1. How is energy being used inefficiently?

2.What should you do to make it efficient?
pls answer thank you :)

In the balanced equation \(2C_{2} H_{6}\) + \(7 O_{2}\) --> \(4 CO_{2}\) + \(6H_{2}O\) if 21 g of \(C_{2} H_{6}\) reacts with 32 g O₂, C₂H6 is the limiting reactant.

To determine the limiting reactant, we need to compare the amount of each reactant to the stoichiometric ratio in the balanced equation.

Let's calculate the number of moles for each reactant using their molar masses:

For \(C_{2} H_{6}\) (ethane):

Molar mass of \(C_{2} H_{6}\) = 2(12.01 g/mol) + 6(1.01 g/mol) = 30.07 g/mol

Number of moles of C₂H6 = 21 g / 30.07 g/mol ≈ 0.698 mol

For O₂ (oxygen):

Molar mass of O₂ = 2(16.00 g/mol) = 32.00 g/mol

Number of moles of O₂ = 32 g / 32.00 g/mol = 1.00 mol

Next, we compare the moles of each reactant to the stoichiometric ratio in the balanced equation:

2 moles of \(C_{2} H_{6}\) react with 7 moles of O₂ to produce 4 moles of CO₂ and 6 moles of H₂O.

From the given amounts, we have:

0.698 mol \(C_{2} H_{6}\) and 1.00 mol O₂.

Using the stoichiometric ratio, we can calculate the expected amount of CO₂ and H₂O produced for each reactant:

For C₂H6:

Expected moles of CO₂ = 0.698 mol C₂H6 * (4 mol CO₂ / 2 mol C₂H6) = 1.396 mol CO₂

For O₂:

Expected moles of CO₂ = 1.00 mol O₂ * (4 mol CO₂ / 7 mol O₂) ≈ 0.571 mol CO₂

Comparing the expected moles, we see that the calculated amount of CO₂ is greater when used \(C_{2} H_{6}\) as the limiting reactant. Therefore, the limiting reactant in this reaction is \(C_{2} H_{6}\).

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Answer: Coefficients

Explanation: The coefficients in a balanced chemical equation tell you how many moles of reactants you'll require and how many moles of product you'll be able to make.

Answer: The coefficient in front of the molecule tells it’s relative number of moles.

Explanation: Quiz

Answer:

After the 4s sublevel is filled with two electrons, ten electrons fill the 3d sublevel, which is slightly higher in energy than the 4s sublevel. After the 3d sublevel is filled, the remaining three electrons will be in orbitals in the 4p sublevel. Thus, the electron configuration for arsenic is [Ar] 4s(2 squared) 3d(10 squared) 4p(3 squared).

The structure for 2-methyl-3-propalhex-2-yne can be shown in the image attached.

We know that a compound is composed of atoms that can be found in the compound. For the organic compound, we can see that we can be able to obtain the structure of the compound from the structure.

The compound as we can see is  2-methyl-3-propalhex-2-yne. The structure of the compound must be able to include a double bond as we can clearly see from the image that is attached to this answer.

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Answer:

Reference point, or frame of reference, a system of geometric axes in relation to which measurements of size, position, or motion can be made. Reference point, a geometrical point used to define the location of another point

Explanation:

Answer:

\(m_{iron}=32.1g\)

Explanation:

Hello!

In this case, since the interaction between hot iron and cold water allows the heat transfer from iron to water and therefore we can write up the following energetic equation:

\(Q_{iron}+Q_{water}=0\)

Whereas the heat terms can be written in terms of mass, specific heat and temperature change:

\(m_{iron}C_{iron}(T_f-T_{iron}) + m_{water}C_{water}(T_f-T_{water}) = 0\)

So we solve for the mass of iron as follows:

\(m_{iron} = \frac{m_{water}C_{water}(T_f-T_{water})}{C_{iron}(T_f-T_{iron}) }\)

Now, we plug in the given data to obtain:

\(m_{iron} = \frac{-100g*0.449\frac{J}{g\°C} (25.7\°C-22.0\°C)}{0.449\frac{J}{g\°C} (25.7\°C-93.3\°C) }\)

\(m_{iron}=32.1g\)

Best regards!

Answer:

the answer is B:

Explanation:

I hope it help.

