Write electronic configuration of first 30 elements in their sub shell
​

The statement that the sodium-potassium pump uses bulk transport to move the sodium and potassium ions is false one. So, the right choice for answer is option (b).

Sodium-Potassium Pump: The sodium-potassium pump is an active transport protein that moves sodium and potassium ions across cell membranes. It is particularly important in neurons, where it establishes ion gradients that facilitate the transmission of nerve impulses. ATP acts as a solute pump, an enzyme that pumps sodium into the cell, exporting three sodium ions and two potassium ions. It is an active transport that results in the hydrolysis of ATP molecules to provide the necessary energy. Fueled by ATP, cells are constantly pumping out sodium ions and potassium ions. Sodium and potassium ions cross the cell membrane at the rate of 3 sodium ions for 2 potassium ions. Therefore, despite bulk transport, the sodium potassium pump uses active transport to move sodium and potassium ions. Thus, correct answer is option (b).

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Answer:

I'm sorry, but I cannot see the observations or the data table you mentioned in your question. However, I can still provide you with some general guidance on how to approach the calculations and answer the questions based on the given information.

4. To calculate the concentration of the NaOH solution, you need to know the mass of NaOH used and the volume of the solution. The formula to calculate concentration is:

Concentration (in mol/L) = (Mass of NaOH (in grams) / molar mass of NaOH) / Volume of solution (in L)

Make sure to convert the mass of NaOH to moles by dividing it by the molar mass of NaOH. The molar mass of NaOH is the sum of the atomic masses of sodium (Na), oxygen (O), and hydrogen (H).

5. The balanced equation for the neutralization reaction between NaOH and HCl is:

NaOH(aq) + HCl(aq) → NaCl(aq) + H2O(l)

(aq) represents an aqueous solution, and (l) represents a liquid.

6a. To calculate the average concentration of HCl in the sample from site B, you need to know the volumes and concentrations of the NaOH and HCl solutions used in the titration. Use the formula:

Concentration of HCl (in mol/L) = (Volume of NaOH solution (in L) * Concentration of NaOH (in mol/L)) / Volume of HCl solution (in L)

Multiply the volume of NaOH solution used by its concentration to find the amount of NaOH used. Then, divide this amount by the volume of HCl solution used to find the concentration of HCl.

6b. To determine the pH of the water at site B, you need to know the concentration of HCl from the previous calculation. The pH can be calculated using the formula:

pH = -log10[H+]

Since HCl is a strong acid, it dissociates completely into H+ ions. Therefore, the concentration of H+ ions is equal to the concentration of HCl. Take the negative logarithm (base 10) of the H+ concentration to find the pH.

To check if the water is safe, compare the calculated pH value to the range provided (pH 4.5-7.5). If the pH falls within this range, the water is considered safe for plant and animal reproduction in an aquatic environment.

6c. Use a similar calculation as in 6a to determine the average concentration of HCl in the sample from site C.

6d. Use the concentration of HCl from 6c to calculate the pH using the formula in 6b. Follow the same procedure to check if the water is safe based on the pH range.

7. To find the most current pH value for the Grand River, you can search for the latest data from reliable sources such as environmental agencies, research institutions, or government websites. Compare this pH value to the pH values obtained in the experiment to assess the difference between them.

Remember, without the specific data and observations, the calculations and comparisons provided here are only general guidelines. It's important to use the actual data from your experiment to obtain accurate results and conclusions.

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Answer:

To determine the limiting reactant and the excess reactant, we need to compare the stoichiometry of the reaction with the amounts of each reactant given.

The balanced chemical equation is:

3 MgCl2 + 2 Na3PO4 -> 6 NaCl + Mg3(PO4)2

Given:

Moles of MgCl2 = 0.75 mol

Moles of Na3PO4 = 0.65 mol

a) To verify the limiting reactant, we need to calculate the moles of Na3PO4 and MgCl2 needed to react completely, based on the stoichiometry of the balanced equation.

From the equation, we can see that:

For every 3 moles of MgCl2, 2 moles of Na3PO4 are required.

