A sample of metal has a mass of 24.54 g, and a volume of 5.02 mL. What is the density of this metal? g/cm

Answer:

12.4 × 10∧3 atoms

Explanation:

Given data:

moles of oxygen molecule= 1.0000 x 10-20 mol

atoms =?

Solution:

32 g O2 = 1 mol = 6.02 × 10∧23

1.0000 x 10∧-20 mol × 6.02 × 10∧23 × 2 = 12.4 × 10∧3 atoms

The energy of the photons emitted by the AM and FM radio stations can be calculated using Planck's equation (E = hf). The calculated energies show that the FM radio station's photons have higher energy than the AM radio station's photons.

To calculate the energy of the photons emitted by radio stations, we use the formula:

E = hf, where E is the energy, h is Planck's constant (6.62607015 × 10⁻³⁴ Js) and f is the frequency.

(For the sake of this calculation, we are going to convert the frequency from kHz and MHz to Hz.)

Comparing these energy values, it's clear that the FM radio station's emitted photons have a higher energy than the AM radio station's photons, due to its higher frequency.

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Answer:

The intermolecular forces between the solute and solvent.

Explanation:

When you are heating a solvent, the intermolecular forces are reduced because the distances between molecules are large. Thus, in a solution where solvent is hot the intermolecular forces between solute and solvent are lower than those solutions where solvent is in room temperature.

The covalent bonds do not change because this mean a chemical reaction that doesn't occur in a solution.

Usually solid solutes melts in a higher temperature than boiling point in solvents. Thus, a compound normally doesn't melt in a hot solvent.

I hope it helps!

When a compound is dissolved in hot ethanol during recrystallization, the intermolecular forces between the solute and solvent are changing on a molecular level. The process of solution formation involves the formation of new intermolecular forces between the solute and solvent molecules, allowing the compound to dissolve in the solvent.

Intermolecular forces are attractive or repulsive forces between molecules. They include London dispersion forces, dipole-dipole interactions, and hydrogen bonding. These forces influence physical properties like boiling points and solubility in molecular substances. The intermolecular forces between the solute and solvent are changing on a molecular level when a compound is dissolved in hot ethanol during recrystallization.

When the compound is dissolved in the hot solvent, the solute-solvent interactions which is also known as solvatio then naturally takes place. In addition to this, these interactions involve the formation of new intermolecular forces between the solute and solvent molecules, which are nearly as strong as the intermolecular forces within the solute and solvent alone. This favorable solution formation process allows the compound to dissolve in the solvent.

Answer: The mass of solid NaOH required is 80 g

Explanation:

Equivalent weight is calculated by dividing the molecular weight by n factor. The equation used is:

[tex]\text{Equivalent weight}=\frac{\text{Molecular weight}}{n}[/tex]

where,

n = acidity for bases = 1 (For NaOH)

Molar mass of NaOH = 40 g/mol

Putting values in above equation, we get:

[tex]\text{Equivalent weight}=\frac{40g/mol}{1eq/mol}=40g/eq[/tex]

Normality is defined as the umber of gram equivalents dissolved per liter of the solution.

Mathematically,

[tex]\text{Normality of solution}=\frac{\text{Number of gram equivalents} \times 1000}{\text{Volume of solution (in mL)}}[/tex]

Or,

[tex]\text{Normality of solution}=\frac{\text{Given mass}\times 1000}{\text{Equivalent mass}\times \text{Volume of solution (in mL)}}[/tex]         ......(1)

We are given:

Given mass of NaOH = ?

Equivalent mass of NaOH = 40 g/eq

Volume of solution = 400 mL

Normality of solution = 5 eq/L

Putting values in equation 1, we get:

[tex]5eq/L=\frac{\text{Mass of NaOH}\times 1000}{40g/eq\times 400mL}\\\\\text{Mass of NaOH}=80g[/tex]

Hence, the mass of solid NaOH required is 80 g

Answer: The correct answer is Option B.