The atomic greatest radius is magnisium

The balanced equation of the redox reactions by the half-reaction method is as follows:

(a) Zn(s) + 4 H+(aq) + NO₃⁻(aq) ⟶ Zn²⁺ (aq) + 2 H₂O(l) + N₂(g)

(b) Zn(s) + 2 OH⁻(aq) + NO₃⁻(aq) ⟶ Zn(OH)₂(aq) + NH₃(aq)

(c) CuS(s) + 6 H⁺(aq) + 2 NO₃⁻(aq) ⟶ Cu²⁺(aq) + S(s) + 2 NO(g) + 3 H₂O(l)

(d) 4 NH₃(aq) + 5 O₂(g) ⟶ 4 NO₂(g) + 6 H₂O(l)

(e) 2 H₂O₂(aq) + 2 MnO₄⁻(aq) ⟶ 2 Mn²⁺(aq) + 5 O₂(g) + 4 H₂O(l)

(f) 3 NO₂ (g) + 2 OH⁻ (aq) ⟶ 3 NO₃⁻ (aq) + NO₂⁻ (aq) + H₂O(l)

(g) 2 Fe³⁺ (aq) + 2 I⁻ (aq) ⟶ 2 Fe²⁺ (aq) + I₂ (aq)

Half reactions refer to the separate reactions that represent the oxidation and reduction processes occurring in the overall redox reaction.

Considering the given redox reactions:

(a) Zn(s) + 4 H+(aq) + NO₃⁻(aq) ⟶ Zn²⁺ (aq) + 2 H₂O(l) + N₂(g)

Half-reaction (oxidation): Zn(s) ⟶ Zn²⁺(aq) + 2 e⁻

Half-reaction (reduction): 4 H+(aq) + NO₃⁻(aq) + 3 e⁻ ⟶ 2 H₂O(l) + N₂(g)

(b) Zn(s) + 2 OH⁻(aq) + NO₃⁻(aq) ⟶ Zn(OH)₂(aq) + NH₃(aq)

Half-reaction (oxidation): Zn(s) + 4 OH⁻(aq) ⟶ Zn(OH)₂(aq) + 2 e⁻

Half-reaction (reduction): NO₃⁻ aq) + 8 H₂O(l) + 6 e⁻ ⟶ NH₃(aq) + 9 OH⁻ (aq)

(c) CuS(s) + 6 H⁺(aq) + 2 NO₃⁻(aq) ⟶ Cu²⁺(aq) + S(s) + 2 NO(g) + 3 H₂O(l)

Half-reaction (oxidation): CuS(s) ⟶ Cu²(aq) + S(s) + 2 e⁻

Half-reaction (reduction): 6 H⁺(aq) + 2 NO₃⁻(aq) + 6 e⁻ ⟶ 2 NO(g) + 3 H₂O(l)

(d) 4 NH₃(aq) + 5 O₂(g) ⟶ 4 NO₂(g) + 6 H₂O(l)

Half-reaction (oxidation): 4 NH₃(aq) ⟶ 4 NO₂(g) + 8 H⁺(aq) + 8 e⁻

Half-reaction (reduction): 5 O₂(g) + 10 H₂O(l) + 10 e⁻ ⟶ 20 OH⁻(aq)

(e) 2 H₂O₂(aq) + 2 MnO₄⁻(aq) ⟶ 2 Mn²⁺(aq) + 5 O₂(g) + 4 H₂O(l)

Half-reaction (oxidation): 2 H₂O₂(aq) ⟶ 4 H⁺(aq) + 4 e⁻ + O₂(g)

Half-reaction (reduction): 2 MnO₄⁻(aq) + 16 H⁺ (aq) + 10 e⁻ ⟶ 2 Mn²⁺ (aq) + 8 H₂O (l)

(f) 3 NO₂ (g) + 2 OH⁻ (aq) ⟶ 3 NO₃⁻ (aq) + NO₂⁻ (aq) + H₂O(l)

Half-reaction (oxidation): 3 NO₂(g) + 6 OH⁻(aq) ⟶ 3 NO₃⁻ (aq) + 3 e⁻ + 3 H₂O(l)

Half-reaction (reduction): 3 NO₂ (g) + 2 e⁻ ⟶ 3 NO₂⁻ (aq)

(g) 2 Fe³⁺ (aq) + 2 I⁻ (aq) ⟶ 2 Fe²⁺ (aq) + I₂ (aq)

Half-reaction (oxidation): 2 Fe³⁺ (aq) ⟶ 2 Fe²⁺ (aq) + 2 e⁻

Half-reaction (reduction): 2 I⁻ (aq) ⟶ I₂ (aq) + 2 e⁻

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According to the Foliated Metamorphic Rock Chart slate, phyllite, schist, and gneiss can all have the same parent rock (shale) is a true statement.

