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Potential energy is the stored or latent energy in an object at rest. It’s fundamental to many physics-related concepts because its laws hold true on any level, from the planetary to the atomic level. The potential energy of an object is measurable.

The question is incomplete. Here is the complete question.

When 2.10 g of a certain molecular compound X are dissolved in 65.0 g of benzene (C₆H₆), the freezing point of the solution is measured to be 3.5°C. Calculate the molar mass of X. If you need any additional information on benzene, use only what you find in the ALEKS Data resource. Also, be sure your answer has a unit symbol, and is rounded to 2 significant digits.

Answer: MM = 47.30 g/mol.

Explanation: There is a relationship between freezing point depression and molality. With this last one, is possible to calculate molar mass or molar weight of a compound.

Freezing Point Depression occurs when a solute is added to a solvent: the freezing point of the solvent decreases when a non-volatile solute is incremented.

Molality or molal concentration is a quantity of solute dissolved in a certain mass, in kg, of solvent. Its symbol is m and it's defined as

[tex]m=\frac{moles(solute)}{kg(solvent)}[/tex]

Freezing point depression and molal are related as the following:

[tex]\Delta T_{f}=K_{f}.m[/tex]

where

[tex]\Delta T_{f}[/tex] is freezing point depression of solution

[tex]K_{f}[/tex] is molal freezing point depression constant

m is molality

Now, to determine molar mass, first, find molality of the mixture:

[tex]\Delta T_{f}=K_{f}.m[/tex]

[tex]m=\frac{\Delta T_{f}}{K_{f}}[/tex]

For benzene, constant is 5.12°C/molal. Then

[tex]m=\frac{3.5}{5.12}[/tex]

m = 0.683 molal

Second, knowing the relationship between molal and moles of solute, determine the last one:

[tex]m=\frac{moles(solute)}{kg(solvent)}[/tex]

[tex]mol(solute)=m.kg(solvent)[/tex]

mol(solute) = 0.683(0.065)

mol(solute) = 0.044 mol

The definition for Molar mass is the mass in grams of 1 mol of substance:

[tex]n(moles)=\frac{m(g)}{MM(g/mol)}[/tex]

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

In the mixture, there are 0.044 moles of X, so its molecular mass is

[tex]MM=\frac{2.1}{0.044}[/tex]

MM = 47.30 g/mol

The molecular compound X has molecular mass of 47.30 g/mol.

Answer:

movement of charged particles through the wire .

Explanation:

When electricity is passed through the wire of electromagnet , moving electrons of the wire produces magnetic field . This magnetic field in increased due to high permeability of soft iron of the electromagnet . It is this magnetic field which creates magnetic force .

Answer: Sol:-

Data provided in the question is :-

Atomic mass of isotope -1 = 64 amu

Atomic mass of isotope -2 = 66 amu

Atomic mass of isotope -3 = 67 amu

Atomic mass of isotope -4 = 68 amu

Atomic mass of isotope - 5 = 70 amu

Percentage abundace of isotope - 1 = 48.89 %

Percentage abundance of isotope -2 = 27.81 %

Percentage abundance of isotope - 3 = 4.11%

Percentage abundance of isotope-4 = 18.57%

Percentage abundance of isotope - 5 = 0.62 %

Formula used :-

Average atomic mass of an element =[ {(atomic mass of isotope-1 * percentage abundance of isotope-1) + ( atomic mass of isotope-2 * percentage abundance of isotope -2) + ( atomic mass of isotope -3 * percantege abundance of isotope-3 ) + ( atomic mass of isotope-4 * percentage abundance of isotope-4) + (atomic mass of isotope-5 * percentage abundance of isotope-5)} / 100]

Calculation :-

Put all the value in the formula :-

Average atomic mass of an element = [{(64 * 48.89) + (66 * 27.81) + (67 * 4.11) + (68 * 18.57) + (70 * 0.62)} / 100] amu

= [{(3128.96) + (1835.46) +(257.37) + (1262.76) + (43.4)} / 100] amu

= {(6528.04) / 100} amu

= 65.2804 amu

Average atomic mass of an element is = 65.2804 amu

Then this mass is approximatly equal to atomic mass of zinc so this element would be zinc

atomic mass of zinc = 65.38 \approx 65.2804 amu

Answer:

Carbon and oxygen are chemically bonded in it.