Therefore, the moles of Na3PO4 required to react with 0.75 mol of MgCl2 would be:

(0.75 mol MgCl2) x (2 mol Na3PO4 / 3 mol MgCl2) = 0.5 mol Na3PO4

Since we have 0.65 mol of Na3PO4, which is greater than the required amount of 0.5 mol, Na3PO4 is the excess reactant.

b) To find the moles of the excess reactant left over, we subtract the moles of Na3PO4 that reacted from the initial moles:

0.65 mol Na3PO4 - 0.5 mol Na3PO4 = 0.15 mol Na3PO4 (left over)

c) To determine the moles of NaCl produced in the reaction, we need to calculate the moles of the limiting reactant (MgCl2) that reacted. From the balanced equation, we know that:

For every 3 moles of MgCl2, 6 moles of NaCl are produced.

Using the stoichiometry, we can calculate the moles of NaCl produced:

(0.75 mol MgCl2) x (6 mol NaCl / 3 mol MgCl2) = 1.5 mol NaCl

Therefore, 1.5 mol of NaCl will be produced in this reaction.

Preform the mathematical operation and then write the answer to the correct number of significant figures.

a) 110 g

b) 9030 cm

c) 6.0 cm

d)9.6 mm

e) 419.8 mL

f) 766.8 m

a) 100 g + 10 g = 110g

b) 903 cm × 10 cm = 9030 cm

c) 0.6030 cm² / 0.1 cm = 6.0 cm

here lowest significant is = 0.1 , so ans will round off till first decimal.

d) 9.5 mm + 0.083 mm = 9.583 mm = 9.6 mm

e) 430.0 mL -10.2111 mL = 419.8 mL

f) 762.4 m + 3.2 m + 1.21 m = 766.8 m

here the lowest significant figure = 3.2 , so the ans will be round off to first decimal figure.

Thus, preform the mathematical operation and then write the answer to the correct number of significant figures.

a) 110 g

b) 9030 cm

c) 6.0 cm

d)9.6 mm

e) 419.8 mL

f) 766.8 m

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Answer:

Four folk love legends: "Cowhand and Weaver Girl", "Meng Jiangnu crying at the Great Wall" (cry Wan Xiliang), "Liang Shanbo and Zhu Yingtai", "The Legend of the White Snake", the author does not know

The development history of the Four ancient Chinese folk legends: China is a country with a vast territory and abundant resources and a long history, which contains extremely rich national cultural heritage. Among them, the four most Chinese ones are the famous "Cowhand and Weaver Girl", "Meng Jiangnu", "Liang Shanbo and Zhu Yingtai" and "The Legend of the White Snake". The most widespread, the most influential.

Brief introduction: the gigolo knit legend began in the book of songs, cable ":" Qi he vega ", "Huan he cow". "Nineteen Ancient Poems · Long Distance Star" has called the cowherd and weaver girl husband and wife. Ying Shao "custom" Yi wen: "Zhinv Tanabata when crossing the river, make magpie bridge, legend has it that the magpie head without reason all Kun, because beam (note: bridge) to cross Zhinv also. The story has been preliminarily formed and combined with the Qixi Festival custom.

The legend of Meng Jiangnu originated from the record of Qi Liang's wife refusing to hang herself in the area of the Marquis of Qi and adhering to the rules of etiquette in the Zhuan. Later, it was added in the Dangong that "Duke Zhuang of Qi attacked Ju and took away (tunnel) and Qi Liang died. The weeping wife who greeted his coffin on the road is the archetype of the story. Han Liu Xiang's Biography of Liannu (IV) : "Qi Liangzhu died in battle, his wife cried in the city, ten days the city collapsed." And Tang (anonymous) "carved jade collection" record "Qin when Yan Qi Liang, married Meng Chao female Zhongzi wife, because good was built the Great Wall officials killed, Zhongzi cry under the Great Wall, the city collapsed. It is known that this legend was popular in the Tang Dynasty, but Meng Zhongzi and Qiliang were renamed Meng Jiangnv and Fan Xiliang in the legend. The story of Butterfly lovers first appeared in Tang's "Ten Taoist and Four Tibetan Records", which recorded the story of Liang and Zhu "having a taste of schoolmates" and "sharing the tomb". There are more detailed records in Xuanbaozhi of late Tang Zhang Reading. To the Ming Dynasty Feng Menglong's "Ancient and modern novels", and added to the British Taiwan confused belt, Liang Shanbo doubts and butterflies.