Explanation:

The chemical equation for the reaction of lead (II) nitrate and ammonium chloride follows:

[tex]Pb(NO_3)_2(aq.)+2NH_4Cl(aq.)\rightarrow PbCl_2(s)+2NH_4NO_3(aq.)[/tex]

Ionic form of the above equation follows:

[tex]Pb^{2+}(aq.)+2NO_3^-(aq.)+2NH_4^+(aq.)+Cl^-(aq.)\rightarrow PbCl_2(s)+2NH_4^+(aq.)+2NO_3^-(aq.)[/tex]

Net ionic equation of any reaction does not include any spectator ions.

Spectator ions are defined as the ions which does not get involved in a chemical equation. They are found on both the sides of the chemical reaction when it is present in ionic form.

As, ammonium and nitrate ions are present on both the sides of the reaction. Thus, it will not be present in the net ionic equation and are spectator ions.

The net ionic equation for the above reaction follows:

[tex]Pb^{2+}(aq.)+2Cl^-(aq.)\rightarrow PbCl_2(s)[/tex]

Solubility rules:

Hence, the correct answer is Option B.

Answer:

4057.85 g/mol

Explanation:

Hello, the numerical procedure is shown in the attached file.

- In this case, since we don't have the density of the protein, we must assume that the volume of the solution is solely given by the benzene's volume, in order to obtain the moles of the solute (protein).

-Van't Hoff factor is assumed to be one.

Best regards.

To calculate the molar mass of the protein, use the formula M = (RT) / (V * P), where M is the molar mass, R is the ideal gas constant, T is the temperature, V is the volume, and P is the osmotic pressure.

To calculate the molar mass of the protein, we can use the formula:

M = (RT) / (V * P)

Where M is the molar mass, R is the ideal gas constant (0.0821 L.atm/mol.K), T is the temperature in Kelvin (24.4 + 273 = 297.4 K), V is the volume in liters (44 mL / 1000 = 0.044 L), and P is the osmotic pressure in atm (50.9 torr / 760 = 0.067 atm).

Substituting the values into the formula:

M = (0.0821 * 297.4) / (0.044 * 0.067) = 25567.14 g/mol

Therefore, the molar mass of the protein is approximately 25567.14 g/mol.

Answer:

Volume of HCl required = 28.4 mL

Explanation:

[tex]Zn(s) + 2HCl (aq) \rightarrow ZnCl_2(aq) + H_2(g)[/tex]

Mass of zinc ore = 4.65 g

% of zinc in zinc ore = 50 %

So, mass of zinc in zinc ore = 0.50 × 4.65 = 2.325 g

No. of moles of Zn = [tex]\frac{2.325}{65.40} = 0.0355\ mol[/tex]

So, as per the reaction coefficient,

1 mol of zinc reacts with 2 mol of HCl

0.0355 mol of zinc reacts with

                                    = 0.0355 × 2 = 0.071 mol of HCl

Molarity of HCl = 2.50 M

Volume of HCl = [tex]\frac{Moles}{Concentration}[/tex]

[tex]Volume\ of\ HCl = \frac{0.071}{2.50} = 0.0284\ L[/tex]

1 L = 1000 mL

0.0284 L = 1000 × 0.0284 = 28.4 mL

Answer:

a. 0.14042

b. 0.221

c. equal to a.

5) 0.579 L

3) 8.1e6 ms

Explanation:

Hello,

In the attached photo you'll find the numerical procedure to get the results.

Best regards.

Answer:

pH of acetic acid solution is 2.88

Explanation:

[tex]pK_{a}=4.75[/tex]

or, [tex]-log(K_{a})=4.75[/tex]

or, [tex]K_{a}=10^{-4.75}=1.78\times 10^{-5}[/tex]

We have to construct an ICE table to determine concentration of [tex]H^{+}[/tex] and corresponding pH. Initial concentration of acetic acid is 0.1 M.

