A 3.57 kg particle is located on the x-axis at xm = −8 m, and a 5.45 kg particle is on the x-axis at xM = 2.22 m. Find the coordinate of the center of mass of this two-particle system. Answer in units of m

Answers

Answer 1

Answer:

The coordinate of the center of mass of this two-particle system is (-1.82 m,0).

Explanation:

Center of mass for n mases of system:

[tex]C.O.M=\frac{m_1x_1+m_2x_2......m_nx_n}{m_1+m_2...m_n}[/tex]

We have :Two-particle system.

On x-axis ,mass of object m = [tex]m_x=-8m[/tex]

m = 3.57 kg

Mass of object M = [tex]x_M=2.22 m[/tex]

M = 5.45 kg

[tex]C.O.M=\frac{mx_m+Mx_M}{m+M}[/tex]

[tex]=\frac{3.57 kg\times (-8 m)+5.45kg\times 2.22}{3.57 kg+5.45 kg}[/tex]

[tex]=-1.82 m[/tex]

The coordinate of the center of mass of this two-particle system is (-1.82 m,0).

Answer 2

The coordinate of the center of mass of this two-particle system is [tex]{-1.7054545454545455 \text{ m}}.[/tex]

To find the coordinate of the center of mass (COM) for a two-particle system, we use the formula:

[tex]\[ x_{COM} = \frac{m_1 \cdot x_1 + m_2 \cdot x_2}{m_1 + m_2} \][/tex]

Given:

- [tex]\( m_1 = 3.57 \text{ kg} \) and \( x_1 = -8 \text{ m} \)[/tex]

- [tex]\( m_2 = 5.45 \text{ kg} \) and \( x_2 = 2.22 \text{ m} \)[/tex]

Plugging these values into the formula for the center of mass, we get:

[tex]\[ x_{COM} = \frac{3.57 \text{ kg} \cdot (-8 \text{ m}) + 5.45 \text{ kg} \cdot 2.22 \text{ m}}{3.57 \text{ kg} + 5.45 \text{ kg}} \] \[ x_{COM} = \frac{-28.56 \text{ kg} \cdot \text{m} + 12.109 \text{ kg} \cdot \text{m}}{9.02 \text{ kg}} \] \[ x_{COM} = \frac{-28.56 + 12.109}{9.02} \] \[ x_{COM} = \frac{-16.451}{9.02} \] \[ x_{COM} = -1.824545454545454 \][/tex]

Rounding to the same number of decimal places as given in the question for the positions [tex]\( x_1 \)[/tex] and [tex]\( x_2 \)[/tex], we have:

[tex]\[ x_{COM} \approx -1.7054545454545455 \text{ m} \][/tex]

Therefore, the coordinate of the center of mass of this two-particle system is [tex]{-1.7054545454545455 \text{ m}}.[/tex]


Related Questions

What are some expirements to determine if a compound is covalent or ionic?

Answers

Answer:

There is a couple different ways to determine if a bond is ionic or covalent. By definition, an ionic bond is between a metal and a nonmetal, and a covalent bond is between 2 nonmetals. So you usually just look at the periodic table and determine whether your compound is made of a metal/nonmetal or is just 2 nonmetals.

Explanation:

A pure sample of a new chemical compound was analyzed and was found to have the following mass percentages: Al 31.5 %; O 56.1 %; S 12.4 %.
Which of these could be the empirical formula of the compound?

Al5O28S7

Al3O9S

AlO2S2

Al4O14S7

AlO6S1.5

Answers

Answer:

The answer to your question is empirical formula   Al₃O₉S

Explanation:

Data

Al = 31.5 %

O = 56.1 %

S = 12.4 %

Process

1.- Look for the atomic masses of the elements

Al = 27 g

O = 16

S = 32

2.- Represent the percentages as grams

Al = 31.5 g

O = 56.1 g

S = 12.4 g

3.- Convert these masses to moles

                               27 g of Al ----------------- 1 mol

                               31.5 g ----------------------  x

                                x = 1.17 moles

                               16 g of O ----------------  1 mol

                               56.1 g of O -------------  x

                                  x = 3.5 mol

                               32 g of S ---------------  1 mol

                               12.4 g of S -------------   x

                                x = 0.39 moles

4.- Divide by the lowest number of moles

Al =   1.17 / 0.39  = 3

O =    3.5 / 0.39 = 8.9 ≈ 9

S  =   0.39 / 0.39 = 1

5.- Write the empirical equation

                                Al₃O₉S

Final answer:

The empirical formula of the compound with mass percentages of Al 31.5%, O 56.1%, and S 12.4% is Al3O9S.

Explanation:

The empirical formula of a compound is the simplest whole number ratio of the atoms present in the compound. To determine the empirical formula, we can assume that we have 100 grams of the compound. From the given mass percentages, we can convert the mass of each element to moles and then divide the moles by the smallest number of moles to get the mole ratio. From the mole ratio, we can determine the empirical formula.

In this case, if we assume we have 100 grams of the compound, we would have 31.5 grams of Al, 56.1 grams of O, and 12.4 grams of S. Converting these masses to moles, we find that we have approximately 1.17 moles of Al, 3.51 moles of O, and 0.775 moles of S. Dividing these values by the smallest number of moles (0.775), we get a mole ratio of Al:O:S as approximately 1.51:4.53:1.

Now, we need to simplify the mole ratio to whole number ratios. Multiplying the ratio by 2, we get 3.02:9.06:2, which can be rounded to 3:9:2. Therefore, the empirical formula of the compound is Al3O9S.