Slate, phyllite, schist, and gneiss are all types of metamorphic rocks that can be formed from shale, which is a sedimentary rock composed of clay and other fine-grained minerals. The specific type of metamorphic rock that is formed from shale depends on the conditions under which the shale undergoes metamorphism, including the temperature, pressure, and presence of fluids.

Slate is a fine-grained metamorphic rock with a uniform, flat surface and a layered structure. It is formed when shale undergoes low-grade metamorphism, which occurs at relatively low temperatures and pressures.

Therefore, Gneiss, on the other hand, is a medium- to coarse-grained metamorphic rock with a banded or wavy texture. It is formed when shale undergoes high-grade metamorphism, which occurs at higher temperatures and pressures.

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While tumor marker test results can be useful, they are not conclusive. A low result does not imply that you do not have cancer or that you are in remission.

CA 15-3, for example, is raised in less than half of patients with early breast cancer and in more than 80% of those with metastatic breast cancer. Three tumor markers, cancer antigen 15-3 (CA 15-3), cancer antigen 27.29 (CA 27.29), and carcinoembryonic antigen (CEA), have been used in breast cancer care to help monitor metastatic breast cancer (advanced disease), but they have not been found to be useful in detecting a breast cancer recurrence or extending lives. If the level falls, the treatment is effective. If it rises, the cancer may be spreading.

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Answer:

Chemistry teacher

Pharmaceuticals

Biochem/biotech

Materials science

Food chemist

Explanation:

Chemistry teaching

Teaching jobs in chemistry can range from teaching in public middle and high schools (bachelor’s degree required) to junior or community college (master’s degree required) to the university level (doctorate required). At the university level, you get to do research along with your teaching responsibilities.

Pharmaceutical/chemical sales

If you like chemistry but don’t enjoy being in the lab, then sales may be the job for you. Somebody has to sell all those chemicals and medicines, and buyers like to deal with somebody who knows their science.

Biochem/biotech

Biochemists and biotechnologists work in research developing new genetic tests, work in the genetic engineering (cloning) area, and are involved in the development of new drugs.

Materials science

Material scientists study the composition and structure of various materials with the goal of developing new products or improving existing ones.

Food chemist

Food and flavor chemists work in the research and development of new foods or keeping food fresh. Chemists test food additives and preservatives, develop new flavors, and analyze food for nutrient levels or the presence of contaminants.

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The value of ΔG at 25 °C for the given reaction is: ΔG = -370.4 kJ/mol + 0 = -370.4 kJ/mol So, the correct answer is -370.4 kJ/mol

To determine the value of ΔG (Gibbs free energy) at 25 °C for the given reaction:

25 (s, rhombic) + 3/2 \(O_2\)(g) → \(2SO_3\)(g)

We can use the equation:

ΔG = ΔG° + RT ln(Q)

where:

ΔG is the standard Gibbs free energy change

ΔG° is the standard Gibbs free energy change under standard conditions

R is the gas constant (8.314 J/(mol·K) or 0.008314 kJ/(mol·K))

T is the temperature in Kelvin (25 °C = 298 K)

Q is the reaction quotient, which is the ratio of the concentrations of the products to the concentrations of the reactants at a given point during the reaction.

Given that ΔG° is -370.4 kJ/mol, we can plug the values into the equation:

ΔG = -370.4 kJ/mol + (0.008314 kJ/(mol·K) * 298 K) * ln(Q)

Now, we need to determine the value of Q. Since all reactants and products are in their standard states, Q = 1, as their concentrations are taken to be 1.

ΔG = -370.4 kJ/mol + (0.008314 kJ/(mol·K) * 298 K) * ln(1)

Since ln(1) = 0, the term (0.008314 kJ/(mol·K) * 298 K) * ln(1) becomes 0.

Therefore, the value of ΔG at 25 °C for the given reaction is:

ΔG = -370.4 kJ/mol + 0 = -370.4 kJ/mol

So, the correct answer is -370.4 kJ/mol.

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The three main components of electrolytic cells are cathode, anode and electrolyte. The negatively charged electrolytic cells are cathode and the positively charged electrolytic cells are anode.