[tex]CH_{3}COOH\rightleftharpoons CH_{3}COO^{-}+H^{+}[/tex]

I(M):  0.1                            0                    0

C(M): -x                              +x                 +x

E(M): 0.1-x                         x                    x

So, [tex]\frac{[CH_{3}COO^{-}][H^{+}]}{[CH_{3}COOH]}=K_{a}[/tex]

or, [tex]\frac{x^{2}}{0.1-x}=1.78\times 10^{-5}[/tex]

or, [tex]x^{2}+(1.78\times 10^{-5}\times x)-(1.78\times 10^{-6})=0[/tex]

So, [tex]x=\frac{-(1.78\times 10^{-5})+\sqrt{(1.78\times 10^{-5})^{2}+(4\times 1\times 1.78\times 10^{-6})}}{2\times 1}[/tex](M)

so, [tex]x=1.33\times 10^{-3}M[/tex]

Hence [tex]pH=-log[H^{+}]=-log(1.33\times 10^{-3})=2.88[/tex]

Answer:

Exit velocity of air is 96.43 m/s.

Explanation:

Given that

[tex]V_1=400\ m/s[/tex]

[tex]T_1=25C[/tex]

[tex]T_2=100C[/tex]

For air

[tex]C_p=1.005\ KJ/kg.K[/tex]

Now from first law of thermodynamics for open system at steady state

[tex]h_1+\dfrac{V_1^2}{2000}+Q=h_2+\dfrac{V_2^2}{2000}+w[/tex]

For diffuser

[tex]h_1+\dfrac{V_1^2}{2000}=h_2+\dfrac{V_2^2}{2000}[/tex]

We know that

[tex]h=C_pT[/tex]

[tex]h_1+\dfrac{V_1^2}{2}=h_2+\dfrac{V_2^2}{2000}[/tex]

[tex]1.005\times 25+\dfrac{400^2}{2000}=1.005\times 100+\dfrac{V_2^2}{2}[/tex]

[tex]V_2=96.43\ m/s[/tex]

So the exit velocity of air is 96.43 m/s.

Answer:

% of mass of electrons in C = 0.0272

Explanation:

Atomic no. of carbon = 6

So, no. of electron = 6

Mass of an electron = [tex]\frac{1}{1836} amu[/tex]

Mass of 6 electrons = [tex]6 \times \frac{1}{1836} = 0.003268 amu[/tex]

Mass of C = 12.0107 u

% of mass of electrons in C = [tex]\frac{Mass\ of\ all\ electrons}{Mass\ of\ C}\times 100[/tex]

% of mass of electrons in C = [tex]\frac{0.003268}{12.0107}\times 100=0.0272 \%[/tex]

Answer:

The final dilution is 1:400

Explanation:

Let's analyze what we are told: we have an initial 1:5 dilution of protein lysate. This means that the initial solution (stock solution) was diluted 5 times. Then, from this dilution the student prepared another dilution taking 2 mL of the first dilution in 8 mL of water. This is the same as saying we took 1 mL of first dilution in 4 mL of water (the ratio is the same), so we now have a second 1:4 dilution of the first dilution (1:5). Finally, the student made a third 1:20 dilution, this means that the second dilution was further diluted 20 times.

So, to calculate the final dilution of protein lysate, we have to multiply all the dilution factors of every dilution prepared: in this case we have a final dilution of 1:20, this means we have a factor dilution of 20. But it was previously diluted 4 times, so we have a factor dilution of 20×4 = 80. However, this dilution was also previously diluted 5 times, so the new dilution factor is 80 × 5 = 400

This means that the final dilution of the compound was diluted a total of 400 times compared to the initial concentration of stock solution.

Explanation:

Orifice meters require to compensate for pressure and temperature when one uses these meters to measure the steam or the gas flow in the pipes with the variable operating pressure as well as temperature conditions.

Normally chemists do not have online density measurement tool and thus to avoid complications, the chemists consider density as constant parameter.