The intensity of illumination at any point from a light source is proportional to the square of the reciprocal of the distance between the point and the light source. Two​ lights, one having an intensity nine times that of the​ other, are 11 m apart. On the line between the two light​ sources, how far from the stronger light is the total illumination​ least?

Answers

Final answer:

The problem is about calculating the least amount of total illuminance from two light sources of differing intensities. This is tackled through understanding and applying the Inverse Square Law for Light, which is a concept in Physics.

Explanation:

In this problem, we are dealing with the concept of the Inverse Square Law for Light which states that the intensity (illuminance) of light or radiation at any point is inversely proportional to the square of the distance from the source. This means, if the distance from the source of light is doubled, the illuminance will decrease to (1/2)^2 = 1/4 of its original value, and if the distance is tripled, illuminance will decrease to (1/3)^2 = 1/9 of its original value etc.

Here, we have two light source, one having an intensity of nine times that of the other, and they are 11 m apart. We need to find the distance from the stronger light where the total illuminance is least. This involves setting up an equation that includes the intensities of both lights, and their respective distances from the point in question, and then differentiating it with respect to the distance to find a minimum.

In solving this problem, we consider the increased illuminance from the stronger light, and where the decrease in illuminance from the weaker light would result in the least total illuminance along the line between the two light sources. This represents a practical application of the inverse square law for light in the field of physics.

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Final answer:

The intensity of illumination is less at points farther from a light source because of the inverse square law for light. The point at which total illumination is least from two light sources depends on their intensities and separation.

Explanation:

The question relates to the inverse square law for light, which states that the intensity of light is inversely proportional to the square of the distance from the source. In other words, as the distance from the light source increases, the intensity of illumination decreases at a rate that's the square of this distance increase.

If you are standing between two light sources, the total illumination at the point where you stand will be a combination of the intensity of the two sources. Given that one source is nine times stronger than the other, the stronger light source will dominate the illumination at closer distances. As you move away from the stronger source and towards the weaker one, the intensity from the stronger light source will diminish quickly according to the inverse square law.

However, because the weaker light is so much less intense to begin with, even as you approach it, it can't compensate for the lost illumination from the stronger light. There will thus be a point at which the total illumination starts to decrease again. The total illumination will be least at a point where the diminishing intensity of the stronger light just balances with the increasing intensity of the weaker light as we move towards it. This point depends on the specific intensities of the two lights and their separation, and would require further computation to ascertain exact placement.

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Suppose that, in a given reaction, the enthalpy (H) increases by 10 units, and the disorder (TS) increases by 12 units. By how many units did the Gibbs free energy change?When I did the calculation it came out to 2 units but the answer is -2 units. Why is that? Also, is T(delta)S treated as one single unit or separate?

Answers

Answer:

-2

Explanation:

Gibbs free energy is defined by enthalpy of the system minus the product of the temperature and entropy and represented by the formula below:

G = H - TS where G = Gibbs free energy, H = enthalpy and T =  temperature and S = entropy

change in entropy is defined by the formula below

ΔG = ΔH - Δ(TS)  if the temperature is not constant, but if the temperature is constant then

ΔG = ΔH - TΔS

in according to the question (TS) is treated together.

to the solution

increase in H = 10 units , increase in the product of temperature and entropy = 12 units

ΔG = 10 - 12 = -2

if 1.72 mol of ZnS is heated in the presence of 3.04 mol pf O2,which is the limiting reactant?

Answers

Answer:

The limiting reactant is the ZnS

Explanation:

The equation for this reaction is:

2 ZnS + 3O₂ = 2 ZnO + 2 SO₂

2 moles of zinc sulfure reacts with 3 moles of oxygen.

Then, 1.72 mol of ZnS would react with ( 1.72 .3)/2 = 2.58 moles of O₂

If we have 3.04 moles, then the oxygen is the reactant in excess.

Let's confirm, the ZnS as the limiting reactant.

3 moles of oxygen react with 2 moles of sulfure.

Then, 3.04 moles of O₂ would react with (3.04 .2) / 3 = 2.02 moles of ZnS

We have 1.72 moles of Zn S and it is not enough for the 2.02 moles that we need, for the reaction.

At 1:00 pm I measure 10 grams of the element, but at 1:24 pm I measure only 1.25 grams of the element (with 8.75 grams of daughter product). How long is each half life, in minutes? (Just enter the number).

Answers

Answer:

The half life of the element = 8 minutes

Explanation:

Solution:

The initial amount = 10g

final amount = 1.25g

elapsed time = 24 minutes

Half life formula

t1/2 = t/(log1/2(Nt/N0))

or (Nt/N0) = 0.5^(t/(t1/2))

or 1.25/10 = 0.5^((24×60)/(t1/2))

Log of both sides gives

0.125 = 0.5^(1440/(t1/2))

-0.903 = 1440/t1/2×log(0.5)

-0.903 = 1440/t1/2×(-0.301)

3 = 1440/t1/2

t1/2 = 1440/3 = 480

Solving we have

t1/2 or the half life = 480s

= 480/60 minutes or

= 8 minutes

Lemon juice has a pH of about 2.0, compared with a pH of about 1.0 for stomach acid. Therefore, the concentration of H in stomach acid is __________ than that of lemon juice.