An electrolytic cell can be defined as the electrochemical device which uses the electrical energy to perform a non-spontaneous redox reaction. They are mainly used for the electrolysis of certain compounds.

Here the anode cell is:

3H₂O (l) + I⁻ (aq) → IO⁻₃ (aq) + 6e⁻ + 6H⁺ (aq)

cathode cell is:

14H⁺ (aq) + Cr₂O₇²⁻ (aq) + 6e⁻ → 2Cr³⁺ (aq) + 7H₂O (l)

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Answer:

Homogenous

Explanation:

Mixture of copper and tin

The volume the balloon will occupy at an altitude where the pressure is 0.600 atm and the temperature is -20.00 C is 6.15 L.

The ideal gas law can be used to solve the given problem, which is as follows:P1V1/T1 = P2V2/T2 Where:P1 is the initial pressure (1.00 atm)P2 is the final pressure (0.600 atm)V1 is the initial volume (4.50 L)V2 is the final volume (what we want to find)T1 is the initial temperature (20.00 C = 293.15 K)T2 is the final temperature (-20.00 C = 253.15 K)The problem does not provide the mass or number of moles of helium gas, but these quantities are not required to solve the problem as they cancel out in the calculation.The problem also does not mention any change in altitude, so we can assume that the balloon rises to an altitude where the pressure and temperature are lower.We can rearrange the equation above to solve for V2, which gives:V2 = (P1V1T2)/(P2T1) Substituting the given values gives:V2 = (1.00 atm × 4.50 L × 253.15 K)/(0.600 atm × 293.15 K)V2 = 6.15 L.

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Answer:

0.45688mL

Explanation:

1) First you should look on the coefficients.
\(2C_8H_{18(l)}\ + 25O_2_{(g)}=\ 16CO_{2(g)}+ 18H_2O_{(g)}\)

2) Do the ratio between C8H18 and O2:

\(C_8H_{18}\ \ \ \ \ \ \ \ \ O_2\\2 : \ \ \ \ \ \ \ \ \ \ \ \ \ 25\\X: \ \ \ \ \ \ \ \ \ \ \ 0.0500moles\\2X0.0500/25= 0.004\\X=0.004 \moles\ of\ C_8H_{18}\)

3) grams= moles X molecular mass

\(0.004*114.22= 0.45688g\)

[1g=1mL] => 0.45688mL

The volume of oxygen gas produced by the decomposition of 3.05 g KCIO₃ according to the given equation is 1.68 liters

Let us begin by obtaining the mole of 3.05 g of KClO₃. Details below:

Mole = mass / molar mass

Mole of KClO₃ = 3.05 / 122.5

Mole of KClO₃ = 0.025 mole

Next, we shall determine the mole of of oxygen gas, O₂ produced from the reaction. Details below:

KClO₃ -> 2KCl + 3O₂

From the balanced equation above,

1 mole of KClO₃ reacted to produced 3 moles of O₂

Therefore,

0.025 mole of KClO₃ will react to produce = 0.025 × 3 = 0.075 mole of O₂

Finally, we shall obtain the volume of oxygen gas, O₂ produced. This is shown below

At STP,

1 mole of O₂ = 22.4 Liters

Therefore,

0.075 moles of O₂ = 0.075 × 22.4

0.075 moles of O₂ = 1.68 liters

Thus, the volume of oxygen gas, O₂ produced is 1.68 liters

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The asbestos exposure during the years 1982 and 1983 was 2.5 fibers per cubic centimeter and 0.8 fibers per cubic centimeter respectively. So asbestos exposure decreased by 1.7 fibers per cubic centimeter during the year 1983.

Breathlessness Persistent, dry cough Chest pain or tightness Lack of a dry, crackling sound in the lungs when you breathe in Wider and rounder fingers and toes are some of the symptoms of asbestos exposure.

The largest group of people exposed to asbestos is those working in the construction industry. Historically, asbestos was also used by pipe fitters and shipyard workers. In addition, asbestos was used by military personnel, auto mechanics, and many other occupations.