In the steam or the gas flow measurement, density of steam or gas changes as the pressure and the temperature change. This significant  change in the density can affect accuracy of measured flow rate if the change is uncompensated and thus, this has to be fixed in order to avoid errors. Therefore, it is important to compensate for the pressure and the temperature when orifice is used to measure the steam or the gas flow.

Answer:

There are 1.041×10²³ atoms in 4.20g of Magnesium.

Explanation:

To find the amount of atoms in 4.20 g of Magnesium we need de molar mass of Mg: 24.305 g/mol

According to Avogadro number there are 6.022×10²³ particles in 1 mol, so the number of atoms of Mg is:

[tex]4.20 g Mg*\frac{1 molMg}{24.305gMg} *\frac{6.022*10^{23}atoms }{1 mol Mg} = 1.041*10^{23}atoms Mg[/tex]

Answer: The number of atoms found in given amount of magnesium is [tex]1.042\times 10^{23}[/tex]

Explanation:

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

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

Given mass of magnesium = 4.20 g

Molar mass of magnesium = 24.31 g/mol

Putting values in above equation, we get:

[tex]\text{Moles of magnesium}=\frac{4.20g}{24.31g/mol}=0.173mol[/tex]

According to mole concept:

1 mole of an element contains [tex]6.022\times 10^{23}[/tex] number of atoms

So, 0.173 moles of an element contains = [tex]0.173\times 6.022\times 10^{23}=1.042\times 10^{23}[/tex] number of atoms

Hence, the number of atoms found in given amount of magnesium is [tex]1.042\times 10^{23}[/tex]

Answer:

NCL3

Explanation:

NCL3

- nitrogen trichloride.

The chemical formula for a compound containing one nitrogen atom for every three chlorine atoms is NCl₃.

The chemical formula NCl₃ represents a compound consisting of one nitrogen atom and three chlorine atoms. In this formula, "N" represents nitrogen, and "Cl" represents chlorine. The subscript "3" indicates that there are three chlorine atoms bonded to each nitrogen atom in the compound.

This arrangement reflects the stoichiometry of the compound, specifying the relative ratios of the elements. Chemical formulas provide a concise way to describe the composition of substances, allowing scientists to understand the types and quantities of atoms present in a compound.

NCl₃, nitrogen trichloride, is a covalent compound with notable chemical properties, and its formula encapsulates the balanced proportion of nitrogen and chlorine within its molecular structure.

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Answer: The mass of hydrogen gas and nitrogen gas produced is 0.45 g and 3.15 g respectively.

Explanation:

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

[tex]\text{Number of moles}=\frac{\text{Given mass}}{\text{Molar mass}}[/tex]      .....(1)

Given mass of water = 4.05 g

Molar mass of water = 18 g/mol

Putting values in equation 1, we get:

[tex]\text{Moles of water}=\frac{4.05g}{18g/mol}=0.225mol[/tex]

For the given chemical reaction:

[tex]2H_2O\rightarrow 2H_2+O_2[/tex]

By Stoichiometry of the reaction:

2 moles of water is producing 2 moles of hydrogen gas

So, 0.225 moles of water will produce = [tex]\frac{2}{2}\times 0.225=0.225mol[/tex] of hydrogen gas.

Now, calculating the mass of hydrogen gas by using equation 1, we get:

Moles of hydrogen gas = 0.225 mol

Molar mass of hydrogen gas = 2 g/mol

Putting values in equation 1, we get:

[tex]0.225mol=\frac{\text{Mass of hydrogen gas}}{2g/mol}\\\\\text{Mass of hydrogen gas}=0.45g[/tex]

By Stoichiometry of the reaction:

2 moles of water is producing 1 mole of nitrogen gas

So, 0.225 moles of water will produce = [tex]\frac{1}{2}\times 0.225=0.1125mol[/tex] of nitrogen gas.