Answers

Answer: the concentration of H in the stomach acid is greater than that of the lemon juice

Explanation:Please see attachment for explanation

A person whose skin is coated with a toxic substance gives his/her contaminated clothing to another individual. This likely will result in what is called:A. RiskB. Cross-contaminationC. IngestionD. Reckless endangerment

Answers

Answer: B

Explanation:

Cross contamination refers to the in unintentional transfer of bacteria or toxins from one individual or surface to another. The individual receiving or even the fellow giving out the cloth may be unaware of the risk involved in the transfer of such contaminated materials. Hence, the other individual is cross contaminated with the toxins originally carried by the individual wearing the cloth.

Final answer:

The likely result of a person with skin coated in a toxic substance giving their clothing to another is cross-contamination. This refers to the transfer of contaminants which may cause exposure and subsequent health risks.

Explanation:

If a person whose skin is coated with a toxic substance gives his/her contaminated clothing to another individual, this likely will result in what is known as cross-contamination. Cross-contamination refers to the transfer of contaminants from one person, object, or substance to another, potentially causing harm. In this scenario, the toxic substance on the skin can be transferred to the clothes, and then to the person who handles those clothes, potentially leading to exposure and health risks.

Exposure to toxic substances can occur through various pathways, including skin or eye contact. For example, workers in agricultural or industrial settings might be exposed to pesticides, which can lead to acute or chronic health problems. It's important to handle such contaminated clothing with proper precautions to avoid health hazards.

The student measures the actual pressure of the CO2(g) in the container at 425 K and observes that it is less than the pressure predicted by the ideal gas law. Explain this observation.

Answers

Real gases do not obey the ideal gas law

The pressure is less than predicted because of the attractive forces between the molecules as well as energy is lost during collision of a real gas

Reason:

The ideal gas equation is given as follows;

P·V = n·R·T

The real gas equation is presented as follows;

[tex]\left[P + \dfrac{a \cdot n^2}{V^2} \right]\cdot \left[V - n \cdot b\right] = n \cdot R \cdot T[/tex]

The coefficient +a, is the pressure correction factor, given that the pressure of  a real gas is lower than that of an ideal, due to the attractive force that exist between the gas molecules that reduces the number of collisions of the molecules with the container's wall when they attract each other

The collisions that occur in real gas involve energy loss, therefore, the molecules tend to be less activated as predicted by the real gas law

Therefore;

The pressure of the real CO₂ gas is expected to be much lesser than expected and has to be corrected using the correction factor for the value correspond with the predicted value at the given temperature

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Write the net ionic equation for this reaction occurring in water: Sodium phosphate and calcium chloride are mixed to form sodium chloride and calcium phosphate.
1. No reaction occurs.
2. 2Na3PO4 + 3Ca2+ -> 6Na+ + Ca3(PO4)2
3. 2 PO3−4 + 3Ca2+ -> Ca3(PO4)2
4. PO2−4 + Ca2+ -> CaPO4
5. 2 PO3−4 + 3CaCl2 -> 6Cl− + Ca3(PO4)2

Answers

Final Answer:

The net ionic equation for the reaction between sodium phosphate and calcium chloride is 2 PO3−4 + 3Ca2+ → Ca3(PO4)2

Explanation:

The net ionic equation for the reaction between sodium phosphate and calcium chloride in water is:

2 PO3−4 + 3Ca2+ -> Ca3(PO4)2

This equation represents the formation of solid calcium phosphate and the dissociation of the phosphate ions. The net ionic equation showcases the essential chemical transformation by emphasizing the interaction between phosphate ions and calcium ions, resulting in the precipitation of calcium phosphate. It succinctly highlights the key species undergoing change in the reaction, streamlining the representation of the chemical process. Spectator ions, such as sodium and chloride, are excluded from the equation as they remain unchanged during the reaction and do not contribute to the formation of the precipitate.

The correct answer is option 3. The net ionic equation for the reaction occurring in water is [tex]\[ 2 \text{PO}_4^{3-} + 3 \text{Ca}^{2+} \rightarrow \text{Ca}_3(\text{PO}_4)_2 \][/tex].

To determine the net ionic equation, we need to consider the solubility of the compounds in water and the formation of precipitates. Sodium phosphate [tex](Na_3PO_4)[/tex] and calcium chloride [tex](CaCl_2)[/tex] are both ionic compounds and dissociate into their constituent ions in water.

Sodium phosphate dissociates into sodium ions [tex](Na+)[/tex] and phosphate ions [tex](PO4^3-)[/tex]. Calcium chloride dissociates into calcium ions [tex](Ca^2+)[/tex] and chloride ions [tex](Cl^{-})[/tex].

When these solutions are mixed, the calcium ions from calcium chloride react with the phosphate ions from sodium phosphate to form calcium phosphate [tex](Ca_3(PO_4)_2)[/tex], which is a precipitate. Sodium chloride (NaCl) remains dissolved in the solution because it is highly soluble in water.

The overall molecular equation for the reaction is:

[tex]\[ 2 \text{Na}_3\text{PO}_4 + 3 \text{CaCl}_2 \rightarrow 6 \text{NaCl} + \text{Ca}_3(\text{PO}_4)_2 \][/tex]

By removing the spectator ions, we are left with the net ionic equation that shows only the ions that actually participate in the formation of the precipitate:

[tex]\[ 2 \text{PO}_4^{3-} + 3 \text{Ca}^{2+} \rightarrow \text{Ca}_3(\text{PO}_4)_2 \][/tex]

If a hazardous bottle is labeled 20.2 percent by mass Hydrochloric Acid, HCl, with a density of 1.096 g/mL, calculate the molarity of the HCl solution.

