576 \approx 0.179 M[/tex]

Therefore, the concentration of butadiene remaining after 10.0 min is approximately 0.179 M.

Final answer:

The molality (m) of an aqueous KCl solution with a mole fraction of KCl, XKCl = 0.175, can be calculated using the formula molality (m) = (moles of solute) / (kilograms of solvent). First, calculate the moles of KCl by multiplying the mole fraction by the mass of water and dividing by the molar mass of KCl. Then, calculate the kilograms of water by dividing the mass of water by 1000. Finally, substitute these values into the molality formula.

Explanation:

The molality (m) of an aqueous KCl solution with a mole fraction (XKCl) of 0.175 can be calculated using the following formula:

molality (m) = (moles of solute) / (kilograms of solvent)

To find the molality, we need to convert the mole fraction into moles of KCl and kilograms of water. The molar mass of KCl is 74.55 g/mol and the molar mass of H2O is 18.02 g/mol.

First, we calculate the moles of KCl:

moles of KCl = XKCl * (mass of water) / (molar mass of KCl)

Next, we calculate the kilograms of water:

kilograms of water = (mass of water) / 1000

Finally, we substitute these values into the formula:

molality (m) = moles of KCl / kilograms of water

Answer: E = 2.455 x 10^5 N/C

Explanation:

q1 = 1.2x10^-7C

q2 = 6.2x10^-8C

Electric field, E = kQ/r²

where k = 9.0x10^9

since the location is (27 - 5)cm from q1

hence electric field, E1 = k*q1/r²

E1= (9x10^9 x 1.2x10^-7)/(0.22)² = 22314.05 N/C

for q2:

E1 = k*q2/r²

E2 at 5cm

E2 = (9x10^9 x 6.2x10^-8)/(0.05)² = 223200 N/C

Hence, the total electric field at 5cm position is

E = E1 + E2

E = 22314.05 + 223200 = 245514.05 N/C

E = 2.455 x 10^5 N/C

Final answer:

The electric field at a point 5.0 cm from a negative charge is calculated using the formula E = kQ/r² with the direction toward the charge.

Explanation:

q1 = 1.2x10⁻⁷ C

q2 = 6.2x10⁻⁸C

Electric field, E = kQ/r²

where k = 9.0x10⁹

since the location is (27 - 5)cm from q1

hence electric field, E1 = k×q1/r²

E1= (9x10⁹ x 1.2x10⁻⁷)/(0.22)² = 22314.05 N/C

for q2:

E1 = k×q2/r²

E2 at 5cm

E2 = (9x10⁹ x 6.2x10⁻⁸)/(0.05)² = 223200 N/C

Hence, the total electric field at 5cm position is

E = E1 + E2

E = 22314.05 + 223200 = 245514.05 N/C

E = 2.455 x 10⁵ N/C

Answer:

As is in the attachment.

Explanation:

The condensed electronic configuration is written in the short form by expressing in terms of the elements of the noble gases.

The attached file is the explanation of the answers.

Explanation:

the correct match can as follows:

1.  helium ⇒ discrete atoms (helium is an inert gas)

2. oxygen ⇒ molecules (oxygen exists in molecular form as O2)

3 Magnesium ⇒ matalic lattice ( Magnesium is a metal FCC crystal Structure :) )

4 Aluminum⇒ covalent network (since it is situated in the middle of the group and posses amphoteric properties too)

Final answer:

To classify the elements: Calcium and Potassium form metallic lattices (category C), Helium exists as discrete atoms (category A), and Sulfur forms molecules (category B). This is due to how the atoms or molecules arrange themselves in solids, often driven by forces between electrons and nuclei.

Explanation:

To classify each element in its elemental form based on the given key:

Calcium (Ca) would fall into category C, Metallic lattice, as it is a metal and metals typically form extended, repeating three-dimensional patterns.

Helium (He) is correctly classified as A, Discrete atoms, because it exists as individual atoms and does not form bonds easily.

Sulfur (S) in its most common form consists of eight-atom rings, making it B, Molecules.

Potassium (K), like calcium, would also be categorized as C, Metallic lattice, due to its metal characteristics, forming a crystal lattice.

Atoms arrange themselves in these structures based on the net attractive forces between electrons and atomic nuclei, and the structural arrangement significantly affects the properties of the material, such as malleability and ductility for metals. Pure metals are crystalline solids with metal atoms packed in a repeating pattern known as a unit cell.

Answer:

Molarity for the solution is 1.05 mol/L

Explanation:

Mole fraction of solute = 0.0194

Solution's density = 1.0627 g/mL

We must know that sum of mole fraction = 1

Mole fraction of solute + Mole fraction of solvent = 1

0.0194 + Mole fraction of solvent = 1

Mole fraction of solvent = 1 - 0.0194 → 0.9806

Molarity is mol of solute in 1L of solution, so we have to determine solution's volume in L

With molar mass we can determine the mass of solute and solvent and then, the solution's mass

0.0194 mol . 101.1 g/ mol = 1.96 g of non electrolyte solute

0.9806 mol . 18 g/mol = 17.65 g of water

Mass of solution = mass of solute + mass of solvent

1.96 g + 17.65 g = 19.6 g (mass of solution)

Solution's density = Solution's mass / Solution's volume

1.0627 g/mL = 19.6 g / Solution's volume

Solution's volume = 19.6 g / 1.0627 g/mol →18.4 mL

Let's convert the mass from mL to L

18.4mL . 1L / 1000 mL = 0.0184 L

We have the moles of solute, so let's determine molarity

mol/L → 0.0194 mol / 0.0184 L =  1.05 M

Answer:

48.32 g of anhydrous MnSO4.