Now, calculating the mass of nitrogen gas by using equation 1, we get:

Moles of nitrogen gas = 0.1125 mol

Molar mass of nitrogen gas = 28 g/mol

Putting values in equation 1, we get:

[tex]0.1125mol=\frac{\text{Mass of nitrogen gas}}{28g/mol}\\\\\text{Mass of nitrogen gas}=3.15g[/tex]

Hence, the mass of hydrogen gas and nitrogen gas produced is 0.45 g and 3.15 g respectively.

Answer:

[tex]93.43\times 10^{-9}[/tex]

Explanation:

Scientific notation is the way of writing numbers which are either large or small. The number is written in the scientific notation when the number is between 1 and 10 and then multiplied by the power of 10. Engineering notation is the same version of the scientific notation but the number can be between 1 and 1000 and in this exponent of the ten is divisible by three.

For example, [tex]1000^2[/tex] is to be written as [tex]10^6[/tex] in engineering notation.

The given number:

0.00000009345 can be written as [tex]93.425\times 10^{-9}[/tex]

Answer upto 4 significant digits = [tex]93.43\times 10^{-9}[/tex]

Answer:

49.3% 81Br and 50.7% 79Br or [tex]\frac{493}{1000}[/tex] 81Br and [tex]\frac{507}{1000}[/tex] 79BR

Step-by-step explanation:

To get the fraction-of-occurrences of the isotopes we must write the following equation. x is the isotopic abundance of 81Br, we can use 1 - x to get the isotopic abundance of 79Br.

(78.918)(1 - x) + (80.916)(x) = 79.903

78.918 - 78.918x + 80.916x = 79.903

1.998x = 0.985

x = 0.493

0.493 × 100 = 49.3% 81Br

100 - 49.3 = 50.7% 79Br

49.3% 81Br and 50.7% 79Br or [tex]\frac{493}{1000}[/tex] 81Br and [tex]\frac{507}{1000}[/tex] 79BR

Answer:

There are 0.1125 g of O₂ less in 1 L of air at 14,000 ft than in 1 L of air at sea level.

Explanation:

To solve this problem we use the ideal gas law:

PV=nRT

Where P is pressure (in atm), V is volume (in L), n is the number of moles, T is temperature (in K), and R is a constant (0.082 atm·L·mol⁻¹·K⁻¹)

Now we calculate the number of moles of air in 1 L at sea level (this means with P=1atm):

1 atm * 1 L = n₁ * 0.082 atm·L·mol⁻¹·K⁻¹ * 298 K

n₁=0.04092 moles

Now we calculate n₂, the number of moles of air in L at an 14,000 ft elevation, this means with P = 0.59 atm:

0.59 atm * 1 L = n₂ * 0.082 atm·L·mol⁻¹·K⁻¹ * 298 K

n₂=0.02414 moles

In order to calculate the difference in O₂, we substract n₂ from n₁:

0.04092 mol - 0.02414 mol = 0.01678 mol

Keep in mind that these 0.01678 moles are of air, which means that we have to look up in literature the content of O₂ in air (20.95%), and then use the molecular weight to calculate the grams of O₂ in 20.95% of 0.01678 moles:

[tex]0.01678mol*\frac{20.95}{100} *32\frac{g}{mol} =0.1125 gO_{2}[/tex]

Answer: The mass of air present in the room is 37.068 kg

Explanation :  Given,

Length of the room = 10.0 ft

Breadth of the room = 11.0 ft

Height of the room = 10.0 ft

To calculate the volume of the room by using the formula of volume of cuboid, we use the equation:

[tex]V=lbh[/tex]

where,

V = volume of the room

l = length of the room

b = breadth of the room

h = height of of the room

Putting values in above equation, we get:

[tex]V=10.0ft\times 11.0ft\times 10.0ft=1100ft^3=31148.53L[/tex]

Conversion used : [tex]1ft^3=28.3168L[/tex]

Now we have to calculate the mass of air in the room.