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Answer:

a. pH = 2.52

b. pH = 8.67

c. pH = 12.83

Explanation:

The equation of the titration between the benzoic acid and NaOH is:

C₆H₅CO₂H + OH⁻ ⇄ C₆H₅CO₂⁻ + H₂O    (1)

a. To find the pH after the addition of 20.0 mL of NaOH we need to find the number of moles of C₆H₅CO₂H and NaOH:

\( \eta_{NaOH} = C*V = 0.250 M*0.020 L = 5.00 \cdot 10^{-3} moles \)

\( \eta_{C_{6}H_{5}CO_{2}H}i = C*V = 0.300 M*0.050 L = 0.015 moles \)  

From the reaction between the benzoic acid and NaOH we have the following number of moles of benzoic acid remaining:

\(\eta_{C_{6}H_{5}CO_{2}H} = \eta_{C_{6}H_{5}CO_{2}H}i - \eta_{NaOH} = 0.015 moles - 5.00 \cdot 10^{-3} moles = 0.01 moles\)

The concentration of benzoic acid is:

\( C = \frac{\eta}{V} = \frac{0.01 moles}{(0.020 + 0.050) L} = 0.14 M \)

Now, from the dissociation equilibrium of benzoic acid we have:

C₆H₅CO₂H + H₂O ⇄ C₆H₅CO₂⁻ + H₃O⁺  

0.14 - x                            x                x

\( Ka = \frac{[C_{6}H_{5}CO_{2}^{-}][H_{3}O^{+}]}{[C_{6}H_{5}CO_{2}H]} \)

\( Ka = \frac{x*x}{0.14 - x} \)

\( 6.5 \cdot 10^{-5}*(0.14 - x) - x^{2} = 0 \)   (2)  

By solving equation (2) for x we have:          

x = 0.0030 = [C₆H₅CO₂⁻] = [H₃O⁺]

Finally, the pH is:

\( pH = -log([H_{3}O^{+}]) = -log (0.0030) = 2.52 \)

b. At the equivalence point, the benzoic acid has been converted to its conjugate base for the reaction with NaOH so, the equilibrium equation is:

C₆H₅CO₂⁻ + H₂O  ⇄  C₆H₅CO₂H + OH⁻     (3)

The number of moles of C₆H₅CO₂⁻ is:

\( \eta_{C_{6}H_{5}CO_{2}^{-}} = \eta_{C_{6}H_{5}CO_{2}H}i = 0.015 moles \)

The volume of NaOH added is:

\( V = \frac{\eta}{C} = \frac{0.015 moles}{0.250 M} = 0.060 L \)

The concentration of C₆H₅CO₂⁻ is:

\( C = \frac{\eta}{V} = \frac{0.015 moles}{(0.060 L + 0.050 L)} = 0.14 M \)

From the equilibrium of equation (3) we have:

C₆H₅CO₂⁻ + H₂O  ⇄  C₆H₅CO₂H + OH⁻  

0.14 - x                              x               x

\(Kb = \frac{[C_{6}H_{5}CO_{2}H][OH^{-}]}{[C_{6}H_{5}CO_{2}^{-}]}\)

\( (\frac{Kw}{Ka})*(0.14 - x) - x^{2} = 0 \)

\( (\frac{1.00 \cdot 10^{-14}}{6.5 \cdot 10^{-5}})*(0.14 - x) - x^{2} = 0 \)

By solving the equation above for x, we have:

x = 4.64x10⁻⁶ = [C₆H₅CO₂H] = [OH⁻]

The pH is:

\( pOH = -log[OH^{-}] = -log(4.64 \cdot 10^{-6}) = 5.33 \)

\( pH = 14 - pOH = 14 - 5.33 = 8.67 \)

c. To find the pH after the addition of 100 mL of NaOH we need to find the number of moles of NaOH:

\( \eta_{NaOH}i = C*V = 0.250 M*0.100 L = 0.025 moles \)

From the reaction between the benzoic acid and NaOH we have the following number of moles remaining:                              

\( \eta_{NaOH} = \eta_{NaOH}i - \eta_{C_{6}H_{5}CO_{2}H} = 0.025 moles - 0.015 moles = 0.010 moles \)                          

The concentration of NaOH is:

\( C = \frac{\eta}{V} = \frac{0.010 moles}{0.100 L + 0.050 L} = 0.067 M \)

Therefore, the pH is given by this excess of NaOH:         

\( pOH = -log([OH^{-}]) = -log(0.067) = 1.17 \)

\( pH = 14 - pOH = 12.83 \)

I hope it helps you!    

