Many objects that conduct electricity or heat are made from this material
A. Metals
B. Wood
C. Plastics
D. Cotton

Here is an example

Explanation:

From n = 5 , the electron can drop to n =

4, 3, 2, 1. Thus, there will be 4

possibilities

Now, the electron that dropped to n = 4 can drop to n =3,2,1 This will give 3 possibilities

The electron at n = 3 can drop to n =2,1 This gives 2 possibilities

Finally, the electron at n = 2 can drop to n = 1 . This gives 1 possibility

Thus in total the number of ways in which the electron can de-excite is :

4 + 3 + 2 + 1 = 10

10 different wavelengths of light can be emitted as the electron loses energy

from n = 5 level

Answer:

Number 3

Explanation:

Answer:

Chlorine is more likely to steal a valence electron from sodium.

Explanation:

Sodium is number 11 on the periodic table with one valence electron. Belonging to the first group, it's one of the alkali metal, which are known to be highly reactive. Chlorine is number 17 with seven valence electrons, and it's in the second-to-last group of halogens--also very reactive.

Considering that elements with one valence electron are just about 100% likely to give up electrons to reach a stable state, sodium would be the element that is more likely to lose its valence electron to chlorine. In other words, chlorine would be the electron thief.

The statement that is true about alloy is that; they can only be composed of metals.

An alloy is a substance that is obtained by the combination of two or more metals. Alloys are produced owing to the fact that they may be stronger or more durable or resistant to corrosion than the individual metals.

There are many allows that find very good use such as the alloy of copper and zinc which we call bronze that could be used as an ornamental. The statement that is true about alloy is that; they can only be composed of metals.

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Answer:

Explanation:

Using the periodic table, complete the table to describe each atom. Type in your answers. - 14514242.

Answer:

then it would still be 101.3

Explanation: because if it's constant that means it remained still

Answer:

1 = c = 0.79 j/g.K

2= c = 1 j/g.K

3 = T2 = 123.1 K

Explanation:

Given data:

Specific heat of metal = ?

Mass of metal = 450 g

Heat loses = 34,500 j

Temperature drop = ΔT  = 97 K

Solution:

Specific heat capacity:

It is the amount of heat required to raise the temperature of one gram of substance by one degree.

Formula:

Q = m.c. ΔT

Q = amount of heat absorbed or released

m = mass of given substance

c = specific heat capacity of substance

ΔT = change in temperature

34500 j = 450 g ×c × 97 K

34500 j = 43650 g.K ×c

c = 34500 j/43650 g.K

c = 0.79 j/g.K

2)

Given data:

Specific heat of metal = ?

Mass of metal = 25 g

Heat loses = 2600 j

Temperature drop = ΔT  = 103 K

Solution:

Specific heat capacity:

It is the amount of heat required to raise the temperature of one gram of substance by one degree.

Formula:

Q = m.c. ΔT

Q = amount of heat absorbed or released

m = mass of given substance

c = specific heat capacity of substance

ΔT = change in temperature

2600 j = 25 g ×c × 103 K

2600  j = 2575 g.K ×c

c = 2600 j/2575 g.K

c = 1 j/g.K

3)

Given data:

Mass of water = 10 g

Heat absorbed  = 840 j

Initial temperature drop = 298 K

Final temperature = ?

Solution:

Formula:

Q = m.c. ΔT

Q = amount of heat absorbed or released

m = mass of given substance

c = specific heat capacity of substance

ΔT = change in temperature

ΔT = T2 - T1

840 j = 10 g ×4.18 j/g.K × [T2 - 103 K]

840 j = 41.8 J/K ×  [T2 - 103 K]

840 J/41.8 J/ K = T2 - 103 K

20.1 k = T2 - 103 K

T2 = 20.1 K + 103 K

T2 = 123.1 K

U has a total of six electrons. This corresponds to carbon (C). A is the second most common element in the atmosphere.

The second most common element in the atmosphere is oxygen (O). E is a noble gas.

Noble gases include helium (He), neon (Ne), argon (Ar), krypton (Kr), xenon (Xe), and radon (Rn). Based on the given options, E could be xenon (Xe).

S is an alkali metal.

Alkali metals include lithium (Li), sodium (Na), potassium (K), rubidium (Rb), cesium (Cs), and francium (Fr). Based on the given options, S could be sodium (Na).

O is a halogen.

Halogens include fluorine (F), chlorine (Cl), bromine (Br), iodine (I), and astatine (At). Based on the given options, O could be bromine (Br).

O has an atomic number larger than V but smaller than W.