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

[tex]1.19g/L=\frac{Mass}{31148.53L}[/tex]

[tex]Mass=37066.7507g=37.068kg[/tex]

Conversion used : (1 kg = 1000 g)

Therefore, the mass of air present in the room is 37.068 kg

Answer: There are 37 kg of air in the room.

Explanation:

To calculate the volume of cuboid (room), we use the equation:

[tex]V=lbh[/tex]

where,

V = volume of cuboid

l = length of room = 11 ft

b = breadth of room =  10 ft

h = height of room= 10 ft

Putting values in above equation, we get:

[tex]V=10\times 11\times 10=1100ft^3=1100\times 28.3L=31130L[/tex]  (Conversion factor: [tex]1ft^3=28.3L[/tex]

To calculate mass of a substance, we use the equation:

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

We are given:

Density of air = 1.19 g/L

Volume of air = volume of room =  31130 L

Putting values in above equation, we get:

[tex]1.19g/L=\frac{\text{Mass of air}}{31130L}\\\\\text{Mass of air}=37000g=37.0kg[/tex]    (1kg=1000g)

Hence, the mass of air is 37 kg.

Answer:

[tex]5.5*10x^{-9}m[/tex]

Explanation:

As you have the diameter of the sphere in nanometers (nm), you need to use de conversion factor to find the diameter in meters (m):

First you should put the quantity that you want to convert with its respective units:

[tex]Diameter=5.5nm[/tex]

Then you put the conversion factor, always you should put the same unit that you want to convert in the denominator:

Diameter = [tex]5.5nm*\frac{1*10^{-9}m}{1nm}[/tex]

And finally, you should multiply and/or divide the quantities:

Diameter = [tex]5.5*10^{-9}m[/tex]

Answer:

a) m  = 2398256.64 Ib/hr

b) Q = 38433.6 ft^3/hr

c) J = = 898560 lb/h-ft2

Explanation:

a) To get the mass flow rate in lb/h we are going to use this formula:

m = ρνA

note: we have to change some units to reach to the final units

when m is the mass flow rate which we need to calculate

ρ is the density of water in lb/ft3 = 62.4 lbs/ft3 it is a constant numbers you can get it from anywhere

ν is the velocity of the water = 4 ft/s we need to change it to ft/h unit so  

when 1 ft/s = 3600 ft/hr (note : 3600 is a result of converting hr to sec 60min*60sec)

so  the velocity of the water = 4 ft/s * 3600 ft/hr= 14400 ft/hr

now , we need to get the area of the pipe in feet also when:

A = π*diameter * length

diameter = 3.068 inches

when 1 inch = 0.0833333333 feet so,

diameter in feet = 3.068 * 0.0833333333 =.300 0.2557ft

and length = 40 * 0.0833333333  = 3.333 ft

so the area of the pipe in feet^2 = 3.14*0.255*3.33 =2.669 ft^2

by substitution in the mass flow rate:

m = 62.4 lbs/ft3 * 14400 ft/hr *2.669 ft^2 = 2398256.64 Ib/hr

b) now to calculate volumetric flow rate in ft /h  we are going to use this formula:

Q = AV

when Q is  the volume flow rate

A is the cross-sectional area filled by water =2.669 ft^2

V is the average velocity of the water = 14400 ft/hr

so , by substitution:

Q = 2.669  ft^2* 14400 ft/hr= 38433.6 ft^3/hr

C) To calculate mass velocity (mass flux) in lb/h-ft  we are going to use this formula:

J = m / A

when J is the mass flux

and m is the mass flow rate which we calculated above = 2398256.64 Ib/hr

and A is the cross section area of the pipe which we calculated above and = 2.669 ft^2

so, by substitution:

J = 2398256.64 Ib/hr /  2.669 ft^2

 = 898560 lb/h-ft2

Answer:

The required volume is 1.6 x 10³mL.