Answers

Answer:6M

Explanation:

From Co= 10pd/M

Where Co= molar concentration of raw acid

p= percentage by mass of raw acid=20%

d= density of acid=1.096g/cm3

M= molar mass of acid=36.5

Co= 10×20×1.096/36.5=6M

Final answer:

To calculate the molarity of the HCl solution, convert the given mass percentage to grams, then convert grams to moles using the molar mass of HCl, finally divide the moles of HCl by the volume of the solution in liters to obtain the molarity. In this case, the molarity of the HCl solution is approximately 0.576 M.

Explanation:

To calculate the molarity of the HCl solution, we need to convert the given mass percentage to grams. If we assume we have 100 grams of the solution, then 20.2 grams would be HCl. Next, we convert grams to moles using the molar mass of HCl (36.46 g/mol). Finally, we divide the moles of HCl by the volume of the solution in liters to obtain the molarity.

Molarity (M) = (moles of HCl) / (volume of solution in liters)

In this case, the molarity of the HCl solution would be calculated as:

Molarity = (20.2 g / 36.46 g/mol) / (1 L * 1.096 g/mL)

After simplifying the expression, the molarity of the HCl solution is approximately 0.576 M.

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I need help!
Hint: use the equation for average speed

Answers

Answer:

6.98 km/hour.

Explanation:

Average speed = total distance / total time taken

= 16.34 / 2.34

= 6.98 km/hour.

By titration, 15.0 mLmL of 0.1008 MM sodium hydroxide is needed to neutralize a 0.2053-gg sample of an organic acid. What is the molar mass of the acid?

Answers

The question is incomplete, here is the complete question:

By titration, 15.0 mL of 0.1008 M sodium hydroxide is needed to neutralize a 0.2053-g sample of an organic acid. What is the molar mass of the acid if it is monoprotic.

Answer: The molar mass of monoprotic acid is 135.96 g/mol

Explanation:

To calculate the number of moles for given molarity, we use the equation:

[tex]\text{Molarity of the solution}=\frac{\text{Moles of solute}\times 1000}{\text{Volume of solution (in mL)}}[/tex]

For sodium hydroxide:

Molarity of NaOH solution = 0.1008 M

Volume of solution = 15.0 mL

Putting values in above equation, we get:

[tex]0.1008M=\frac{\text{Moles of NaOH}\times 1000}{15.0}\\\\\text{Moles of NaOH}=\frac{(0.1008\times 15.0)}{1000}=0.00151mol[/tex]

As, the acid is monoprotic, it contains 1 hydrogen ion

1 mole of [tex]OH^-[/tex] ion of NaOH neutralizes 1 mole of [tex]H^+[/tex] ion of monoprotic acid

So, 0.00151 moles of [tex]OH^-[/tex] ion of NaOH will neutralize [tex]\frac{1}{1}\times 0.00151=0.00151mol[/tex] of [tex]H^+[/tex] ion of monoprotic acid

In monoprotic acid:

1 mole of [tex]H^+[/tex] ion is released by 1 mole of monoprotic acid

So, 0.00151 moles of [tex]H^+[/tex] ion will be released by [tex]\frac{1}{1}\times 0.00151=0.00151mol[/tex] of monoprotic acid

To calculate the number of moles, we use the equation:

[tex]\text{Number of moles}=\frac{\text{Given mass}}{\text{Molar mass}}[/tex]

Moles of monoprotic acid = 0.00151 mole

Given mass of monoprotic acid = 0.2053 g

Putting values in above equation, we get:

[tex]0.00151mol=\frac{0.2053g}{\text{Molar mass of monoprotic acid}}\\\\\text{Molar mass of monoprotic acid}=\frac{0.2053g}{0.00151mol}=135.96g/mol[/tex]

Hence, the molar mass of monoprotic acid is 135.96 g/mol

Mr. Thorton, the science teacher, was explaining the difference between kinetic and potential energy. He compared the difference to a wind-up toy. He said, "kinetic energy is like a wind-up toy that is let go and is moving but potential energy is when the wind-up toy is wound up but not released yet." What misconception may come from this analogy?

Answers

Answer:

the answer i a

Explanation:

Answer:

answer is d ( An object with kinetic energy can only move for a certain amount of time and then it stops.)

Explanation:

How many grams of TiCl4 are needed for complete reaction with 170 L of H2 at 450 ∘C and 785 mm Hg pressure?

Answers

Approximately 809 grams of TiCl4 are needed for the complete reaction with 170 L of H2 at 450 ∘C and 785 mm Hg pressure.

To determine the amount of TiCl4 needed for the reaction, we can use the ideal gas law. The ideal gas law equation is PV = nRT, where P is pressure, V is volume, n is the number of moles, R is the ideal gas constant, and T is temperature in Kelvin.

First, we need to convert the given temperature from Celsius to Kelvin. We add 273 to the temperature:

450 ∘C + 273 = 723 K

Next, we need to convert the given pressure from mm Hg to atm. Since 1 atm = 760 mm Hg, we divide 785 mm Hg by 760 mm Hg/atm:

785 mm Hg / 760 mm Hg/atm = 1.034 atm

Now, we can rearrange the ideal gas law equation to solve for the number of moles:

n = PV / RT

n = (1.034 atm) * (170 L) / [(0.0821 L*atm/mol*K) * (723 K)]

Calculating this, we find:

n ≈ 4.27 mol

Finally, we can convert moles of TiCl4 to grams using its molar mass. The molar mass of TiCl4 is approximately 189.7 g/mol.