Based on the periodic table, the atomic number of oxygen (O) is 8, which is larger than the atomic number of vanadium (V) (23) and smaller than the atomic number of tungsten (W) (74).

The charge on an L ion is +2.

The charge of +2 indicates that L must lose two electrons to form the ion. Based on the given options, L could be calcium (Ca).

C has five electrons in its outer energy fever.

Carbon (C) has four valence electrons, not five. This contradicts the given statement, so we need to revisit the deductions.

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Valence electrons affects ion formation by transferring from one atom to another atom.

Valence electrons are the outermost electrons in an atom and are involved in the formation of ions. Ions are atoms or molecules that have a net electric charge due to a surplus or deficiency of electrons. When an atom forms an ion, it either gains or loses electrons to achieve a more stable electron configuration. The number of valence electrons in an atom determines how easily it can gain or lose electrons to form ions. Atoms with few valence electrons, such as alkali metals, tend to lose those electrons easily to form positive ions. Atoms with many valence electrons, such as halogens, tend to gain electrons easily to form negative ions. The number of valence electrons also affects the charge of the ions that are formed. Atoms with fewer valence electrons will form ions with a higher charge, while atoms with more valence electrons will form ions with a lower charge.

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The mutated amino acid at position 229 is arginine (Arg), and its substitution can potentially disrupt the 3D structure and alter the function of the protein due to changes in side chain properties and interactions.

Structure of the Amino Acids:

The wild-type amino acid at position 229 is not specified, so the structure cannot be provided.

The mutated amino acid is arginine (Arg), which has a side chain containing a positively charged guanidinium group.

Importance of the Position:

The specific position 229 in the protein sequence may be functionally significant, such as being involved in protein-protein interactions, binding sites, catalytic activity, or structural stability.

Without detailed knowledge of the protein, its function, and its structural context, it is difficult to determine the exact importance of this specific position.

Effects of the Mutation on Structure and Function:

The substitution of an amino acid at position 229 from the original to arginine can have various effects on protein structure and function.

Arginine's larger and positively charged side chain may introduce steric clashes or alter electrostatic interactions within the protein structure.

The mutation can potentially disrupt local or global protein folding, stability, or conformational changes, affecting its overall 3D structure.

The functional consequences of the mutation depend on the specific role of the amino acid at position 229, which can include changes in protein-protein interactions, enzymatic activity, substrate binding, or signal transduction pathways.

It is crucial to analyze the protein's structural context, available experimental data, and computational modeling techniques to gain a more accurate understanding of the specific effects of the mutation on the protein's structure and function in the given context.

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The relationship between C2 and Q2 is dependent on the specific values of C1, ΔV, and the desired total charge in the circuit.

To analyze the relationship between C2 and Q2, we need to consider the basic equation of a capacitor, which states that Q = C × ΔV, where Q is the charge stored in the capacitor, C is the capacitance, and ΔV is the voltage across the capacitor.

1. Initially, with both capacitors (C1 and C2) in the circuit, let's assume the charge on C1 is Q1 and the charge on C2 is Q2.

2. Since Kylie removed C2, the only capacitor remaining in the circuit is C1, so the total charge in the circuit remains the same, Q1 + Q2.

3. Since C1 and ΔV remain constant, Q1 also remains constant.

4. Therefore, as C2 varies, Q2 must vary to compensate for the changes in C2, while still maintaining the sum of charges as Q1 + Q2.

In summary, as Kylie varies C2 while keeping C1 and ΔV constant, the charge Q2 on C2 will change to maintain the total charge in the circuit, but the exact relationship between C2 and Q2 cannot be determined without further information.

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The amide produced from pentanoic acid and ammonia is pentanamide. To form this amide, pentanoic acid reacts with ammonia in a process called amidation.

Here's a step-by-step explanation:

1. Pentanoic acid (C4H9COOH) has a carboxyl group (-COOH) at one end.

2. Ammonia (NH3) has a lone pair of electrons on the nitrogen atom, making it a nucleophile.

3. The nitrogen atom in ammonia attacks the carbonyl carbon atom in pentanoic acid, forming a tetrahedral intermediate.

4. The intermediate collapses, expelling a hydroxyl group (-OH) from the carboxyl group and forming a new bond between the nitrogen and carbonyl carbon atoms.

5. The product is pentanamide (C4H9CONH2), an amide formed from pentanoic acid and ammonia.

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d) The concentration of carbonic acid will increase if the pressure of carbon dioxide is increased in the container.

According to Le Chatelier's principle, if the pressure of carbon dioxide is increased in the container, the system will shift in a direction that reduces the pressure. In this case, it means the equilibrium will shift towards the formation of more carbonic acid. The increase in pressure will cause more CO₂ molecules to dissolve in water, leading to an increase in the concentration of carbonic acid.