Answer:

the answer i believe is increased nutritional value of crops

Explanation:

Answer: increased nutritional value of crops

Explanation:) edge 2022

The total pressure in a mixture of gases is equal to the partial pressure of all the gases added together

According to Dalton's Law of Partial Pressure,  The total pressure exerted by a mixture of gases which are non reacting is equal to the sum of the partial pressure of each individual gas that make up the mixture

Mathematically, Ut can be written as

P{total}= p1+p2, ....+ pn

where p1, p2, …, pn are partial pressures of each gas component.

For example :A gaseous mixture consisting of Helium and Nitrogen gas . If the pressure exerted by He is 3 atm while Nitrogeb exerted is 9 atm. The total pressure will be:

P{total}= p1+p2, ....+ pn

P{total}= 3+9

P{total}= 12

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To determine the empirical formula of the unknown sample, we need to use the given composition and the molar masses of the elements to find the number of moles of each element in the sample. Then, we can divide the number of moles of each element by the smallest number of moles to get the simplest, whole-number ratio of the elements in the compound.

First, we can calculate the number of moles of carbon in the sample:

mass of carbon = 0.3751 x 5.91 g = 2.216 g

moles of carbon = 2.216 g / 12.01 g/mol = 0.1845 mol

Next, we can calculate the number of moles of hydrogen and oxygen in the sample using Avogadro's number:

moles of hydrogen = 1.4830 x 10^23 atoms / 6.022 x 10^23 mol^-1 = 0.2461 mol

moles of oxygen = 1.2966 x 10^23 atoms / 6.022 x 10^23 mol^-1 = 0.2154 mol

Now, we can divide each number of moles by the smallest number of moles, which is 0.1845:

0.1845 mol C / 0.1845 mol = 1.000

0.2461 mol H / 0.1845 mol = 1.333

0.2154 mol O / 0.1845 mol = 1.167

These ratios are not whole numbers, so we need to multiply each ratio by a common factor to get the simplest, whole-number ratio. The simplest ratio is obtained by dividing each ratio by the smallest ratio:

1.000 / 1.000 = 1

1.333 / 1.000 = 4/3

1.167 / 1.000 = 7/6

We can then multiply each ratio by 6 to obtain whole numbers:

6 x 1 = 6

6 x 4/3 = 8

6 x 7/6 = 7

Therefore, the empirical formula of the unknown sample is C6H8O7.

Answer:

fe9ufeohdwbkdwsdjvdwihdwkbfw

The parts of the neuron include (A) dendrites.

As it is given in the description, Synapses are also found on dendrites.

Therefore, the parts of the neuron include (A) dendrites.

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Combining "hydro" and "genes" in the context of the Latin roots, we can interpret it as "water origin" or "water-related origin."

The term "hydro" in Latin means "water." It is derived from the Greek word "hydor," which also signifies water. "Hydro" is commonly used as a prefix in scientific terms related to water or hydrogen. In the context of chemistry, "hydro" often denotes a compound or process involving water or hydrogen atoms. For example, hydrocarbon refers to organic compounds composed solely of hydrogen and carbon atoms.

On the other hand, "genes" in Latin is derived from the Greek word "genos," which translates to "origin" or "birth." In biology, the term "genes" refers to the segments of DNA that contain instructions for the development and functioning of living organisms. Genes determine various traits, such as physical characteristics, behavior, and susceptibility to certain diseases. They play a fundamental role in the transmission of hereditary information from one generation to another.

This interpretation aligns with the understanding that hydrogen, which is essential for life and abundant in water, plays a vital role in various biological and chemical processes.

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Explanation:

The electron density number as well as the sublevel letter are used to describe valence electrons in an atom. The third total energy and subbasement p, for example, is denoted by 3p. The electron configuration of oxygen, for example, is 1s^2 2s^2 2p^4, which means the first two electrons will couple up in the 1s orbital, while the following two protons will pair up in the 2s orbital.

The sample atom is Carbon with electron configuration; 1s² 2s² 2p².

The principal energy level of an electron refers to the shellp in which the electron is located relative to the atom's nucleus. In this case only 2 energy levels exist in a carbon atom; which are energy level 1 and 2

The sublevels exist within a principal energy and the electron configuration of an atom is described with consideration of energy sublevels. The sublevels in a carbon atom are;

The orbitals in this configuration are: 1s 2s 2px 2py 2pz in which case; each orbital can accommodate 2 electrons each.

Ultimately, the location of an element on the periodic table with respect to group and period are used to determine the valency and no. of energy levels in the atom of that Element.

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