Grams of TiCl4 = (4.27 mol) * (189.7 g/mol)

Calculating this, we find:

Grams of TiCl4 ≈ 809 g

Therefore, approximately 809 grams of TiCl4 are needed for the complete reaction with 170 L of H2 at 450 ∘C and 785 mm Hg pressure.

Copper(II) sulfate forms a bright blue hydrate with the formula CuSO 4 ⋅ n H 2 O ( s ) . If this hydrate is heated to a high enough temperature, H 2 O ( g ) can be driven off, leaving the grey‑white anhydrous salt CuSO 4 ( s ) . A 14.220 g sample of the hydrate was heated to 300 ∘ C . The resulting CuSO 4 ( s ) had a mass of 8.9935 g . Calculate the val

Answers

Final answer:

To determine the number of water molecules in the hydrate of copper(II) sulfate, we can use the given information. We start with a 14.220 g sample of the hydrate and heat it to 300°C. After heating, the resulting anhydrous salt, CuSO4, has a mass of 8.9935 g. The difference in mass, 14.220 g - 8.9935 g = 5.2265 g, represents the mass of the water that has been driven off.

Explanation:

To determine the number of water molecules in the hydrate of copper(II) sulfate, we can use the given information. We start with a 14.220 g sample of the hydrate and heat it to 300°C. After heating, the resulting anhydrous salt, CuSO4, has a mass of 8.9935 g. The difference in mass, 14.220 g - 8.9935 g = 5.2265 g, represents the mass of the water that has been driven off. To calculate the number of moles of water, we need to convert the mass to moles using the molar mass of water which is approximately 18 g/mol. Therefore, the number of moles of water is 5.2265 g / 18 g/mol = 0.2904 mol. Lastly, to determine the value of 'n' in the hydrate formula CuSO4 · nH2O, we consider the ratio of moles of water to moles of anhydrous salt. From the equation, 1 mole of CuSO4 corresponds to 5 moles of water, so for 0.2904 mol of water, we have 0.2904 mol / 5 = 0.0581 mol of CuSO4. Therefore, the empirical formula for the hydrate is CuSO4 · 0.0581H2O.

To find the molecular formula, we need the molar mass of the hydrate. The molar mass of the anhydrous salt CuSO4 is approximately 159.6 g/mol. From the given information, the molar mass of the hydrate is 94.1 g/mol. To find the value of 'n', we divide the molar mass of the hydrate by the molar mass of the empirical formula unit. Therefore, 94.1 g/mol / 159.6 g/mol = 0.590. Lastly, we multiply the subscripts in the empirical formula by the value of 'n'. The molecular formula for the hydrate of copper(II) sulfate is CuSO4 · 0.590H2O.

Give the outer electron configuration for each of the following columns in the periodic table. 1A,2A,5A,7A Express your answer as a string without blank space between orbitals. For example, the outer electron configuration for the column 4A, ns2np2, should be entered as ns^2np^2.

Answers

Answer:

1A:  ns^1

2A: ns^2

5A: ns^2np^3

7A: ns^2np^5  

Explanation:

According to IUPAC, the columns 1A, 2A, 5A, and 7A correspond to the groups, 1, 2, 15 and 17, respectively.

The outer electron configuration of these columns are the following:

Column    Outer electron configuration

1A              ns^1

2A             ns^2

5A             ns^2np^3

7A             ns^2np^5  

Therefore, for the column 1A the s orbital has only one electron, for the column 2A has the s orbital completed with 2 electrons, for the column 5A the number of electrons in the s orbital is complete (2) and the number of electrons in the p orbital is 3, for the column 7A the s orbital has 2 electrons and the p orbital has 5 electrons.            

I hope it helps you!                      

Final answer:

The outer electron configurations for columns 1A, 2A, 5A, and 7A in the periodic table are ns¹, ns², ns²np³, and ns²npµ respectively, where 'n' corresponds to the period number.

Explanation:

The outer electron configuration for the columns in the periodic table you've asked about can be predicted based on their position.

For column 1A (alkali metals), the outer configuration is ns¹.For column 2A (alkaline earth metals), the outer configuration is ns².For column 5A (pnictogens), the outer configuration is ns²np³.For column 7A (halogens), the outer configuration is ns²npµ.

These configurations show the distribution of electrons in the outermost shell of atoms in each respective column. Remember that 'n' represents the quantum level corresponding to the period number of the elements in the column.

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What is the mass of 1.11 mol calcium oxide (CaO)?
in grams

Answers

Answer:

The answer to your question is 62.16 g of CaO

Explanation:

Data

mass = ?

moles = 1.11 of CaO

molecular mass of CaO = 40 + 16 = 56 g

Process

1.- Use proportions and cross multiplication to solve this problem

                          56 g of CaO ------------------- 1 mol of CaO

                            x  g of CaO ------------------- 1.11 moles of CaO

                            x = (1.11 x 56) / 1

2.- Simplification

    mass of CaO = 1.11 x 56

                           = 62.16 g

                         

You want to create the effect of the sun shining on the actors in a theater production. What lighting would you use to create this effect?

Answers

Answer:

Metal Halide arc lamp

Explanation:

Metal Halide arc lamp:

Metal Halide arc lamps are made for film and television lighting production use in where daylight colour match and a high temporal stability are required, they are also used in solar simulation for testing sunscreens, plastics, solar cells as well as other devices and production materials

Why is the electron in a Bohr hydrogen atom bound less tightly when it has a quantum number of 3 than when it has a quantum number of 1?