Explanation:

When we want to prepare a dilute solution from a concentrated one, we can use the dilution rule to find out the required volume to dilute. This rule states:

C₁ . V₁ = C₂ . V₂

where,

C₁ and V₁ are the concentration and volume of the concentrated solution

C₂ and V₂ are the concentration and volume of the dilute solution

In this case, we want to find out V₁:

C₁ . V₁ = C₂ . V₂

[tex]V_{1} = \frac{C_{2}.V_{2}}{C_{1}} = \frac{1.5M \times1.11.10^{4}mL }{10.2M} =1.6\times10^{3} mL[/tex]

Answer:

ghdtgfgrdvreeeegrwwggegteefewqrwefrwerrftrtdsfgyuytfgererdf

Explanation:

The given data is as follows.

     [tex]V_{1}[/tex] = 1.50 L,    [tex]T_{1}[/tex] = 159 K,

      [tex]P_{1}[/tex] = 5.00 atm,   [tex]V_{2}[/tex] = 0.2 L,

       [tex]T_{2}[/tex] = ?,       [tex]P_{2}[/tex] = 50.0 atm  

And, according to ideal gas equation,  

               [tex]\frac{P_{1}V_{1}}{T_{1}} = \frac{P_{2}V_{2}}{T_{2}}[/tex]

Hence, putting the given values into the above formula to calculate the value of final temperature as follows.          

 [tex]\frac{5.0 atm \times 1.50 L}{159 K} = \frac{5 atm \times 0.2 L}{T_{2}}[/tex]

            [tex]T_{2}[/tex] = 21.27 K

Thus, we can conclude that the final temperature is 21.27 K .

Answer:

The expression to calculate the mass of the reactant is [tex]m = \frac{1.080kJ}{31.2kJ/g}[/tex]

Explanation:

The amount of heat released is equal to the amount of heat released per gram of reactant times the mass of the reactant. To keep to coherence between units we need to transform 1,080 J to kJ. We do so with proportions:

[tex]1,080J.\frac{1kJ}{10^{3}J } =1.080kJ[/tex]

Then,

[tex]1.080kJ=31.2\frac{kJ}{g} .m\\m = \frac{1.080kJ}{31.2kJ/g}[/tex]

Answer:

1. The expression is: [tex]m=\frac{E}{\Delta _rH}[/tex]

2. The computed mass is: [tex]m=0.0346g[/tex]

Explanation:

Hello,

In this case, we know the so called enthalpy of reaction whose symbol and value is shown below:

[tex]\Delta _rH=31.2\frac{kJ}{g}[/tex]

In addition, we know that the energy released by the involved reactant is:

[tex]E=1080 J[/tex]

Therefore, the expression to compute the required mass, based on the given units is:

[tex]m=\frac{E}{\Delta _rH}[/tex]

Finally, the computed mass turns out:

[tex]m=\frac{1080J*\frac{1kJ}{1000J} }{31.2\frac{kJ}{g}} \\m=0.0346g[/tex]

Best regards.

Answer:

Significant hydrogen bonding is possible in [tex]NH_{2}Cl[/tex]

Explanation:

The only substance from above in which significant hydrogen bonding possible is chloramine ( NH2Cl )

Organic compounds are substances which contain carbon and hydrogen. Some few groups of organic compounds include:

So therefore, the only substance from above in which significant hydrogen bonding possible is chloramine ( NH2Cl )

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Answer:

Due to the low difference of electronegativity in carbon and nitrogen, these will form a covalent bond, and this covalent union of C-N is one of the most common bonds in the organic chemistry and biological systems. This type of bonds can be found in amines, amides, imines, etc. Also, a single atom of C and a single atom of N can form a cyanide, that is a triple covalent bond between the atoms.

Answer:

It is necessary 1 meter of thickness.

Explanation:

You need to know the equation (P/A)=(k*T)/(L), where P/A is the heat flux, k is the thermal conductivity, T the temperature drop, and L the unknown thickness. So if rearrange the variables you will have that L = (k*T)/(P/A), and substituting the terms L = (0.2*400)/(200) = 1 meter

Final answer:

The necessary thickness of insulation to achieve a temperature drop of 400°C with a heat flux of 200 W/m^2, given a thermal conductivity of 0.2 W/m°C, is 0.4 meters.