Explanation:

Equation of dehydration reaction:

MnSO4 •4H2O --> MnSO4 + 4H2O

Molar mass = 55 + 32 + (4*16) + 4((1*2) + 16)

= 223 g/mol

Mass of MnSO4 • 4H2O = 71.6 g

Number of moles = mass/molar mass

= 71.6/223

= 0.32 mol.

By stoichiometry, since 1 mole of MnSO4 •4H2O is dehydrated to give 1 mole of anhydrous MnSO4

Number of moles of MnSO4 = 0.32 mol.

Molar mass = 55 + 32 + (4*16)

= 151 g/mol.

Mass = 151 * 0.32

= 48.32 g of anhydrous MnSO4.

Answer: The concentration of acetic acid and sodium acetate (acetate ion) is 0.06 M and 0.34 M respectively

Explanation:

We are given:

Concentration of buffer solution having acetic acid and sodium acetate = 0.4 M

Let the concentration of acetic acid be x M

So, the concentration of sodium acetate will be = (0.4 - x) M

To calculate the concentration of acid for given pH, we use the equation given by Henderson Hasselbalch:

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

[tex]pH=pK_a+\log(\frac{[CH_3COONa]}{[CH_3COOH]})[/tex]

where,

[tex]pK_a[/tex] = negative logarithm of acid dissociation constant of acetic acid = 4.76

[tex][CH_3COONa]=0.4-x[/tex]

[tex][CH_3COOH]=x[/tex]

pH = 5.5

Putting values in above equation, we get:

[tex]5.5=4.76+\log(0.4-x}{x})\\\\x=0.062M[/tex]

So, concentration of acetic acid = x = 0.06 M

Concentration of sodium acetate = (0.4 - x) = (0.4 - 0.06) = 0.34 M

Hence, the concentration of acetic acid and sodium acetate (acetate ion) is 0.06 M and 0.34 M respectively

Answer : The concentration of AB after 14.0 s is, 0.29 M

Explanation :

The expression used for second order kinetics is:

[tex]kt=\frac{1}{[A_t]}-\frac{1}{[A_o]}[/tex]

where,

k = rate constant = [tex]0.20M^{-1}s^{-1}[/tex]

t = time = 14.0 s

[tex][A_t][/tex] = final concentration = ?

[tex][A_o][/tex] = initial concentration = 1.50 M

Now put all the given values in the above expression, we get:

[tex]0.20\times 14.0=\frac{1}{[A_t]}-\frac{1}{1.50}[/tex]

[tex][A_t]=0.29M[/tex]

Therefore, the concentration of AB after 14.0 s is, 0.29 M

Answer: The number of electrons present in the fourth energy level are 32

Explanation:

To calculate the number of electrons present in a particular energy level, we use the equation:

[tex]\text{Number of electrons}=2n^2[/tex]

where,

n = principle quantum number

Calculating the number of electrons for fourth energy level:

n = 4

Putting values in above equation, we get:

[tex]\text{Number of electrons}=2(4)^2\\\\\text{Number of electrons}=32[/tex]

Hence, the number of electrons present in the fourth energy level are 32

Answer:

Explanation:

The final net force will be in the Z- direction. Let's find out the z component of the force on the differential volume of charge is:

df = dqEcosθz

[tex]E = \frac{1}{4\pi epsilon} \frac{Qr}{R^{3} }[/tex]

dq = ρdV =  [tex]\frac{3Q}{4\pi R^{3} }[/tex][tex]r^{2}[/tex]dr.sinθdθdΦ

integrate it over half ball,

[tex]F_{z} = \int\limits^._V {df_{x}dV} =\frac{1}{4\pi epsilon } \frac{Q}{R^{3} } \frac{3Q}{4\pi R^{3} }\int\limits^R_0 {\int\limits^\frac{\pi }{2} _{0} {\int\limits^\frac{\pi }{2} _0 {r^{3} } \, dr } \, } \,[/tex].sinθcosθdθdΦ.( these are part of the integral, i was unable to write it in equation format).

    = [tex]\frac{3Q^{2} }{32\pi epsilonR^{2} } \int\limits^\frac{\pi }{2} _b {} \,[/tex]  sinθcosθdθ

   = [tex]\frac{3Q^{2} }{64\pi epsilon R^{2} }[/tex]

[tex]F = \frac{3Q^{2} }{64\pi epsilon R^{2} } z[/tex]

The net force that the southern hemisphere of a uniformly charged solid sphere exerts on the northern hemisphere is (kQ²)/(2R²), where k is the electrostatic constant, Q is the total charge on the sphere, and R is the radius of the sphere.

The net force that the southern hemisphere of a uniformly charged solid sphere exerts on the northern hemisphere can be found by considering the electric field at the surface of the sphere. Since the charge distribution is spherically symmetric, the electric field at the surface of the sphere will only have a radial component. Using Gauss's law, we can determine that the electric field at the surface is given by E = kQ/R², where k is the electrostatic constant, Q is the total charge on the sphere, and R is the radius of the sphere.

To find the force, we can multiply the electric field by the charge on the northern hemisphere. The charge on the northern hemisphere can be calculated as half the total charge on the sphere, Q/2. Therefore, the net force is given by F = (kQ²)/(2R²).

he could not preidct it bud