Explanation:

The other answer choices do not apply for compounds, but rather for mixtures instead.

Answer:

Q= Magnesium R= Chlorine

Explanation:

The element Q should be magnesium and R is chlorine.

An ionic compound is a compound that is formed by the combination of a metal and non-metal. Such bonds forms when there is a transfer of electrons from the metals to the non-metals. This leaves a net positive charge on the metal and a negative charge on the non-metal.

The electrostatic attraction leads to the formation of the bond.

 To solve this problem, the hypothetical compound is QR₂

       Mg                        Cl

     2 8 2                    2 8 7

So, Mg transfers 2 electrons to two atoms of chlorine.

 This leads to the formation of the compound MgCl₂

Answer:

Explanation:

The missing incomplete resonance structure is attached in the image below. From there, we can see the addition of the nonbonding electrons and its' formal charge which makes the resonance structure a complete resonance structure. The others two resonance structure that can be derived from the complete structure is also shown in the image. Out of these three structures, the structure that contributes most to the hybrid is the structure with the negative charge on the oxygen.

Answer:

Molarity: 0.21M

Molality: 0.20m

Explanation:

...dissolves 3.9g of aniline (C6H5NH2) in 200.mL of a solvent with a density of 1.05 g/mL...

To solve this question, we need to find the moles of aniline in 3.9g using its molar mass. Then, we need to find the kg and Liters of solution in order to find molarity (Moles/L solution) and molality (Moles/kg of solvent):

Moles aniline:

Molar mass:

6C: 6* 12.01g/mol = 72.06g/mol

7H: 7*1.008g/mol = 7.056g/mol

N: 1*14.007g/mol = 14.007g/mol

72.06g/mol+7.056g/mol+14.007g/mol = 93.123g/mol

Moles of 3.9g: 3.9g * (1mol / 93.123g) = 0.04188moles

Liters solution:

200mL * (1L / 1000mL) = 0.200L

kg solvent:

200mL * (1.05g/mL) * (1kg/1000g) = 0.210L

Molarity:

0.04188mol / 0.200L = 0.21M

Molality:

0.04188mol / 0.210L =0.20m

Answer: Photosynthesizing organisms use carbon dioxide and water to produce glucose.

Explanation:

Photosynthesis is a phenomenon in which green plants containing chlorophyll use sunlight as a source of energy to convert carbon dioxide and water to form glucose and oxygen.

Photosynthesis is the process used by plants, algae and certain bacteria to convert energy from sunlight and turn it into chemical energy in the form of glucose which is used a s a source of energy by many organisms.

[tex]6CO_2+6H_2O\overset{sunlight}\rightarrow C_6H_{12}O_6+6O_2[/tex]

The question is incomplete. Here is the complete question.

In the Haber reaction, patented by German chemist Fritz Haber in 1908, dinitrogen gas combines with dihydrogen gas to produce gaseous ammonia. This reaction is now the first step taken to make most of the world's fertilizer. Suppose a chemical engineer studying a new catalyst for the Haber reaction finds that 505. liters per second of dinitrogen are consumed when the reaction is run at 172.°C and 0.88 atm. Calculate the rate at which ammonia is being produced. Give your answer in kilogram per second. Be sure your answer has the correct number of significant digits.

Answer: Rate = 0.41 kg/s

Explanation: The balanced Haber reaction is

[tex]N_{2}+3H_{2}\rightarrow2NH_{3}[/tex]

As all the components are gases, we can use Ideal Gas Law, which relates Pressure (P), Volume (V), Temperature (T) and Moles (n) in the following formula:

PV = nRT

where

R is gas constant and, in this case, is R = 0.082 L.atm.K⁻¹mol⁻¹

T is in Kelvin

Converting Celsius in Kelvin:

T = 273 + 172

T = 445 K

Calculating moles

[tex]n=\frac{PV}{RT}[/tex]

[tex]n=\frac{0.88(505)}{0.082(445)}[/tex]

n = 12.18 moles

According to the balanced equation, for 1 mol of dinitrogen gas consumed, 2 moles of ammonia is produced.