The story of White Snake came into being the latest. The origin of the story is as follows. It originates from the Three Towers of West Lake. To the Ming Dynasty Feng Menglong's "Lady White Forever Town Leifeng Pagoda" (" Warning Word "), the story has been preliminarily finalized.

Do these things right away when a chemical splashes on your eye.Use water to irrigate your eye.For the at least 20 minutes, use pristine, lukewarm tap water.Use soap and water to wash your hands.To ensure that no chemicals or soap are still on your hands, thoroughly rinse them.Remove your contacts.

Safety Glasses, goggles, and shields — These products shield the face and eyes from chemical spills.Additionally, they protect them against gases, vapors, and dust.Skin Protection - Clothing items like gloves, boots, & coveralls should be used in order to protect the skin.

Chemical splashes are incidents where dangerous substances unintentionally spill, project, aerosolize, or otherwise disseminate inside a laboratory setting.

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The micromoles of silver(II) oxide (AgO) the chemist has added to the flask is 3.5805 × 10⁻⁶ µmol (rounded to 2 significant digits).Note: 1 µmol = 10⁶ mol, where µ means micro. 1 µmol = 0.000001 mol.

Given,Volume of silver(II) oxide (AgO) solution, V = 385.0 mL = 0.385L Concentration of silver(II) oxide (AgO) solution, C = 9.3 × 10⁻⁵ mol/LNumber of micromoles of silver(II) oxide (AgO) added,N = VC = 0.385 L × 9.3 × 10⁻⁵ mol/L= 3.5805 × 10⁻⁶ molNumber of micromoles of silver(II) oxide (AgO) added, N = 3.5805 × 10⁻⁶ µmol (rounded to 2 significant digits).

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Answer:

-6.31°C

Explanation:

To solve this problem, we can use the principle of heat transfer:The heat gained by the lead (q_lead) equals the heat lost by the water (q_water).The formula to calculate heat transfer is:

q = m * c * ΔTWhere:

q = heat transfer

m = mass

c = specific heat

ΔT = change in temperatureFor the lead:

q_lead = m_lead * c_lead * ΔT_leadFor the water:

q_water = m_water * c_water * ΔT_waterGiven values:

m_lead = 322 g

c_lead = 0.138 J/gºC

ΔT_lead = T_final - T_initial_lead (unknown)

m_water = 264 g

c_water = 4.18 J/gºC (specific heat of water)

ΔT_water = T_final - T_initial_water = 46°C - 25°C = 21°CSince the heat gained by the lead is equal to the heat lost by the water, we can set up the equation:

m_lead * c_lead * ΔT_lead = m_water * c_water * ΔT_waterSubstituting the given values:

322 g * 0.138 J/gºC * ΔT_lead = 264 g * 4.18 J/gºC * 21°CSimplifying the equation:

44.436 J/ºC * ΔT_lead = 2325.12 J/ºCDividing both sides of the equation by 44.436 J/ºC:

ΔT_lead = 2325.12 J/ºC / 44.436 J/ºC ≈ 52.31°CFinally, we can find the initial temperature of the lead:

T_initial_lead = T_final - ΔT_lead

T_initial_lead = 46°C - 52.31°C ≈ -6.31°CTherefore, the initial temperature of the lead was approximately -6.31°C.

Answer:

Expiremental group for -first one.    Second one-2

Explanation:

I Had Those QuestIons On mY tEsttt!!

When we have an acid-base reaction, the acid will be that substance that donates protons, in the form of H+ ions. The base will be the one that accepts the H+ ions. Now, the base that accepts the protons becomes a potential proton donor, thus becoming a conjugate acid. This is the case with H3O+.

H2O accepts H+ ions, so it is a base and when it becomes H3O+ it becomes a potential H+ proton donor, that is, a conjugate acid.

so, the answer will be B. Conjugate acid

There are approximately 3.225 × 10^24 molecules of H2 in 120 liters of H2.

To determine the number of molecules of H2 in 120 liters of H2, we need to use Avogadro's number and the ideal gas law.