Explanation:

The necessary thickness of insulation for a specific temperature drop and heat flux. By using the equation of thermal conduction Q = (kAΔT)/d, where Q is the heat flux, k is the thermal conductivity, A is the area, ΔT is the temperature difference, and d is the thickness, we can solve for the thickness d.

To start, we know that k (thermal conductivity) is 0.2 W/m°C, the temperature drop ΔT is 400°C, and the heat flux Q is 200 W/m^2. Plugging these values into the equation gives us d = (kAΔT)/Q. Since the area A is not specified, it will cancel out in this case, allowing us to solve for d directly. Rearranging the equation yields d = (kΔT)/Q, which equals (0.2 W/m°C * 400°C)/200 W/m^2, resulting in d = 0.4 m. Therefore, the thickness required for this insulation is 0.4 meters.

Answer:

Dispersion system is a system in which certain particles are scattered in a continuous liquid or solid medium. The two phases present in this system are the dispersed particles and dispersion medium. These phases may or may not be present in the same state.

In a dispersion system, the particles that are dispersed are known as the dispersed particles and the medium in which the particles are dispersed is known as the dispersion medium.

Answer:

The sample of lithium occupies the largest volume.

Explanation:

Given the densities for the four elements, we have the expression [tex]d=\frac{m}{V}[/tex] that shows the relationship between mass and Volume to express the density of an element.

For each element we have:

[tex]d_{lithium}=\frac{m_{lithium}}{V_{lithium}}=0.53g/mL[/tex]

[tex]d_{gold}=\frac{m_{gold}}{V_{gold}}=19.3g/mL[/tex]

[tex]d_{aluminum}=\frac{m_{aluminum}}{V_{aluminum}}=2.70g/mL[/tex]

[tex]d_{lead}=\frac{m_{lead}}{V_{lead}}=11.3g/mL[/tex]

The problem says that all the samples have the same mass, so:

[tex]m_{lithium}=m_{gold}=m_{aluminum}=m_{lead}=m[/tex]

it means that m is a constant

Now, solving for the Volume in each element and with m as a constant, we have:

[tex]V_{lithium}=\frac{m}{d_{lithium}}[/tex]

[tex]V_{lithium}=\frac{1}{0.53\frac{g}{mL}} *m[/tex]

[tex]V_{lithium}=1.88\frac{mL}{g}*m[/tex]

[tex]V_{gold}=\frac{m}{d_{gold}}[/tex]

[tex]V_{gold}=\frac{1}{19.3\frac{g}{mL}} *m[/tex]

[tex]V_{gold}=5.18*10^{-2}\frac{mL}{g}*m[/tex]

[tex]V_{aluminum}=\frac{m}{d_{aluminum}}[/tex]

[tex]V_{aluminum}=\frac{1}{2.70\frac{g}{mL}} *m[/tex]

[tex]V_{aluminum}=3.70*10^{-1}\frac{mL}{g}*m[/tex]

[tex]V_{lead}=\frac{m}{d_{lead}}[/tex]

[tex]V_{lead}=\frac{1}{11.3\frac{g}{mL}} *m[/tex]

[tex]V_{lead}=8.85*10^{-2}\frac{mL}{g}*m[/tex]

If we assume m = 1g, we find that:

[tex]V_{lithium}=1.88mL[/tex]

[tex]V_{gold}=5.18*10^{-2}mL[/tex]

[tex]V_{aluminum}=3.70*10^{-1}mL[/tex]

[tex]V_{lead}=8.85*10^{-2}mL[/tex]

So we can see that the sample of lithium occupies the largest volume with 1.88mL

Note that m only can take positive values, so if you change the value of m, always will be the lithium which occupies the largest volume.