With 12.18 moles of dinitrogen, the reaction will result in

2(12.18) = 24.36 moles of ammonia

Molar mass of ammonia is M = 17.031 g/mol.

In 24.36 moles, there are

[tex]m=n.M[/tex]

m = 24.36.17.031

m = 414.87 grams

Since it's asking in kilograms: m = 0.41 kg.

In the beginning, it is said that dinitrogen gas is consumed at a rate of liters per second. So, the production rate of ammonia will be 0.41 kg/s.

Answer:

0.171 M

Explanation:

Step 1: Given data

Step 2: Calculate the moles of solute

The molar mass of H₃PO₄ is 97.99 g/mol.

3.35 g × 1 mol/97.99 g = 0.0342 mol

Step 3: Convert "V" to liters

We will use the conversion factor 1 L = 1000 mL.

200 mL × 1 L/1000 mL = 0.200 L

Step 4: Calculate the molarity of the solution

We will use the definition of molarity.

M = moles of solute / liters of solution

M = 0.0342 mol/0.200 L = 0.171 M

Answer:

1.2×10² mmole of Na₂S₂O₃

Explanation:

From the question given above, the following data were obtained:

Volume = 0.6 L

Molarity = 0.2 mol/L

Mole of Na₂S₂O₃ =?

Molarity is simply defined as the mole of solute per unit litre of water. Mathematically, it is expressed as:

Molarity = mole /Volume

With the above formula, we can obtain the number of mole of Na₂S₂O₃ in the solution as illustrated below:

Volume = 0.6 L

Molarity = 0.2 mol/L

Mole of Na₂S₂O₃ =?

Molarity = mole /Volume

0.2 = Mole of Na₂S₂O₃ / 0.6

Cross multiply

Mole of Na₂S₂O₃ = 0.2 × 0.6

Mole of Na₂S₂O₃ = 0.12 mole

Finally, we shall convert 0.12 mole to millimole (mmol). This can be obtained as follow:

1 mole = 1000 mmol

Therefore,

0.12 mole = 0.12 mole × 1000 mmol / 1 mole

0.12 mole = 120 = 1.2×10² mmole

Thus, the chemist added 1.2×10² mmole of Na₂S₂O₃

Answer:

21.13 L

Explanation:

Step 1: Write the balanced equation

2 C₂H₆(g) + 7 O₂(g) ⇒ 4 CO₂(g) + 6 H₂O(g)

Step 2: Determine the appropriate volume ratio

Since all the gases are in the same container at the same temperature and pressure, the volume ratio is equal to the molar ratio, because the volume depends on the number of moles. The volume ratio of O₂(g) to H₂O(g) is 7:6.

Step 3: Determine the volume of H₂O produced from 24.65 L of O₂

24.65 L O₂ × 6 L H₂O/7 L O₂ = 21.13 L H₂O

Answer:

2.5 moles of oxygen are produced.

Explanation:

Given data:

Number of moles of O₂ produced = ?

Number of moles of HgO decomposed = 5 mol

Solution:

Chemical equation:

2HgO      →     2Hg + O₂

now we will compare the moles of HgO and O₂.

           HgO         :          O₂

            2              :          1

            5              :         1/2×5 = 2.5

Thus, from 5 moles of HgO 2.5 moles of oxygen are produced.