Avogadro's number states that there are 6.022 x 10^23 molecules in one mole of any substance.

Given that the volume is 120 liters, we need to calculate the number of moles of H2 using the ideal gas law:

PV = nRT

Where:

P is the pressure of the gas (not given)

V is the volume of the gas (in liters) = 120 L

n is the number of moles of gas (unknown)

R is the ideal gas constant (0.0821 L·atm/(mol·K))

T is the temperature of the gas (not given)

Since the pressure and temperature are not provided, we cannot calculate the number of moles of H2 directly.

However, we can use the relationship between volume and moles at STP ( molecules  and Pressure) to estimate the number of moles. At STP, one mole of any ideal gas occupies 22.4 liters.

Number of moles (n) = V/22.4 = 120 L / 22.4 L/mol = 5.36 moles (approx.)

Now, we can use Avogadro's number to calculate the number of molecules:

Number of molecules = Number of moles × Avogadro's number

Number of molecules = 5.36 mol × 6.022 × 10^23 molecules/mol

Number of molecules ≈ 3.225 × 10^24 molecules.

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Answer:

single-celled, flagella, fresh water, decaying, psuedepodia, cillia, complex, ditches, algae

Explanation:

Euglena are single-celled protists that live in fresh water and move with flagella. Amoeba are found in fresh water and salt water around a lot of dead and decaying material. They move with a psuedepodia. Paramecium are found in fresh water and move with a cillia. This is a single celled organism but is more complex than other organisms. Volvos are found in ponds, ditches and puddles. They are commonly known as algae.

Hope this helped, however this is definitely not chemistry XD

Answer:

\(\boxed {\boxed {\sf 142.2 \ K}}\)

Explanation:

This problem asks us to find the temperature change necessary to make the volume change. We use Charles's Law which states that temperature is directly proportional to the volume of a gas. The formula is:

\(\frac{V_1}{T_1}=\frac{V_2}{T_2}\)

We know the balloon originally had a volume of 6.5 liters and a temperature of 280 Kelvin. Then, the temperature was cooled so the new volume is 3.3 liters. However, the exact new temperature is unknown.

Substitute all known values into the formula.

\(\frac {6.5 \ L}{280 \ K}=\frac{3.3 \ L}{T_2}\)

Now, solve the new temperature (T₂). First, cross multiply. Multiply the 1st numerator by the 2nd denominator. Then, multiply the 1st denominator by the 2nd numerator.

\(280 \ K * 3.3 \ L = 6.5 \ L * T_2\)

Multiply the left side.

\(924 \ K *L=6.5 \ L *T_2\)

We must isolate the variable. Currently, it is being multiplied by 6.5 liters. The inverse of multiplication is division. Divide both sides by 6.5 L.

\(\frac { 924 \ K*L}{6.5 \ L}= \frac{ 6.5 \ L *T_2}{ 6.5 \ L}\)

\(\frac { 924 \ K*L}{6.5 \ L}= T_2\)

The units of liters (L) cancel.

\(\frac { 924 \ K}{6.5 }= T_2\)

\(142.153846 \ K= T_2\)

Let's round to the nearest tenth place. The 5 in the hundredth place tells us to round the 1 up to a 2.

\(142.2 \ K= T_2\)

The temperature must be cooled to approximately 142.2 Kelvin.

Secondary consumers are organisms that primarily feed on herbivores or other primary consumers.

They occupy the next trophic level above the primary consumers in a food web. They obtain energy by consuming the primary consumers and play an important role in regulating the population of herbivores.

Examples of commonly observed secondary consumers include:

Carnivorous mammals: Animals such as wolves, lions, and tigers that feed on herbivores like deer, zebras, or gazelles.

Birds of prey: Species like eagles, hawks, and owls that consume small mammals, reptiles, or other birds.

Carnivorous fish: Fish like pike, barracuda, or bass that prey on smaller fish or aquatic invertebrates.

Predatory insects: Insects such as spiders, mantises, or dragonflies that feed on other insects, including herbivorous insects.

In a specific food web, the identification of secondary consumers would depend on the specific organisms present and their feeding interactions. It would be necessary to analyze the trophic relationships among the organisms in the food web to determine the secondary consumers accurately.