The reaction CO₂ + H₂O ⇌ H₂CO₃ is an equilibrium reaction, and changing the pressure affects the equilibrium position. By increasing the pressure of CO₂, the equilibrium will favor the formation of more carbonic acid to alleviate the increased pressure. Therefore, the concentration of carbonic acid will increase as a result of the increased CO₂ pressure. The other options (a, b, c, e) are not accurate as they do not take into account the impact of the increased CO₂ pressure on the equilibrium.

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Therefore, the estimated boiling point of the sugar solution is 103.00°C.

The boiling point of a solution is elevated due to the presence of solute particles, which lowers the vapor pressure of the solution. We can estimate the boiling point of the sugar solution using the following equation:

ΔTb = kb x molality

where ΔTb is the boiling point elevation, kb is the boiling point elevation constant of water, and molality is the concentration of the solution in moles of solute per kilogram of solvent.

First, we need to calculate the molality of the sugar solution:

mass of sucrose = 2000 g

moles of sucrose = mass / molar mass = 2000 g / 342 g/mol = 5.85 mol

mass of water = 1000 g

molality = moles of sucrose / mass of water (in kg) = 5.85 mol / 1 kg = 5.85 m

Now we can calculate the boiling point elevation:

ΔTb = kb x molality = 0.514 °C/mol kg x 5.85 m = 3.00 °C

The boiling point of the sugar solution is therefore:

boiling point = normal boiling point of water + ΔTb

boiling point = 100°C + 3.00°C = 103.00°C

Therefore, the estimated boiling point of the sugar solution is 103.00°C.

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Answer:

La masa molar de la sustancia disuelta es de aproximadamente 138,473 g/mol

Explanation:

Los parámetros dados son;

La masa de benceno en el recipiente = 100 gramos

La masa de benceno que se evapora del benceno puro = 1.273 gramos

La masa del soluto no volátil disuelto en 100 gramos de benceno = 10 gramos

La masa de benceno que se evapora de la mezcla = 1.205 gramos

Según la ley de Raoult, tenemos;

\(P_ {solucion} = P_ {solvente} \times X _ {solvente}\)

Dónde;

\(P_ {solucion}\) = La presión de vapor de la solución.

\(P_ {solvente}\) = La presión de vapor del solvente

\(X_ {solvente}\) = La fracción molar del solvente

Por lo tanto;

\(X_ {solvente} = \dfrac {P_ {solucion}} {P_ {solvente}}\)

Según la ley de los gases ideales, tenemos;

\(P \times V = n \times R \times T = \dfrac{m}{M} \times R \times T\)

Dónde;

P = La presión del gas

V = El volumen ocupado por el gas

m = La masa del gas

M = La masa molar del gas

R = La constante universal de los gases

T = La temperatura del gas

n = El número de moles del gas

Por lo tanto;

\(P = \dfrac{m \times R \times T}{M \times V}\)

Donde R, T, M y V son constantes, tenemos P ∝ m

\(\therefore X_ {solvente} = \dfrac {P_ {solucion}} {P_ {solvente}} = \dfrac {Masa \ evaporada \ de \ solucion} {Masa \ evaporada \ de \ solvente} = \dfrac {1.205 \ g} {1.273 \ g} \approx 0.9466\)

\(X_ {solvente}\) ≈ 0.9466

\(X_ {solvente} = \dfrac {El \ numero \ de \ moles \ del \ solvente} {El \ total \ numero \ de \ moles \ de \ la \ solucion}\)

El número de moles del solvente, C₆H₆ = (Masa de C₆H₆) / (La masa molar de C₆H₆) = 100 g / (78.11 g.mol) ≈ 1.28025 moles

El número de moles del solvente, C₆H₆ ≈ 1.28025 moles

El número total de moles = El número de moles del solvente, C₆H₆ + El número de moles del soluto no volátil, \(n_ {soluto}\)

\(\therefore X_ {solvente} = 0.9466 = \dfrac {1.28025 \ moles} {1.28025 \ moles + n_ {soluto}}\)

0.9466 × (1.28025 moles + \(n_ {soluto}\)) = 1.28025 moles

1.21189 moles + 0.9466 × \(n_ {soluto}\) = 1.28025 moles

\(n_ {soluto}\) = (1.28025 - 1.21189)/0.9466 moles ≈ 0.07221635 moles

\(\therefore \dfrac {Masa \ de \ soluto} {Molar \ masa \ de \ soluto} = \dfrac {10 \ g} {Molar \ masa \ de \ soluto} \approx 0.07221635 \ moles\)

\(\ Masa \ molar \ de \ soluto = \dfrac {10 \ g} {0.07221635 \ moles} \approx 138.473 \ g / moles\)

La masa molar de la sustancia disuelta (el soluto) ≈ 138,473 g/mol

Answer:

In Australia, it's warmer than most of the United States during winter.