Answer:

% Free space in water = [tex]\frac{9.945* 10^{-7} }{1*10^{-6} }[/tex]×100 = 99.45%

% Free space in ice  = [tex]\frac{9.98* 10^{-7} }{1*10^{-6} }[/tex]×100 = 99.8%

Explanation:

As given ,

Density for ice at 0⁰C = 0.917 g/ml

Density for water at 0⁰C = 0.999 g/ml

Radii of H atoms = 37 pm

Radii of O atoms = 66 pm

Now,

Consider 1 ml of water = 1 cm²

As , we know that mass of water in 1 cm² = 0.999 g

Moles of water = [tex]\frac{0.999}{18} = 0.056[/tex]

Volume of H₂O = 1.624×[tex]10^{-31}[/tex] m²

Now,

Volume occupied by water = 0.056×6.022×[tex]10^{23}[/tex]× 1.624×[tex]10^{-31}[/tex] m²

                                              = 5.48×[tex]10^{-9}[/tex] m²

⇒Volume occupied by water = 5.48×[tex]10^{-9}[/tex] m²

Now,

Free space = 1×[tex]10^{-6}[/tex]  - 5.48×[tex]10^{-9}[/tex] = 9.95×[tex]10^{-7}[/tex] m²

% Free space = [tex]\frac{9.945* 10^{-7} }{1*10^{-6} }[/tex]×100 = 99.45%

Now,

Consider 1 ml of ice  = 1 cm²

S.I unit of ice = 1×[tex]10^{-6}[/tex] m²

As , we know that mass of water in 1×[tex]10^{-6}[/tex] m² = 0.917 g

Moles of ice = [tex]\frac{0.917}{18} = 0.012[/tex]

Volume of H₂O = 6.022×[tex]10^{23}[/tex] ×0.012

Volume of ice unit = [tex]\frac{4}{3} \pi (37*10^{-12})^{3} *2 + \frac{4}{3} \pi (66*10^{-12})^{3} = 1.624*10^{-31}m^{3}[/tex]

Now,

Volume occupied by water = 0.012×6.022×[tex]10^{23}[/tex]× 1.624×[tex]10^{-31}[/tex] m²

                                              = 1.17×[tex]10^{-9}[/tex] m²

⇒Volume occupied by water = 1.17×[tex]10^{-9}[/tex] m²

Now,

Free space = 1×[tex]10^{-6}[/tex]  - 1.17×[tex]10^{-9}[/tex] = 9.98×[tex]10^{-7}[/tex] m²

% Free space = [tex]\frac{9.98* 10^{-7} }{1*10^{-6} }[/tex]×100 = 99.8%

Answer:

3.3 moles of H₂O.

Explanation:

We'll begin by writing the balanced equation for the reaction. This is illustrated below:

4NH₃ + 5O₂ —> 6H₂O + 4NO

From the balanced equation above,

4 moles of NH₃ reacted to produce 6 moles of H₂O.

Finally, we shall determine the number of mole of H₂O produced by the reaction of 2.2 moles of NH₃. This can be obtained as follow :

From the balanced equation above,

4 moles of NH₃ reacted to produce 6 moles of H₂O.

Therefore, 2.2 moles of NH₃ will react to produce = (2.2 × 6)/4 = 3.3 moles of H₂O.

Thus, 3.3 moles of H₂O were obtained from the reaction.

Answer:

An Alkane

Explanation:

Formula of ethane is CH3-CH3

Answer:

Ethane is the second member of

An alkane

Answer:

light

Explanation:

light is plasma, which is a state of matter

Answer:

Fluorine (F)

Explanation:

The poorest conductor of electricity from the given choices is fluorine. This is because fluorine is a non - metal.

Like other non - metals, fluorine does not conduct electricity.

Only metals are known to conduct electricity and heat readily.

Semi - metals like silicon will conduct electricity under specific condition.

The free mobile electrons in metals makes it easy for them propagate electricity

Answer:

[tex]\%N=26.2\%\\\\\%H=7.5\%\\\\\%Cl=66.3\%[/tex]

Explanation:

Hello!