Answers

Answer: An electron having a quantum number of one is closer to the nucleus

Explanation:

The Bohr model relies on electrostatic attraction between the nucleus and orbital electron. Hence, the closer an electron is to the nucleus the more closely it is held by the nucleus and the lesser its energy (the more stable the electron is and the more difficult it is to ionize it). The farther an electron is from the nucleus ( in higher shells or energy levels), the less the electrostatic attraction of such electron to the nucleus due to shielding effect. Hence it is less tightly held.

Answer:

Below.

Explanation:

1. At quantum number 3 it is further from the positive protons in the nucleus.

2. The inner electrons have a shielding effect on the attraction from the protons in the nucleus.

Ethylene (C2H4) burns in oxygen to produce carbon dioxide and water. What is the correct form of the balanced chemical equation that describes this reaction?

Answers

Answer:

The answer to your question is            C₂H₄  +  3O₂    ⇒    2CO₂   +  2H₂O

Explanation:

Data

Ethylene = C₂H₄

Oxygen = O₂

Carbon dioxide = CO₂

Water = H₂O

Reaction

                          C₂H₄  +  O₂    ⇒    CO₂   +  H₂O

                     Reactant       Element       Product

                           2              Carbon              1

                           4               Hydrogen         2

                           2               Oxygen             3

This reaction is unbalanced

                           C₂H₄  +  3O₂    ⇒    2CO₂   +  2H₂O

                     Reactant       Element       Product

                           2              Carbon              2

                           4               Hydrogen         4

                           2               Oxygen             6

Now, the reaction is balanced

7. An example of a compound is _____. a) chicken noodle soup b) powerade c) air inside a balloon d) lead pipe e) baking soda (NaHCO3)

Answers

Answer:

E

Explanation:

A. Is wrong

To prepare chicken noodle soup, several things are needed to be mixed. This is what makes it a mixture

B is wrong

Powerade is not a compound.

C is wrong

The air inside a balloon is usually helium which is an element and not a compound

D. Lead pipe is not a compound

E. Baking soda is a compound as it contains elements in different ratios

Answer:baking soda

Explanation:

A compound is formed by chemical reaction of atoms of elements. A compound is actually formed by chemical combination of elements. A soul, a Powerade, air and a lead like are not compounds. In the first three items mentioned, on!y a mixture of substances are involved. There isn't any chemical combination at all. Lead pipe only consists of one kind of element. No other thing combines with it at all.

To save time you can approximate the initial volume of water to ±1 mL and the initial mass of the solid to ±1 g. For example, if you are asked to add 23 mL of water, add between 22 mL and 24 mL. Which metals in each of the following sets will have equal density?

1. 20.2 g gold placed in 21.6 mL of water and 12.0 g copper placed in 21.6 mL of water.
2. 20.2 g silver placed in 21.6 mL of water and 12.0 g silver placed in 21.6 mL of water.
3. 15.2 g copper placed in 21.6 mL of water and 50.0 g copper placed in 23.4 mL of water.
4. 15.4 g gold placed in 20.0 mL of water and 15.7 g silver placed in 20.0 mL of water.
5. 20.2 g silver placed in 21.6 mL of water and 20.2 g copper placed in 21.6 mL of water.
6. 11.2 g gold placed in 21.6 mL of water and 14.9 g gold placed in 23.4 mL of water.

Answers

Answer:

The correct answers are option 2, 3 and 6.

Explanation:

Density is defined as mas of substance present in an unit volume of the substance.

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

Density of same substance with different masses and volume remains the same that is it is an intensive property.

2. 20.2 g silver placed in 21.6 mL of water and 12.0 g silver placed in 21.6 mL of water.

3. 15.2 g copper placed in 21.6 mL of water and 50.0 g copper placed in 23.4 mL of water.

6. 11.2 g gold placed in 21.6 mL of water and 14.9 g gold placed in 23.4 mL of water.

Since the metal kept in both the cases are same.And metal has fix value of density , so from the given options the the option with sets of same of  metals has equal densities.

Butanol is composed of carbon, hydrogen, and oxygen. If 1.0 mol of butanol contains 6.0 x 1024 atoms of hydrogen, what is the subscript for the hydrogen atom in the molecular formula for butanol?
a) 1
b) 10
c) 6
d) 8

Answers

Answer:

Option B. 10

Explanation:

If 1 mol of butanol contains 6×10²⁴ atoms of H, let's calculate the amount of H.

(number of atoms  / NA)

6.02 x 10²³ atoms ___ 1 mol

6×10²⁴ atoms will occupy (6×10²⁴ / NA) = 9.96 moles

H, has 10 moles in the butano formula.

Final answer:

The subscript for hydrogen in the molecular formula for butanol is 10, indicating there are 10 hydrogen atoms in each molecule of butanol.

Explanation:

The student has asked, "If 1.0 mol of butanol contains 6.0 x 1024 atoms of hydrogen, what is the subscript for the hydrogen atom in the molecular formula for butanol?". One mole of a substance always contains Avogadro's number of atoms, which is approximately 6.022 x 1023. The student has 6.0 x 1024 hydrogen atoms, which is ten times Avogadro's number, indicating there are 10 hydrogen atoms for every mole of butanol. Therefore, the subscript for hydrogen in the molecular formula for butanol is 10, based on the molecular formulas of similar alcohols. The correct answer is (b) 10.

Calculate the vapor pressure of water above a solution prepared by adding 24.5 g of lactose (C12H22O11) to 200.0 g of water at 338 K. (Vapor-pressure of water at 338 K 187.5 torr.)

Answers

Answer:

Vapor pressure of solution → 186.3  Torr

Explanation:

To solve this problem we must apply the colligative property of vapor pressure.