Explanation:

Among the given options, the tests that are left-tailed are H0:μ≥18, Ha:μ<18, H0:μ≥11.3, Ha:μ<11.3, and H0:μ≥3.7, Ha:μ<3.7.

In these tests, the null hypothesis (H0) states that the population mean (μ) is greater than or equal to a specific value, while the alternative hypothesis (Ha) suggests that the population mean is less than that value.

A left-tailed test is used when the alternative hypothesis suggests that the population parameter is less than a certain value.

This indicates a left-tailed test, where the critical region is in the left tail of the distribution. These tests focus on detecting a significant decrease or difference in the population mean.

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Complete question :

Which of the hypothesis tests listed below is a left-tailed test? Select all correct answers.

Select all that apply: H0:μ≥18, Ha:μ<18 H0:μ≤19.3, Ha:μ>19.3 H0:μ=8, Ha:μ≠8 H0:μ≥11.3, Ha:μ<11.3 H0:μ≥3.7, Ha:μ<3.7

Answer: If you're looking for the type of bond, you need to find the difference in electromagnetically. The electromagnetically of Br is 2.96 and Si is 1.9. Subtracting the two will give you the difference of the previously mentioned (1.06).

For the bond to be POLAR the difference should fall between 0.5 and 1.6

For the bond to be NON-POLAR the difference should be 1.6

A liquid-gas system at equilibrium is disturbed by adding or removing vapor from the system (at constant temperature). The correct statements for the vapor pressure regarding this situation are A, B, and D.

A. Adding vapor will cause a temporary increase in vapor pressure: When the vapor is added to the system, the total vapor pressure increases, and the vapor pressure in the system is greater than the original equilibrium vapor pressure until the system re-equilibrates.

B. Adding or removing vapor will result in a new equilibrium vapor pressure: The equilibrium vapor pressure will be affected by the addition or removal of vapor. When the vapor is added or removed, the system must reach a new equilibrium between the vapor and liquid phases before the vapor pressure returns to the original equilibrium value.

D. Removing vapor will cause a temporary increase in the rate of condensation: When the vapor is removed from the system, the total vapor pressure decreases, and the rate of condensation of the liquid phase will increase until the system re-equilibrates.

Statement C. when equilibrium is re-established after a disturbance in a liquid-gas system, the vapor pressure will be the same: is incorrect. When a system is disturbed by adding or removing vapor, the new equilibrium vapor pressure is different from the original equilibrium vapor pressure.

Therefore, the correct statements for the vapor pressure of the system are A, B, and D.

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A characteristic of a material that can be seen or measured without altering its chemical makeup is said to have a physical property. The physical asset has the quality of being "malleable" among the possibilities.

The term "malleability" describes a material's capacity to take on diverse shapes without breaking. By pressing on the substance and monitoring how it responds, this attribute may be evaluated and measured.

Chemical properties refer to how a substance behaves in a chemical process, especially its ignition and burning potential. The option "does not burn" refers to this behavior. Option "becomes moldy rapidly" also refers to a substance's propensity to experience a biological transformation through decomposition and is therefore a chemical attribute. Option "easy to digest" refers to the ability of the substance to be broken down in the digestive system, making it a biological property rather than a physical one.

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Answer:

because it does not cost much and can be found easily.  due to their low resistance and excellent conductivity.

Explanation:

We will prepare the stock solution of 0.5 m CuSO₄  by using the formulae M1V1 = M2V2

1. Prepare 5 ml of 0.1 m

  Use the formulae M1V1 = M2V2

  Where M1 = 0.5 m

              M2 = 0.1 m

              V2 = 5 ml

              V1 = ?

  Put the values

              0.5 × V1 = 0.1 × 5

               V1 = 0.5 / 0.5

                V1 = 1 ml

1 ml of 0.5 m CuSO₄ solution is diluted up to 5 ml to get 0.1 m CuSO₄  solution

1. Prepare 5 ml of 0.1 m

  Use the formulae M1V1 = M2V2

  Where M1 = 0.5 m

              M2 = 0.1 m

              V2 = 5 ml

              V1 = ?