In this case, since the calculation of the percent composition of an element in a chemical compound is computing considering its atomic mass, subscript in the formula and molecular mass of the compound it is; for nitrogen, hydrogen and chlorine we have that ammonium chloride has a molar mass of 53.49 g/mol so the percent compositions are:

[tex]\%N=\frac{14.01*1}{53.49}*100\% =26.2\%\\\\\%H=\frac{1.01*4}{53.49}*100\% =7.5\%\\\\\%Cl=\frac{35.45*1}{53.49}*100\% =66.3\%[/tex]

Best regards!

Answer:

I believe the correct answer is cold surfaces.

Explanation:

Answer:

C

Explanation:

Answer:

well heat travels by conduction, convection, and radiation but I think it's 2.

Explanation:

heat travels to colder things trying to make a balanced temperature for both of the objects.

Answer: inside

Explanation:

Answer:

direct and indirect contact

Explanation:

if you touch a doorknob right after an infected person than you make be exposed to the disease.

The ocean is not a part of Earth's layers.

Answer:

Ocean

Explanation:

Answer:

Explanation:

To find the concentration; let's first compute the average density and the average atomic weight.

For the average density [tex]\rho_{avg}[/tex]; we have:

[tex]\rho_{avg} = \dfrac{100}{ \dfrac{C_{Fe} }{\rho_{Fe}} + \dfrac{C_v}{\rho_v} }[/tex]

The average atomic weight is:

[tex]A_{avg} = \dfrac{100}{ \dfrac{C_{Fe} }{A_{Fe}} + \dfrac{C_v}{A_v} }[/tex]

So; in terms of vanadium, the Concentration of iron is:

[tex]C_{Fe} = 100 - C_v[/tex]

From a unit cell volume [tex]V_c[/tex]

[tex]V_c = \dfrac{n A_{avc}}{\rho_{avc} N_A}[/tex]

where;

[tex]N_A[/tex] = number of Avogadro constant.

SO; replacing [tex]V_c[/tex] with [tex]a^3[/tex] ; [tex]\rho_{avg}[/tex] with [tex]\dfrac{100}{ \dfrac{C_{Fe} }{\rho_{Fe}} + \dfrac{C_v}{\rho_v} }[/tex] ; [tex]A_{avg}[/tex] with [tex]\dfrac{100}{ \dfrac{C_{Fe} }{A_{Fe}} + \dfrac{C_v}{A_v} }[/tex] and

[tex]C_{Fe}[/tex] with [tex]100-C_v[/tex]

Then:

[tex]a^3 = \dfrac { n \Big (\dfrac{100}{[(100-C_v)/A_{Fe} ] + [C_v/A_v]} \Big) } {N_A\Big (\dfrac{100}{[(100-C_v)/\rho_{Fe} ] + [C_v/\rho_v]} \Big) }[/tex]

[tex]a^3 = \dfrac { n \Big (\dfrac{100 \times A_{Fe} \times A_v}{[(100-C_v)A_{v} ] + [C_v/A_Fe]} \Big) } {N_A \Big (\dfrac{100 \times \rho_{Fe} \times \rho_v }{[(100-C_v)/\rho_{v} ] + [C_v \rho_{Fe}]} \Big) }[/tex]

[tex]a^3 = \dfrac { n \Big (\dfrac{100 \times A_{Fe} \times A_v}{[(100A_{v}-C_vA_{v}) ] + [C_vA_Fe]} \Big) } {N_A \Big (\dfrac{100 \times \rho_{Fe} \times \rho_v }{[(100\rho_{v} - C_v \rho_{v}) ] + [C_v \rho_{Fe}]} \Big) }[/tex]

Replacing the values; we have:

[tex](0.289 \times 10^{-7} \ cm)^3 = \dfrac{2 \ atoms/unit \ cell}{6.023 \times 10^{23}} \dfrac{ \dfrac{100 (50.94 \g/mol) (55.84(g/mol)} { 100(50.94 \ g/mol) - C_v(50.94 \ g/mol) + C_v (55.84 \ g/mol) } }{ \dfrac{100 (7.84 \ g/cm^3) (6.0 \ g/cm^3 } { 100(6.0 \ g/cm^3) - C_v(6.0 \ g/cm^3) + C_v (7.84 \ g/cm^3) } }[/tex]