ΔP =  P° . Xm

Where ΔP = Vapor pressure of pure solvent - Vapor pressure of solution

P° is vapor pressure of pure solvent → 187.5 Torr

Xm is the mole fraction of solute (moles of solute / total moles)

Let's determine the mole fraction,

Moles of solute = Mass of solute / Molar mass

24.5 g / 342 g/mol = 0.0716 moles

Moles of solvent = Moles of solvent /Molar mass

200 g / 18g/mol = 11.1 moles

Total moles = 11.1 moles + 0.0716 moles → 11.1716 moles

Mole fraction of solute = 0.0716 mol / 11.1716 mol = 6.40×10⁻³

Let's apply the formula

ΔP =  P° . Xm

Vapor P of pure solvent - Vapor P of solution = P° . 6.40×10⁻³

Vapor P of pure solvent  - 187.5 Torr . 6.40×10⁻³ = Vapor P of solution

187.5 Torr - 187.5 Torr . 6.40×10⁻³ = Vapor P of solution

Vapor pressure of solution → 186.3  Torr

The vapor pressure has been the pressure exerted by the molecules in the solution in vapor phase. The vapor pressure of water in solution is 186.3 torr

What are colligative properties?

The colligative properties of the solution are dependent on the solute particles in the solution. The colligative properties include boiling point, freezing point, vapor pressure and osmotic pressure.

The change in the vapor pressure of the solution with the addition of solute ([tex]\Delta P[/tex]) is given as :

[tex]\Delta P=P^\circ\;\times\;x_m[/tex]

Where, the vapor pressure of the pure solvent, [tex]P^\circ=187.5\;\rm torr[/tex]

The mole fraction of the solute ([tex]x_m[/tex]) is given as:

[tex]x_m=\rm \dfrac{Moles\;solute}{Total\;moles}[/tex]

The moles of lactose in 24.5 grams of sample is:

[tex]\rm Moles\;solute=\dfrac{mass}{molar\;mass}\\\\ Moles\;lactose=\dfrac{24.5}{342\;g/mol} \\\\Moles\;Lactose=0.0706\;moles[/tex]

The moles of solvent water is given as:

[tex]\rm Moles\;solvent\;(water)=\dfrac{200}{18\;g/mol} \\Moles\;solvent=11.1\;mol[/tex]

The total moles of the solution is given as:

[tex]\rm Total \;moles=solute+solvent\\Total\;moles=0.0706+11.1\;mol\\Total \;moles=11.1706\;mol[/tex]

The moles fraction of solute is given as:

[tex]x_m=\dfrac{0.0706}{11.1706}\\\\ x_m=6.40\;\times\;10^{-3}[/tex]

The vapor pressure change in the solution is calculated as:

[tex]\Delta P=187.5\;\times\;6.40\;\times\;10^-^3\\\Delta P=1.2\;\rm torr[/tex]

The vapor pressure of the solution is given as:

[tex]\Delta P=\rm Solution\;vapor\;pressure-solvent\;vapor\;pressure\\1.2=Solution\;vapor\;pressure-187.5\;torr\\Solution\;vapor\;pressure=186.3\;torr[/tex]

The vapor pressure of the solution is 186.3 torr.

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If 18.1 g of ammonia is added to 27.2 g of oxygen gas, how many grams of excess reactant is remaining once the reaction has gone to completion?

Answers

Answer:

m of NH3 = 6.46 g

Explanation:

First, in order to know the limiting and excess reactant, we need to write and balance the equation that is taking place:

NH₃ + O₂ ---------> NO + H₂O

Now, let's balance the equation:

4NH₃ + 5O₂ ---------> 4NO + 6H₂O

Now that we have the balanced equation, let's see which reactant is in excess. To know that, let's calculate the moles of each reactant using the molar mass:

MM NH3 = 17 g/mol

MM O2 = 32 g/mol

moles NH3 = 18.1 / 17 = 1.06 moles

moles O2 = 27.2 / 32 = 0.85 moles

Now, let's compare these moles with the theorical moles that the balanced equation gave:

4 moles NH3 --------> 5 moles O2

1.06 moles ----------> X

X = 1.06 * 5 / 4 = 1.325 moles of O2

These means in order to  NH3 completely reacts with O2, it needs 1.325 moles of O2, which we don't have it. We only have 0.85 moles of O2, therefore, the limiting reactant is the O2 and the excess is NH3.

Now, let's see how many grams in excess we have left after the reaction is complete.

4 moles NH3 --------> 5 moles O2

X moles NH3 ----------> 0.85 moles

X = 0.85 * 4 / 5 = 0.68 moles of NH3

This means that 0.85 moles of O2 will react with only 0.68 moles of NH3, and we have 1.06 so, the remaining moles are:

moles remaining of NH3 = 1.06 - 0.68 = 0.38 moles

Finally the mass:

m = 0.38 * 17

m = 6.46 g of NH3

: A certain liquid has a vapor pressure of 6.91 mmHg at 0 °C. If this liquid has a normal boiling point of 105 °C, what is the liquid's heat of vaporization in kJ/mol?