  Put the values

              0.5 × V1 = 0.1 × 5

               V1 = 0.5 / 0.5

                V1 = 1 ml

1 ml of 0.5 m CuSO₄ solution is diluted up to 5 ml to get 0.1 m CuSO₄  solution

2. Prepare 5 ml of 0.2 m

  Use the formulae M1V1 = M2V2

  Where M1 = 0.5 m

              M2 = 0.2 m

              V2 = 5 ml

              V1 = ?

  Put the values

              0.5 × V1 = 0.2 × 5

               V1 = 1 / 0.5

                V1 = 2 ml

2 ml of 0.5 m CuSO₄ solution is diluted up to 5 ml to get 0.2 m CuSO₄  solution

3. Prepare 5 ml of 0.3 m

  Use the formulae M1V1 = M2V2

  Where M1 = 0.5 m

              M2 = 0.3 m

              V2 = 5 ml

              V1 = ?

  Put the values

              0.5 × V1 = 0.3 × 5

               V1 = 1.5 / 0.5

                V1 = 3 ml

3 ml of 0.5 m CuSO₄ solution is diluted up to 5 ml to get 0.3 m CuSO₄  solution

4. Prepare 5 ml of 0.4 m

  Use the formulae M1V1 = M2V2

  Where M1 = 0.5 m

              M2 = 0.4 m

              V2 = 5 ml

              V1 = ?

  Put the values

              0.5 × V1 = 0.4 × 5

               V1 = 2 / 0.5

                V1 = 4 ml

4 ml of 0.5 m CuSO₄ solution is diluted up to 5 ml to get 0.4 m CuSO₄  solution

5. Prepare 5 ml of 0.5 m

   The stock solution is already a 0.5 m solution so no need for any addition.

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The molarity of the solution containing 6.7 grams of Li2O in 3500 ml was found to be 0.064 M.

To calculate the molarity of the solution, we need to first convert the volume of the solution from milliliters to liters. 3500 ml = 3.5 L.

Then, we need to calculate the number of moles of Li2O present in the solution. The molar mass of Li2O is 29.88 g/mol. So, 6.7 g of Li2O is equal to 0.224 moles.

Finally, we can calculate the molarity using the formula: Molarity = moles of solute / liters of solution.

Therefore, the molarity of the solution is 0.064 M (0.224 moles / 3.5 L).

The molarity of a solution is a measure of the number of moles of solute per liter of solution. To calculate the molarity of the given solution, we first converted the volume from milliliters to liters. Then, we determined the number of moles of Li2O present in the solution by using the molar mass of Li2O. Finally, we applied the formula to calculate the molarity of the solution. The molarity of the solution was found to be 0.064 M.

The molarity of a solution can be determined by calculating the number of moles of solute present in the solution and dividing it by the volume of the solution in liters. In this case, the molarity of the solution containing 6.7 grams of Li2O in 3500 ml was found to be 0.064 M.

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Answer:

9 protons

Explanation:

By looking at the periodic table, you will see that Fluorine has 9 protons. Since the number of electrons equal the number of protons, Fluorine has 9 electrons as well. Meanwhile, it's mass number of 19, minus 10 neutrons, gives you 9 protons or electrons. Hence, the atom would be Fluorine.

hope u make me brainlesst ʘ‿ʘ

Answer:

i think so for this i will 9 protons . atomic mass is 19. 10 neutrons

If 10.0 grams of ammonium chloride is added to 1000 mL of water, the temperature would drop by 20.9°C.

At normal temperature, aluminium and chlorine do not react quickly, but when heated, the reaction is ferocious. Aluminum chloride, a white powder that emits fumes into the atmosphere and reacts with water to produce aluminium oxide and HCl gas, is the end product.

q = m × c × ΔT

Since the water is the solvent in this case, we can assume that the specific heat capacity of the solution is the same as the specific heat capacity of water (4.184 J/g·°C).

m = 5.00 g

ΔT = -4.2°C

q = m × c × ΔT

q = (5.00 g) × (4.184 J/g·°C) × (-4.2°C)

q = -879.36 J

Now, we can use the amount of heat released by the 5.00 grams of ammonium chloride to find the temperature change for 10.0 grams of ammonium chloride in 1000 mL of water:

m = 10.0 g

V = 1000 mL = 1000 g (since 1 mL of water has a mass of 1 g)

q = -879.36 J (from the first part)

We can solve for ΔT:

q = m × c × ΔT

ΔT = q / (m × c)

ΔT = (-879.36 J) / (10.0 g × 4.184 J/g·°C)

ΔT = -20.9°C

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Answer:

it is salt

Explanation:

hope it helps

Answer:

ANY KENITIC PARTICLE

Particles in both liquids and gases collectively called fluids

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Answer:

the height of the hill is 30 feet tall