[tex]2.41 \times 10^{-23} = \dfrac{2}{6.023 \times 10^{23} } \dfrac{ \dfrac{100 *50*55.84}{100*50.94 -50.94 C_v +55.84 C_v} }{\dfrac{100 * 7.84 *6}{600-6C_v +7.84 C_v} }[/tex]

[tex]2.41 \times 10^{-23} (\dfrac{4704}{600+1.84 C_v})=3.2 \times 10^{-24} ( \dfrac{284448.96}{5094 +4.9 C_v})[/tex]

[tex]\mathbf{C_v = 9.1 \ wt\%}[/tex]

Answer:

Partial pressures:

PCl₅ = 0.558 atm

PCl₃ = 0.22 atm

Cl₂ = 0.22 atm

Explanation:

From the given information:

The number of moles of PCl₅ associated with the evaporation is:

[tex]n_{PCl_5}= \dfrac {weight \ of \ PCl_5} {M.Wt. \ of \ PCl_5}[/tex]

[tex]n_{PCl_5}= \dfrac {2.69 \ g} {208.5 \ g/mol}[/tex]

[tex]n_{PCl_5}= 0.013 \ mol[/tex]

Temperature of the gas = 250° C = (250 + 273.15) K

= 523.15 K

Using the Ideal gas equation to determine the pressure exerted by the completely vaporized PCl₅

PV = nRT

[tex]P = \dfrac{nRT}{V}[/tex]

[tex]P = \dfrac{0.0013 \ mol \times 0.082 \ Latm^0 K^{-1} . mol ^{-1} \times 523.15 \ K}{1.0 \ L}[/tex]

P = 0.558 atm

Thus, at  250° C, decomposition of PCl₅ occurs.

In the container, PCl₅  decomposes to PCl₃ and Cl₂.

i.e.

[tex]PCl_{5(g)} \to PCl_{3(g)}+ Cl_{2(g)}[/tex]

Using Dalton's Law:

[tex]P_{total } =P_1 + P_2+P_3 +...[/tex]

[tex]P_1 = P_{Total} \times X_1[/tex]

where;

X = mole fraction

Then, the total no. of moles in the container is:

[tex]n = \dfrac{PV} {RT}[/tex]

[tex]n = \dfrac{1\ atm \times 1.0\ L}{0.0821 \ L \ atm \ K^{-1}.mol \times 523.15\ K}[/tex]

n = 0.023 mol

Now, the container contains a total amount of 0.023 mol where initially 0.013 mol are that of PCl₅ and remaining 0.005 mol of PCl₃ and 0.005 mol of Cl₂.

Thus, the partial pressure of  PCl₃  is:

[tex]P__{PCL_3} }= P_{total} \times \dfrac{no. \ of \ moles \ of PCl_5}{total \ no. \ of \ moles}[/tex]

[tex]P__{PCL_3}} = 1 \ atm \times \dfrac{0.005}{0.023}[/tex]

[tex]P__{PCL_3}} = 0.22 \ atm[/tex]

Thus, since the no of moles of PCl₃ and Cl₂ are the same, then the partial pressure for Cl₂ is = 0.22 atm

Answer:

V₂ = 104.76 mL

Explanation:

Given data:

Initial volume = 100.0 mL

Initial temperature = 21°C (21 + 273.15 K = 294.15 K)

Final temperature = 35°C (35 + 273.15 K = 308.15 k)

Final volume = ?

Solution:

Charles Law:

According to this law, The volume of given amount of a gas is directly proportional to its temperature at constant number of moles and pressure.

Mathematical expression:

V₁/T₁ = V₂/T₂

V₁ = Initial volume

T₁ = Initial temperature

V₂ = Final volume  

T₂ = Final temperature

Now we will put the values in formula.

V₁/T₁ = V₂/T₂

V₂ = V₁T₂/T₁  

V₂ =100.0 mL × 308.15 K / 294.15 K

V₂ = 30815 mL.K /294.15 K

V₂ = 104.76 mL