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Answer:

Explanation: Water is Hydrogen and 2 oxygen 2 being the 2 oxygen molecules then the other molecules like cool-aid or Magnesium

Examples

C6H12O6 + H2O

NaCI + H2O

Hope this helps!

Answer:

26.8224 Meters per Second.      

Answer: Molarity when 25.0 g of the compound \(NaClO_{3}\)is placed in 85.0 mL of solution is 294.12 M.

Explanation:

Given: Mass = 25.0 g

Volume  = 85.0 mL (1 mL = 0.001 L) = 0.085 L

Molarity is the number of moles of a substance divided by volume in liter.

Hence, molarity of given solution is calculated as follows.

\(Molarity = \frac{mass}{Volume (in L)}\)

Substitute the values into above formula as follows.

\(Molarity = \frac{mass}{volume (in L)}\\= \frac{25.0 g}{0.085 L}\\= 294.12 M\)

Thus, we can conclude that molarity when 25.0 g of the compound \(NaClO_{3}\)is placed in 85.0 mL of solution is 294.12 M.

The concentration of A after 30 seconds when the given reaction obeys the rate law rate = k[A]²[B].

We use the initial concentration of A and B and the rate constant of the reaction to find the rates at these concentrations. Using the integrated rate law for a second-order reaction, we find the concentration of A after 30 seconds to be 0.0934 M.

Given reaction obeys the rate law, rate=k[A]²[B].

Here, the initial concentration of A= 0.10 M,

initial concentration of B = 0.05 M, and

rate constant, k = 2.0 × 10⁻⁴ M⁻¹s⁻¹

We have to find the concentration of A, after 30 seconds.

To find the concentration of A, we need to know the rate at 0.10 M and 0.05 M. Therefore, we have to calculate the rates at these concentrations.

rate1 = k[A]²[B]

= (2.0 × 10⁻⁴ M⁻¹s⁻¹)(0.10 M)²(0.05 M)

= 1.0 × 10⁻⁷ M/srate2

= k[A]²[B] = (2.0 × 10⁻⁴ M⁻¹s⁻¹)(0.09 M)²(0.04 M)

= 6.48 × 10⁻⁸ M/s

Using the integrated rate law for a second-order reaction: [A] = [A]₀ - kt where [A]₀ = initial concentration of A, k = rate constant, and t = time in seconds.

We know [A]₀ = 0.10 M and k = 2.0 × 10⁻⁴ M⁻¹s⁻¹.

Substituting the values in the above equation, we get: [A] = [A]₀ - kt= 0.10 M - (2.0 × 10⁻⁴ M⁻¹s⁻¹)(30 s)≈ 0.0934 M

Therefore, the concentration of A in the vessel after 30 seconds is 0.0934 M.

This question requires us to calculate the concentration of A after 30 seconds when the given reaction obeys the rate law rate = k[A]²[B].

We are given the initial concentration of A and B and the rate constant of the reaction. To find the concentration of A after 30 seconds, we need to calculate the rates at the initial concentrations of A and B.

Using the integrated rate law for a second-order reaction, we can find the concentration of A at any given time. We substitute the given values in the formula and solve for [A]. We get the concentration of A as 0.0934 M after 30 seconds. This calculation is based on the assumption that no other reaction is important.

The concentration of A after 30 seconds when the given reaction obeys the rate law rate = k[A]²[B]. We use the initial concentration of A and B and the rate constant of the reaction to find the rates at these concentrations. Using the integrated rate law for a second-order reaction, we find the concentration of A after 30 seconds to be 0.0934 M. This calculation assumes that no other reaction is important.

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A unit cell can be visualized as a building block that, when replicated in all directions, creates the entire crystal lattice. Unit cells can have different shapes, the most common types of unit cells include simple cubic, body-centered cubic, and face-centered cubic etc.

(a) Definition of a unit cell: A unit cell is the basic repeating unit of a crystal lattice. It represents the smallest portion of a crystal that, when stacked together, can generate the entire crystal structure.

(b) Sketch of a body-centered cubic (BCC) unit cell is represented in the image below. In the BCC unit cell, atoms are located at the eight corners of the cube and one atom is positioned at the center of the cube.