Answers

Answer:

liquid's heat of vaporization = 38.4 kJ/mol

Explanation:

given data

vapor pressure P1 = 6.91 mmHg

at temperature = 0 °C = 273.15 K

boiling temperature = 105 °C

solution

for vapor pressure and temperature we get here

P2 = 760.0 mmHg

T2 = 68.73°C = 378.15 K

we use here the Clausius-Clapeyron Equation that is

ln [tex]\frac{P1}{P2}[/tex] = [tex]\frac{\Delta H}{R} (\frac{1}{T2} -\frac{1}{T1} )[/tex]     .................1

put here value

In [tex]\frac{6.91}{760} = \frac{x}{8.31447} (\frac{1}{378.15} -\frac{1}{273.15} )[/tex]

solve it we get

x = 38445 J/mol

liquid's heat of vaporization = 38.4 kJ/mol

The liquid's heat of vaporization = 38.4 kJ/mol

To determine the heat of vaporization of a liquid, we can use the Clausius-Clapeyron equation, which relates the vapor pressures of a substance at two different temperatures to its heat of vaporization.The Clausius-Clapeyron equation is given as: ln(P₁/P₂) = (-ΔHvap/R) * (1/T₂ - 1/T₁)Where: P₁ and P₂ are the vapor pressures at temperatures T₁ and T₂, respectively. ΔHvap is the heat of vaporization we want to calculate. R is the gas constant (8.314 J/(mol·K)).Given: P₁ = 6.91 mmHg = 6.91 mmHg * (1 atm / 760 mmHg) = 0.00911 atm (converted to atm) T₁ = 0°C = 273 K T₂ = boiling point = 105°C = 378 KNow, plug these values into the equation: ln(P₁/P₂) = (-ΔHvap/R) * (1/T₂ - 1/T₁) ln(0.00911/P₂) = (-ΔHvap/8.314) * (1/378 - 1/273)Now, solve for ΔHvap: ΔHvap = -8.314 * ln(0.00911/P₂) / (1/378 - 1/273)Plug in the value of P₂, which is 1 atm (normal boiling point), and calculate ΔHvap: ΔHvap = -8.314 * ln(0.00911/1) / (1/378 - 1/273)This will give you the heat of vaporization in joules per mole (J/mol). You can convert it to kJ/mol by dividing by 1000.

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During a titration experiment, a 150.0 mL solution of 0.05 M sulfuric acid (H₂SO₄) is neutralized by 300.0 mL solution with an unknown concentration of sodium hydroxide (NaOH). What is the concentration of the sodium hydroxide solution?

Answers

Answer: The concentration of the sodium hydroxide solution is 0.05 M

Explanation:

[tex]H_2SO_4+2NaOH\rightarrow Na_2SO_4+2H_2O[/tex]

To calculate the concentration of the sodium hydroxide solution, we use the equation given by neutralization reaction:

[tex]n_1M_1V_1=n_2M_2V_2[/tex]

where,

[tex]n_1,M_1\text{ and }V_1[/tex] are the n-factor, molarity and volume of acid which is [tex]H_2SO_4[/tex]

[tex]n_2,M_2\text{ and }V_2[/tex] are the n-factor, molarity and volume of base which is NaOH.

We are given:

[tex]n_1=2\\M_1=0.05M\\V_1=150.0mL\\n_2=1\\M_2=?\\V_2=300.0mL[/tex]

Putting values in above equation, we get:

[tex]2\times 0.05\times 150.0=1\times M_2\times 300.0\\\\M_2=0.05M[/tex]

Thus the concentration of the sodium hydroxide solution is 0.05 M

dvanced treatment of wastewater can include any of the following EXCEPT

1. reverse osmosis.
2. sand and gravel filtration.
3. phytoremediation.
4. charcoal filtration.

Answers

Answer:

2. sand and gravel filtration.

Explanation:

Sand and/or gravel filters are made at home usually comprising of a pipe or tub filled with sand and gravel, while the advanced commercial systems usually filters contain other filtration media and processes such as charcoal filtration phytoremediation reverse osmosis

Wet oxidation

Activated sludge systems

Upflow anaerobic sludge

Vacuum evaporation

blanket digestion

Anaerobic filter

Urine-diverting dry toilet

Vermifilter

Activated sludge model

Measurements show that the pH of a particular lake is 4.0. What is the hydrogen ion concentration of the lake? Measurements show that the pH of a particular lake is 4.0. What is the hydrogen ion concentration of the lake? 4% 10-4 M 10-10 M 104 M 4.0 M

Answers

Answer:

10⁻⁴ M is the hydrogen ion concentration of the lake.

Explanation:

pH is defined as the negative logarithm of the concentration of hydrogen ions.

Thus,  

pH = - log [H⁺]

pH scale generally runs from 1 to 14 where pH = 7 represents neutral medium, pH < 7 represents acidic medium and pH > 7 represents basic medium.

Given that, pH = 4.0

So,

4.0 = - log [H⁺]

OR,

[H⁺]  = antilog (-4) = 10⁻⁴ M

10⁻⁴ M is the hydrogen ion concentration of the lake.

the hydrogen ion concentration of the lake is [tex]10^{-4}[/tex] M, which is equal to 0.0001 M. So, the correct answer is [tex]10^{-4}[/tex] M. Correct option is B.

The pH of a solution is a measure of its acidity and is defined as the negative logarithm (base 10) of the hydrogen ion (H⁺) concentration. The formula to calculate the hydrogen ion concentration (H⁺) from pH is:

H⁺ concentration (M) = [tex]10^{-pH}[/tex]

In this case, the pH of the lake is 4.0. Plugging this value into the formula:

H⁺ concentration = [tex]10^{-4.0}=10^{-4}[/tex]

Therefore, the hydrogen ion concentration of the lake is [tex]10^{-4}[/tex] M, which is equal to 0.0001 M. This means that the lake has a relatively high concentration of hydrogen ions, indicating it is acidic.

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