(c) Coordination number in a BCC unit cell: The coordination number is the number of nearest neighboring atoms surrounding a central atom. In a BCC unit cell, each atom is in contact with eight nearest neighboring atoms: one in the center of the unit cell and one at each of the eight corners.

(d) Calculation for the atomic packing factor (APF) of a BCC unit cell:

The APF is calculated as the volume occupied by the atoms in the unit cell divided by the total volume of the unit cell.

For a BCC unit cell, the APF can be calculated as follows:

APF = (Number of atoms in the unit cell * Volume of each atom) / Volume of the unit cell

In a BCC unit cell, there are two atoms (one at the center and one at the corner), and the volume of each atom can be approximated as (\(4/3\)) * π * \((radius)^3\). The volume of the unit cell can be calculated as the cube of the edge length.

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The first step is to calculate the volume of the metal cylinder. This can be found by subtracting the final volume of water (48.3 mL) from the initial volume of water (25.2 mL), giving a volume of 23.1 mL. This volume is equivalent to 23.1 cm³ because 1 mL = 1 cm³.

The next step is to calculate the density of the metal. We know the mass of the metal cylinder (101.356 g), so we can divide this by its volume (23.1 cm³): density = mass/volume density = 101.356 g/23.1 cm³density = 4.39 g/cm³. We were given the mass of an unknown metal cylinder as 101.356 g and the initial volume of water as 25.2 mL. Upon immersion of the metal cylinder in the graduated cylinder, the water level rose to 48.3 mL. To determine the density of the metal cylinder, we need to find the volume of the cylinder. We can calculate the volume of the metal cylinder by finding the difference between the final and initial volumes of water. This gives us a volume of 23.1 mL. This volume is equivalent to 23.1 cm³ since 1 mL is equivalent to 1 cm³.To calculate the density of the metal cylinder, we divide the mass of the cylinder by its volume. This gives us a density of 4.39 g/cm³. Therefore, the unknown metal cylinder has a density of 4.39 g/cm³.

The density of the metal cylinder can be found by dividing its mass by its volume. In this case, we were given the mass of the cylinder and the initial and final volumes of water, so we used the difference between these volumes to find the volume of the cylinder. Dividing the mass by the volume gave us a density of 4.39 g/cm³ for the metal cylinder.

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Compared to a completely amorphous polymer of the same composition, a semi-crystalline polymer is expected to show improved stability to higher operating temperatures.

The glass liquid transition, or glass transition, is the slow and reversible transition in amorphous materials or in amorphous areas within semi crystalline substances from a tough and relatively brittle glassy kingdom into a viscous or rubbery state as the temperature is increased.

Polymers are large but single chain-like molecules wherein the repeating unit derived from small molecules called monomers are covalently linked. Structurally, they may be characterised by many repeating molecular gadgets which shape linear chains or a go-linked network.

The maximum significant distinction among amorphous and semi-crystalline polymers is within the molecular shape. As cited, amorphous polymers are random, entangled chains, at the same time as semi-crystalline polymers are established.

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Answer:

\(50ml=0,05l\\n_H_2SO_4=2.0,05=0,1mol\\mH_2SO_4=0,1.98=9,8g\)

Explanation:

The best evidence that the moon has been extensively heated is the lack of volatiles.

The following are the characteristics of the Moon

:Mass of the moon.Lack of iron core.

Oxygen isotopes like the Earth.

Lack of volatiles.

Volatiles are materials with low boiling points that exist in solid or liquid form at the Earth's surface. Water and carbon dioxide are two examples of volatile materials. The lack of volatiles on the moon is a strong indication that the moon was subjected to high temperatures. It also indicates that volatiles have been expelled from the moon's surface due to the loss of gas molecules that occurred as a result of the heat. As a result, the absence of volatiles is the best evidence that the Moon has been extensively heated.

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Answer:

Yes

Explanation:

Sulfur and Sodium will have the same number of electron shells.

  Electronic configuration of

           k   l  m

  Na = 2  8  1

   S =  2  8  6

We see that the have the same energy level.

Also, they belong to the same period on the periodic table. Elements in the same period will have equal number of shell.